Chapter 3: Graphing Lines
Chapter 3: Graphing Lines 3.1 Rectangular Coordinate System Part A: Ordered Pairs
Give the coordinates of points A, B, C, D and E. 1.
PROBLEM:
SOLUTION: The coordinates of A are The coordinates of B are The coordinates of C are The coordinates of D are The coordinates of E are
x = 3 and y = 5 . x = − 2 and y = 3 . x = − 5 and y = 0 . x = 1 and y = −3 . x = − 3 and y = −4 .
ANSWER: The coordinates of A are (3, 5). The coordinates of B are (–2, 3). The coordinates of C are (–5, 0). The coordinates of D are (1, –3). The coordinates of E are (–3, –4).
2.
PROBLEM:
ANSWER: The coordinates of A are x = 2 and y = 4 , (2, 4). The coordinates of B are x = 0 and y = 0 , ( 0, 0 ) .
The coordinates of C are x = − 5 and y = −2 , ( −5, −2 ) . The coordinates of D are x = 0 and y = −4 , ( 0, −4 ) . The coordinates of E are x = 5 and y = −5 , ( 0, −5 ) . 3.
PROBLEM:
SOLUTION: The coordinates of A are x = 0 and y = 6 . The coordinates of B are x = − 4 and y = 3 . The coordinates of C are x = − 8 and y = 0 .
The coordinates of D are x = − 6 and y = −6 . The coordinates of E are x = 8 and y = −9 . ANSWER: The coordinates of A are (0, 6). The coordinates of B are (–4, 3). The coordinates of C are (–8, 0). The coordinates of D are (–6, –6). The coordinates of E are (8, –9).
4.
PROBLEM:
ANSWER: The coordinates of A are x = 0 and y = 9 , ( 0,9 ) .
The coordinates of B are x = 6 and y = 9 , ( 6,9 ) . The coordinates of C are x = − 2 and y = 3 , ( −2,3) . The coordinates of D are x = 4 and y = −3 , ( 4, −3) . The coordinates of E are x = − 8 and y = −6 , ( −8, −6 ) . 5.
PROBLEM:
SOLUTION: The coordinates of A are The coordinates of B are The coordinates of C are The coordinates of D are The coordinates of E are
x = − 10 and y = 25 . x = 30 and y = 20 . x = 0 and y = 10 . x = 15 and y = 0 . x = 25 and y = −10 .
ANSWER: The coordinates of A are (–10, 25). The coordinates of B are (30, 20). The coordinates of C are (0, 10). The coordinates of D are (15, 0). The coordinates of E are (25, –10).
6.
PROBLEM:
ANSWER: The coordinates of A are x = 0 and y = 15 , ( 0,15 ) .
The coordinates of B are x = 10 and y = 15 , (10,15 ) . The coordinates of C are x = − 10 and y = 5 , ( −10,5 ) . The coordinates of D are x = 40 and y = 0 , ( 40, 0 ) . The coordinates of E are x = − 10 and y = −15 , ( −10, −15 ) . Graph the given set of ordered pairs.
7.
PROBLEM: {( −4, 5) , ( −1, 1) , ( −3, −2 ) , ( 5, −1)} SOLUTION: The coordinates x = − 4 and y = 5 , indicate that we should mark a point 4 units left of and 5 units above the origin.
The coordinates x = − 1 and y = 1 , indicate that we should mark a point 1 unit left of and 1 unit above the origin. The coordinates x = − 3 and y = −2 , indicate that we should mark a point mark a point 3 units left of and 2 units below the origin. The coordinates x = 5 and y = −1 , indicate that we should mark a point 5 units right of and 1 unit below the origin. ANSWER:
8.
PROBLEM: {( −15, −10 ) , ( −5, 10 ) , (15,10 ) , ( 5, −10 )} SOLUTION: The coordinates x = − 15 and y = −10 , indicate that we should mark a point 15 units left of and 10 units below the origin. The coordinates x = − 5 and y = 10 , indicate that we should mark a point 5 units left of and 10 units above the origin. The coordinates x = 15 and y = 10 , indicate that we should mark a point mark a point 15 units right of and 10 units above the origin. The coordinates x = 5 and y = −10 , indicate that we should mark a point 5 units right of and 10 units below the origin. ANSWER:
9.
PROBLEM: {( −2, 5) , (10, 0 ) , ( 2, −5) , ( 6, −10 )} SOLUTION: The coordinates x = − 2 and y = 5 , indicate that we should mark a point 2 units left of and 5 units above the origin. The coordinates x = 10 and y = 0 , indicate that we should mark a point 10 units right of the origin. The coordinates x = 2 and y = −5 , indicate that we should mark a point 2 units right on the x-axis and 5 below the origin. The coordinates x = 6 and y = −10 , indicate that we should mark a point 6 units right of and 10 units below the origin. ANSWER:
10.
PROBLEM: {( −8, 3) , ( −4, 6 ) , ( 0, − 6 ) , ( 6, 9 )} SOLUTION: The coordinates x = − 8 and y = 3 , indicate that we should mark a point 8 units left of and 3 units above the origin. The coordinates x = − 4 and y = 6 , indicate that we should mark a point 4 units left of and 6 units above the origin. The coordinates x = 0 and y = −6 , indicate that we should mark a point 6 units below the origin on the y-axis. The coordinates x = 6 and y = 9 , indicate that we should mark a point 6 units right of and 9 units above the origin.
ANSWER:
11.
PROBLEM: {( −10, 5 ) , ( 20, −10 ) , ( 30,15 ) , ( 50, 0 )} SOLUTION: The coordinates x = − 10 and y = 5 , indicate that we should mark a point 10 units left of and 5 units above the origin. The coordinates x = 20 and y = −10 , indicate that we should mark a point 20 units right of and 10 units below the origin. The coordinates x = 30 and y = 15 , indicate that we should mark a point 30 units right of and 15 units above the origin. The coordinates x = 50 and y = 0 , indicate that we should mark a point 50 units right of the origin on the x-axis. ANSWER:
12.
PROBLEM: ⎧⎛ 5 1 ⎞ ⎛ 1 1 ⎞ ⎛ 2 ⎞ ⎛ 5 ⎞⎫ ⎨⎜ − , − ⎟ , ⎜ − , ⎟ , ⎜ , −1⎟ , ⎜ ,1⎟ ⎬ ⎠ ⎝ 3 ⎠⎭ ⎩⎝ 3 2 ⎠ ⎝ 3 2 ⎠ ⎝ 3 SOLUTION: The coordinates x = −
1 5 and y = − , indicate that we should mark 2 3
2 1 a point 1 units left of and unit below the origin. 3 2
The coordinates x = −
left of and
1 1 1 and y = , indicate that we should mark a point unit 2 3 3
1 unit above the origin. 2
The coordinates x =
2 2 and y = −1 , indicate that we should mark a point unit to 3 3
the right and 1 unit below the origin The coordinates x =
2 5 and y = 1 , indicate that we should mark a point 1 units 3 3
right of and 1 unit above the origin. ANSWER:
13.
PROBLEM: ⎧⎛ 3 4 ⎞ ⎛ 2 4 ⎞ ⎛ ⎫ 2⎞ ⎨⎜ − , − ⎟ , ⎜ , ⎟ , ⎜1, − ⎟ , ( 0,1) ⎬ 3⎠ ⎩⎝ 5 3 ⎠ ⎝ 5 3 ⎠ ⎝ ⎭ SOLUTION:
The coordinates x = −
4 3 3 and y = − , indicate that we should mark a point unit 3 5 5
1 units below the origin. 3 4 2 2 The coordinates x = and y = , indicate that we should mark a point unit 3 5 5 1 right of and 1 units above the origin. 3 2 The coordinates x = 1 and y = − , indicate that we should mark a point 1 unit 3 2 right of and unit below the origin. 3 The coordinates x = 0 and y = 1 , indicate that we should mark a point 1 unit above the origin on the y-axis. left of and 1
ANSWER:
14.
PROBLEM: {( −3.5, 0 ) , ( −1.5, 2 ) , ( 0,1.5) , ( 2.5, −1.5)} SOLUTION: The coordinates x = − 3.5 and y = 0 , indicate that we should mark a point 3.5 units left of the origin on the x-axis. The coordinates x = − 1.5 and y = 2 , indicate that we should mark a point 1.5 units left of and 2 unit above the origin. The coordinates x = 0 and y = 1.5 , indicate that we should mark a point 1.5 units above the origin on the y-axis. The coordinates x = 2.5 and y = −1.5 , indicate that we should mark a point 2.5 units right of and 1.5 units below the origin. ANSWER:
15.
PROBLEM: {( −0.8, 0.2 ) , ( −0.2, −0.4 ) , ( 0, −1) , ( 0.6, −0.4 )} SOLUTION: The coordinates x = − 0.8 and y = 0.2 , indicate that we should mark a point 0.8 unit left of and 0.2 unit above the origin. The coordinates x = − 0.2 and y = −0.4 , indicate that we should mark a point 0.2 unit left of and 0.4 unit below the origin.
The coordinates x = 0 and y = −1 , indicate that we should mark a point 1 unit below the origin on the y-axis. The coordinates x = 0.6 and y = −0.4 , indicate that we should mark a point 0.6 unit right and 0.4 unit below the origin. ANSWER:
16.
PROBLEM: {( −1.2, −1.2 ) , ( −0.3, −0.3) , ( 0, 0 ) , ( 0.6, 0.6 ) , (1.2,1.2 )} SOLUTION: The coordinates x = − 1.2 and y = −1.2 , indicate that we should mark a point 1.2 units left of and 1.2 units below the origin. The coordinates x = − 0.3 and y = −0.3 , indicate that we should mark a point 0.3 unit left of and 0.3 unit below the origin. The coordinates x = 0 and y = 0 , indicate that the point is marked on the origin. The coordinates x = 0.6 and y = 0.6 , indicate that we should mark a point 0.6 unit right and 0.6 unit above the origin. The coordinates x = 1.2 and y = 1.2 , indicate that we should mark a point 1.2 units right and 1.2 units above the origin. ANSWER:
State the quadrant in which the given point lies.
17.
PROBLEM: (−3, 2) SOLUTION: The point is plotted in quadrant two (QII) because the x-coordinate is negative and the y-coordinate is positive. ANSWER: Quadrant two (QII)
18.
PROBLEM: (5, 7) SOLUTION: The point is plotted in quadrant one (QI) because the x-coordinate is positive and the y-coordinate is positive. ANSWER: quadrant one (QI)
19.
PROBLEM: (−12, −15) SOLUTION: The point is plotted in quadrant three (QIII) because the x-coordinate is negative and the y-coordinate is negative. ANSWER: Quadrant three (QIII)
20.
PROBLEM: (7, −8) SOLUTION: The point is plotted in quadrant four (QIV) because the xcoordinate is positive and the y-coordinate is negative. ANSWER: quadrant four (QIV)
21.
PROBLEM: (−3.8, 4.6) SOLUTION: The point is plotted in quadrant two (QII) because the x-coordinate is negative and the y-coordinate is positive. ANSWER: Quadrant two (QII)
22.
PROBLEM: (17.3, 1.9) SOLUTION: The point is plotted in quadrant one (QI) because the x-coordinate is positive and the y-coordinate is positive. ANSWER: quadrant one (QI)
23.
PROBLEM: ( − 18 , − 85 ) SOLUTION: The point is plotted in quadrant three (QIII) because the x-coordinate is negative and the y-coordinate is negative. ANSWER: Quadrant three (QIII)
24.
PROBLEM: ( 34 , − 14 ) SOLUTION: The point is plotted in quadrant four (QIV) because the xcoordinate is positive and the y-coordinate is negative. ANSWER: quadrant four (QIV)
25.
PROBLEM:
x > 0 and y < 0
SOLUTION: The point is plotted in quadrant four (QIV) because the x-coordinate is positive and the y-coordinate is negative. ANSWER: Quadrant four (QIV)
26.
PROBLEM:
x < 0 and y < 0
SOLUTION: The point is plotted in quadrant three (QIII) because the xcoordinate is negative and the y-coordinate is negative. ANSWER: quadrant three (QIII)
27.
PROBLEM:
x < 0 and y > 0
SOLUTION: The point is plotted in quadrant two (QII) because the x-coordinate is negative and the y-coordinate is positive. ANSWER: Quadrant two (QII)
28.
PROBLEM:
x > 0 and y > 0
SOLUTION: The point is plotted in quadrant one (QI) because the x-coordinate is positive and the y-coordinate is positive. ANSWER: quadrant one (QI) The average price of a gallon of regular unleaded gasoline in US cities is given in the following line graph. Use the graph to answer the following questions. (Data: Bureau of Labor Statistics)
Scale: x-axis:5 years = 1unit y-axis: $1 = 5 units 29.
PROBLEM: What was the average price of a gallon of unleaded gasoline in 2004? SOLUTION: Year 2004 corresponds to 7 on the x-coordinate. The corresponding y-value is 9. This value corresponds to $1.8. ANSWER: In the year 2004, the average price of a gallon of unleaded gasoline is $1.8.
30.
PROBLEM: What was the average price of a gallon of unleaded gasoline in 1976? SOLUTION: Year 1976 corresponds to 0 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.6. ANSWER: In the year 1976, the average price of a gallon of unleaded gasoline is $0.6.
31.
PROBLEM: In what years was the average price of a gallon of unleaded gasoline $1.20? SOLUTION: $1.20 corresponds to y = 6. This occurs in three data points, when x = 1, x = 2, and x = 5. These values correspond to 1980, 1984, and 1996 respectively. ANSWER: The average price of a gallon of unleaded gasoline was $1.20 in 1980, 1984, and 1996.
32.
PROBLEM: What is the price increase of a gallon on gasoline from 1980 to 2008? SOLUTION: Year 1980 corresponds to 1 on the x-coordinate. The corresponding y-value is 6. This value corresponds to $1.2. Year 2008 corresponds to 8 on the x-coordinate. The corresponding y-value is 16. This value corresponds to $3.2. From 1980 to 2008, the price per gallon of gasoline increases from $1.2 to $3.2. ANSWER: Hence the increase in the price is $3.2 – $1.2 = $2.0.
33.
PROBLEM: What was the percentage increase in the price of a gallon of unleaded gasoline from 1976 to 1980? SOLUTION: Year 1976 corresponds to 0 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.6. Year 1980 corresponds to 1 on the x-coordinate. The corresponding y-value is 6. This value corresponds to $1.2. From 1976 to 1980, the price per gallon of gasoline increases from $0.6 to $1.2. Hence the increase in the price is $1.2 – $0.6 = $0.6.
Increase in price × 100 Initial price $0.6 = × 100 $0.6 = 100%
% Increase in price =
ANSWER: The percentage increase in the price of a gallon of unleaded gasoline from 1976 to 1980 is 100%.
34.
PROBLEM: What was the percentage increase in the price of a gallon of unleaded gasoline from 2000 to 2008? SOLUTION: Year 2000 corresponds to 6 on the x-coordinate. The corresponding y-value is 7.5. This value corresponds to $1.5. Year 2008 corresponds to 8 on the x-coordinate. The corresponding y-value is 16. This value corresponds to $3.2. From 2000 to 2008, the price per gallon of gasoline increases from $1.5 to $3.2. Hence the increase in the price is $3.2 – $1.5 = $1.7. Increase in price % Increase in price = × 100 Initial price $1.7 × 100 = $1.5 = 113% ANSWER: The percentage increase in the price of a gallon of unleaded gasoline from 2000 to 2008 is 113%.
The average price of all purpose white flour in US cities from 1980 to 2008 is given in the following line graph. Use the graph to answer the following questions. (Data: Bureau of Labor Statistics)
Scale: x-axis: 1 unit = 4 years y-axis: 1 unit = $0.1 35.
PROBLEM: What was the average price per pound of all purpose white flour in 2000? SOLUTION: Year 2000 corresponds to 6 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.3. ANSWER: In the year 2000, the average price per pound of all purpose white flour is $0.3.
36.
PROBLEM: What was the average price per pound of all purpose white flour in 2008? SOLUTION: Year 2008 corresponds to 7 on the x-coordinate. The corresponding y-value is 5. This value corresponds to $0.5. ANSWER: In the year 2008, the average price per pound of all purpose white flour is $0.5.
37.
PROBLEM: In what year did the price of flour average $0.25 per pound? SOLUTION: $0.25 corresponds to y = 2.5. This occurs in one data point, when x = 3. This value corresponds to the year 1992. ANSWER: The price of the flour averages $0.25 per pound in the year 1992.
38.
PROBLEM: In what years did the price of flour average $0.20 per pound? SOLUTION: $0.20 corresponds to 2 on the y-coordinate. This data point occurs in three places, when x = 0, x = 1, and x = 2. These values correspond to 1980, 1984, and 1988 respectively. ANSWER: The price of flour averages $0.20 per pound in the years 1980, 1984, and 1988.
39.
PROBLEM: What was the percentage increase in flour from the year 2000 to 2008? SOLUTION: Year 2000 corresponds to 5 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.3. Year 2008 corresponds to 7 on the x-coordinate. The corresponding y-value is 5. This value corresponds to $0.5. From 2000 to 2008, the price per pound of flour increases from $0.3 to $0.5. Hence the increase in the price is $0.5 – $0.3 = $0.2. Increase in price % Increase in price = × 100 Initial price $0.2 = × 100 $0.3 = 67% ANSWER: The percentage increase in the price per pound of flour from 2000 to 2008 is 67%.
40.
PROBLEM: What was the percentage increase in flour from the year 1992 to 2000? SOLUTION: Year 1992 corresponds to 4 on the x-coordinate. The corresponding y-value is 2.5. This value corresponds to $0.25. Year 2000 corresponds to 5 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.3. From 1992 to 2000, the price per pound of flour increases from $0.25 to $0.3. Hence the increase in the price is $0.3 – $0.25 = $0.05. Increase in price % Increase in price = × 100 Initial price $0.05 = × 100 $0.25 = 20% ANSWER: The percentage increase in the price per pound of flour from 1992 to 2000 is 20%.
Given the following data create a line graph.
41.
PROBLEM: The percentage of total high school graduates who enroll in college. (Source: Digest of Education Statistics) Year Percentage 1969 36% 1979 40% 1989 47% 1999 42% ANSWER:
42.
PROBLEM: The average daily temperature given in degrees Fahrenheit in May. Exam 8:00 am 12:00 pm 4:00 pm 8:00 pm 12:00am 4:00 am
Temperature 60 72 75 67 60 55
ANSWER:
Calculate the area of the shape formed by connecting the following set of vertices.
43.
PROBLEM: {(0, 0), (0, 3), (5, 0), (5, 3)} SOLUTION:
When the vertices are connected a rectangle is formed. Area of rectangle = length × breadth Length = 5 – 0 = 5 units. Breadth = 3 – 0 = 3 units. ANSWER: Area = 5 units × 3 units = 15 sq. units
44.
PROBLEM: {(−1, −1), (−1, 1), (1, −1), (1, 1)} SOLUTION:
When the vertices are connected a square is formed. Area of square = (side)2 side = 1 – (–1) = 2 units. ANSWER: Area = (2 units)2 = 4 sq.units
45.
PROBLEM: {(−2, −1), (−2, 3), (5, 3), (5, −1)} SOLUTION:
When the vertices are connected a rectangle is formed. Area of rectangle = length × breadth Length = 5 – (– 2) = 7 units. Breadth = 3 – (–1) = 4 units. ANSWER: Area = 7 units × 4 units = 28 sq. units
46.
PROBLEM: {(−5, −4), (−5, 5), (3, 5), (3, −4)} SOLUTION:
When the vertices are connected a rectangle is formed. Area of rectangle = length × breadth Length = 3 – (– 5) = 8 units. Breadth = 5 – (–4) = 9 units. ANSWER: Area = 8 units × 9 units = 72 sq.units
47.
PROBLEM: {(0, 0), (4, 0), (2, 2)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 4 – 0 = 4 units. Height = 2 – 0 = 2 units. ANSWER: 1 Area = × 4 units × 2 units = 4 sq.units 2
48.
PROBLEM: {(−2, −2), (2, −2), (0, 2)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 2 – (– 2) = 4 units. Height = 2 – (– 2) = 4 units. 1 ANSWER: Area = × 4 units × 4 units = 8 sq.units 2
49.
PROBLEM: {(0, 0), (0, 6), (3, 4)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 6 – 0 = 6 units. Height = 3 – 0 = 3 units. ANSWER: 1 Area = × 6 units × 3 units = 9 sq.units 2
50.
PROBLEM: {(−2, 0), (5, 0), (3, −3)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 5 – (– 2) = 7 units. Height = 0 – (– 3) = 3 units. 1 1 ANSWER: Area = × 7 units × 3 units = 10 sq.units 2 2
Part B: Distance Formula Calculate the distance between the given two points.
51.
PROBLEM: (−5, 3) and (−1, 6) SOLUTION: x2 = −1, x1 = −5, y2 = 6, y1 = 3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣ −1 − ( −5 ) ⎤⎦ + ( 6 − 3 )
=
[ −1 + 5] + ( 6 − 3 )
=
[ 4] + ( 3)
2
2
2
2
2
2
= 16 + 9 =
25
= 5 units
ANSWER: 5 units
52.
PROBLEM: (6, −2) and (−2, 4) SOLUTION: x2 = −2, x1 = 6, y2 = 4, y1 = −2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
( −2 − 6 )
2
+ ⎡⎣ 4 − ( −2 ) ⎤⎦
=
( −2 − 6 )
2
+ [ 4 + 2]
=
( −8 )
=
64 + 36
2
2
+ [6]
2
= 100 = 10 units
ANSWER: 10 units
2
53.
PROBLEM: (0, 0) and (5, 12) SOLUTION: x2 = 5, x1 = 0, y2 = 12, y1 = 0 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 5 − 0 ) + (12 − 0 )
=
( 5 ) + (12 )
=
25 + 144
2
2
2
2
2
= 169 = 13 units
ANSWER: 13 units
54.
PROBLEM: (−6, −8) and (0, 0) SOLUTION: x2 = 0, x1 = −6 , y2 = 0, y1 = −8 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣ 0 − ( −6 ) ⎤⎦ + ⎣⎡ 0 − ( −8 ) ⎦⎤
=
( −6 )
2
2
+ ( −8 )
2
= 36 + 64 = 100 = 10 units
ANSWER: 10 units
2
55.
PROBLEM: (−7, 8) and (5, −1) SOLUTION: x2 = 5, x1 = −7 , y2 = −1, y1 = 8 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣5 − ( −7 ) ⎤⎦ + ( −1 − 8 )
=
[ 5 + 7 ] + ( −1 − 8 )
=
[12]
2
2
2
+ ( −9 )
2
2
2
= 144 + 81 =
225
= 15 units
ANSWER: 15 units
56.
PROBLEM: (−1, −2) and (9, 22) SOLUTION: x2 = 9, x1 = −1 , y2 = 22, y1 = −2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣9 − ( −1) ⎤⎦ + ⎡⎣ 22 − ( −2 ) ⎤⎦
=
[9 + 1]
=
[10]
2
2
2
+ ( 22 + 2 )
+ [ 24 ]
2
= 100 + 576 =
676
= 26 units
ANSWER: 26 units
2
2
57.
PROBLEM: (−1, 2) and (−7/2, −4) SOLUTION: x2 = −7/2 , x1 = −1 , y2 = −4, y1 = 2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
=
2 ⎡ 7 ⎤ ⎢⎣ − 2 − ( −1) ⎥⎦ + ( −4 − 2 )
=
2 ⎡ 7 ⎤ ⎢⎣ − 2 + 1⎥⎦ + ( −4 − 2 )
=
2 ⎡ 5 ⎤ ⎢⎣ − 2 ⎥⎦ + ( −6 )
=
25 + 36 4
2
2
169 4 13 = units 2 =
ANSWER: 13 units 2
58.
PROBLEM: ⎛ 1 1⎞ ⎜ − , ⎟ and ⎝ 2 3⎠
⎛ 5 11 ⎞ ⎜ ,− ⎟ ⎝2 3 ⎠
SOLUTION: x2 = d=
5 1 11 1 , x1 = − , y2 = − , y1 = 2 2 3 3
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
=
⎡ 5 ⎛ 1 ⎞⎤ ⎛ 11 1 ⎞ ⎢ 2 − ⎜ − 2 ⎟⎥ + ⎜ − 3 − 3 ⎟ ⎝ ⎠⎦ ⎝ ⎠ ⎣
=
⎡5 1⎤ ⎛ 11 1 ⎞ ⎢⎣ 2 + 2 ⎥⎦ + ⎜⎝ − 3 − 3 ⎟⎠
=
⎡6⎤ ⎛ 12 ⎞ ⎢⎣ 2 ⎥⎦ + ⎜⎝ − 3 ⎟⎠
=
[3]
2
2
2
+ ( −4 )
2
= 9 + 16 =
25
= 5 units
ANSWER: 5 units
2
2
2
59.
PROBLEM: ⎛ 1 2⎞ ⎛ 1⎞ ⎜ − , ⎟ and ⎜1,− ⎟ ⎝ 3 3⎠ ⎝ 3⎠ SOLUTION: 1 1 2 x2 =1, x1 = − , y2 = − , y1 = 3 3 3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
=
⎡ ⎛ 1 ⎞⎤ ⎛ 1 2⎞ ⎢1 − ⎜ − 3 ⎟ ⎥ + ⎜ − 3 − 3 ⎟ ⎠⎦ ⎝ ⎠ ⎣ ⎝
=
⎡ 1⎤ ⎛ 1 2⎞ ⎢1 + 3 ⎥ + ⎜ − 3 − 3 ⎟ ⎣ ⎦ ⎝ ⎠
=
⎡ 3 + 1⎤ ⎛ 1 2⎞ ⎢⎣ 3 ⎥⎦ + ⎜⎝ − 3 − 3 ⎟⎠
=
⎡4⎤ ⎛ −3 ⎞ ⎢⎣ 3 ⎦⎥ + ⎝⎜ 3 ⎠⎟
=
16 9 + 9 9
=
25 9
2
2
2
=
5 units 3
ANSWER: 5 units 3
2
2
2
2
60.
PROBLEM: ⎛1 3⎞ ⎜ , − ⎟ and ⎝2 4⎠
⎛3 1⎞ ⎜ , ⎟ ⎝2 4⎠
SOLUTION: x2 = d=
3 1 1 3 , x1 = , y2 = , y1 = − 2 2 4 4
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
⎡ 1 ⎛ 3 ⎞⎤ ⎛3 1⎞ = ⎜ − ⎟ + ⎢ − ⎜ − ⎟⎥ ⎝2 2⎠ ⎣ 4 ⎝ 4 ⎠⎦ 2
⎛3 1⎞ ⎡1 3⎤ = ⎜ − ⎟ + ⎢ + ⎥ ⎣4 4⎦ ⎝2 2⎠ 2
⎛2⎞ ⎛4⎞ = ⎜ ⎟ +⎜ ⎟ ⎝2⎠ ⎝4⎠ =
(1)
=
2 units
2
+ (1)
2
2
2
ANSWER: 2 units
61.
PROBLEM: (1, 2) and (4, 3) SOLUTION: x2 = 4, x1 = 1, y2 = 3, y1 = 2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 4 − 1)
=
( 3)
2
2
+ (3 − 2)
+ (1)
= 9+1 = 10 units
ANSWER: 10 units
2
2
2
2
62.
PROBLEM: (2,−4) and (−3, −2) SOLUTION: x2 = −3, x1 = 2, y2 = −2, y1 = −4 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( −3 − 2 )
=
( −3 − 2 )
=
( −5 )
=
25 + 4
=
29 units
2
+ ⎡⎣ −2 − ( −4 ) ⎦⎤ + [ −2 + 4 ]
2
2
+ ( 2)
2
2
2
ANSWER: 29 units
63.
PROBLEM: (−1, 5) and (1, −3) SOLUTION: x2 = 1, x1 = −1, y2 = −3, y1 = 5 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣1 − ( −1) ⎤⎦ + ( −3 − 5 )
=
[1 + 1]
=
[ 2]
=
68
=
4 × 17
2
2
2
+ ( −3 − 5 )
+ ( −8 )
= 2 17 units
ANSWER: 2 17 units
2
2
2
2
64.
PROBLEM: (1, −7) and (5, −1) SOLUTION: x2 = 5, x1 = 1, y2 = −1, y1 = −7 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 5 − 1)
=
( 5 − 1)
=
( 4)
2
+ ⎡⎣ −1 − ( −7 ) ⎤⎦
2
2
2
2
+ [ −1 + 7 ]
2
+ (6)
2
= 16 + 36 = 52 =
4 × 13
= 2 13 units
ANSWER: 2 13 units
65.
PROBLEM: (−7, −3) and (−1, 6) SOLUTION: x2 = −1, x1 = −7, y2 = 6, y1 = −3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣ −1 − ( −7 ) ⎤⎦ + ⎡⎣ 6 − ( −3 ) ⎦⎤
=
[ −1 + 7 ]
=
[6]
2
2
2
+ [ 6 + 3]
2
+ [9 ]
= 36 + 81 = 117 = 13 × 9 = 3 13 units
ANSWER: 3 13 units
2
2
66.
PROBLEM: (0, 1) and (1, 0) SOLUTION: x2 = 1, x1 = 0, y2 = 0, y1 = 1 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
(1 − 0 )
=
(1)
2
2
+ ( 0 − 1)
+ (1)
2
2
2
= 1+1 =
2 units
ANSWER: 2 units
67.
PROBLEM: (−0.2, −0.2) and (1.8, 1.8) SOLUTION: x2 = 1.8, x1 = −0.2, y2 = 1.8, y1 = −0.2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣1.8 − ( −0.2 ) ⎤⎦ + ⎡⎣1.8 − ( −0.2 ) ⎤⎦
=
[1.8 + 0.2]
= =
2
2
[ 2.0]
2
4+4
= 8 = 2.8 units
ANSWER: 2.8 units
+ [1.8 + 0.2 ]
2
+ [ 2.0 ]
2
2
68.
PROBLEM: (1.2, −3.3) and (2.2, −1.7) SOLUTION: x2 = 2.2, x1 = 1.2, y2 = −1.7, y1 = −3.3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 2.2 − 1.2 )
=
( 2.2 − 1.2 )
=
(1.0 )
2
2
2
+ ⎡⎣ −1.7 − ( −3.3 ) ⎤⎦
2
+ [ −1.7 + 3.3]
2
+ [1.6 ]
2
2
= 1.0 + 2.56 = 3.56 = 1.9 units
ANSWER: 1.9 units Show that the three points form a right triangle.
69.
PROBLEM: (−3, −2), (0, −2), and (0,4) SOLUTION: Points: (−3, −2) and (0, −2)
a = ⎡⎣0 − ( −3) ⎤⎦ + ⎡⎣ −2 − ( −2 ) ⎤⎦ 2
=
[3] + [0] 2
2
= 9 =3 Points: (0, −2) and (0, 4)
b=
= =
( 0 − 0)
[4
+ ⎡⎣ 4 − ( −2 ) ⎤⎦
2
2
+ 2]
( 6)
2
2
= 36 = 6 Points: (0, 4) and (−3, −2)
2
( −3 − 0 ) + ( −2 − 4 ) 2
c=
( −3)
= =
+ ( −6 )
2
2
2
9 + 36
= 45
Now we check to see if a 2 + b 2 = c 2 . ANSWER: a 2 + b2 = c2
( 3) + ( 6 ) 2
2
=
(
9 + 36 = 45 45 = 45 45 = 45
45
)
2
70.
PROBLEM: (7,12), (7, −13), and (−5, −4) SOLUTION: Points: (7, 12) and (7, −13)
( 7 − 7 ) + ( −13 − 12 ) 2
a=
( 0)
=
2
+ ( −25)
2
2
= 625 = 25 Points: (7, −13) and (−5, −4) b=
( −5 − 7 )
2
+ ⎡⎣ −4 − ( −13) ⎤⎦
=
( −5 − 7 )
2
+ [ −4 + 13]
=
( −12 )
2
2
2
+ [ 9]
2
= 144 + 81 = 225 = 15 Points: (−5, −4) and (7, 12) c = ⎡⎣7 − ( −5) ⎤⎦ + ⎡⎣12 − ( −4 ) ⎤⎦ 2
[ 7 + 5]
2
=
+ (12 + 4 )
(12 ) + (16 ) 2
=
2
2
2
= 144 + 256 = 400 = 20 Now we check to see if b 2 + c 2 = a 2 . b2 + c2 = a 2
(15 ) + ( 20 ) 2
2
= ( 25 )
225 + 400 = 625 625 = 625
ANSWER:
2
71.
PROBLEM: (−1.4, 0.2), (1, 2) and (1, −3) SOLUTION: Points: (−1.4, −0.2) and (1, 2)
a = ⎡⎣1 − ( −1.4 ) ⎤⎦ + ( 2 − 0.2 ) 2
=
2 2 [1 + 1.4] + ( 2 − 0.2 )
=
[ 2.4]
2
+ (1.8 )
2
2
= 5.76 + 3.24 = 9 =3 Points: (1, 2) and (1, −3)
(1 − 1) + ( −3 − 2 ) 2
b=
2
[ −5]
2
=
= 25 = 5 Points: (1, −3) and (−1.4, 0.2)
( −1.4 − 1)
c=
2
+ ⎡⎣0.2 − ( −3) ⎤⎦
2
( −1.4 − 1) + [0.2 + 3] 2
=
2
( −2.4 ) + [3.2] 2
=
2
= 5.76 + 10.24 = 16 =4
Now we check to see if a 2 + c 2 = b 2 . ANSWER: a 2 + c2 = b2
( 3) + ( 4 ) 2
2
= ( 5)
9 + 16 = 25 25 = 25
2
72.
PROBLEM: (2,−1), (−1,2), and (6,3) SOLUTION: Points: (2,−1) and ( −1,2)
a=
( −1 − 2 )
2
+ ⎡⎣ 2 − ( −1) ⎤⎦
=
( −1 − 2 ) + [ 2 + 1]
=
( −3) + [3]
2
2
2
2
= 9+9 = 18 Points: (−1,2) and (6,3)
2
b = ⎡⎣6 − ( −1) ⎤⎦ + ( 3 − 2 ) 2
=
[6 + 1]
=
[7]
2
+ (3 − 2)
+ (1)
2
2
2
2
= 49 + 1 = 50 Points: (6,3) and (2,−1) c=
( 2 − 6 ) + ( −1 − 3) 2
=
( −4 )
=
16 + 16
2
+ ( −4 )
2
2
= 32 Now we check to see if a 2 + c 2 = b 2 . a 2 + c2 = b2
( 18 ) + ( 2
32
) =( 2
18 + 32 = 50 50 = 50
ANSWER:
50
)
2
73.
PROBLEM: (−5,2), (−1, −2), and (−2,5) SOLUTION: Points: (−5,2) and (−1, −2)
a = ⎡⎣ −1 − ( −5) ⎤⎦ + ( −2 − 2 ) 2
=
2 2 [ −1 + 5] + ( −2 − 2 )
=
2 2 [ 4] + ( −4 )
2
= 16 + 16 = 32 Points: (−1, −2) and (−2,5) b = ⎡⎣ −2 − ( −1) ⎤⎦ + ⎡⎣5 − ( −2 ) ⎤⎦ 2
=
2
[ −2 + 1] + [5 + 2] 2
2
=
[ −1] + [7]
=
1 + 49
2
2
= 50 Points: (−2,5) and (−5,2) c = ⎡⎣ −5 − ( −2 ) ⎤⎦ + ( 2 − 5) 2
= =
2
2 2 [ −5 + 2] + ( 2 − 5) 2 2 [ −3] + ( −3)
= 9+9 = 18 Now we check to see if a 2 + c 2 = b 2 . ANSWER: a 2 + c2 = b2
(
32
) + ( 18 ) = ( 2
2
32 + 18 = 50 50 = 50
50
)
2
74.
PROBLEM: (1,−2), (2,3), and (−3,4) SOLUTION: Points: (1,−2) and ( 2,3)
a=
( 2 − 1)
2
+ ⎡⎣3 − ( −2 ) ⎤⎦
=
( 2 − 1) + [3 + 2]
=
(1) + [5]
2
2
2
2
= 1 + 25 = 26 Points: (2,3) and (−3,4)
2
( −3 − 2 ) + ( 4 − 3) 2
b=
( −5)
= =
2
+ (1)
2
2
25 + 1
= 26
Points: (−3,4) and (1,−2) c = ⎡⎣1 − ( −3) ⎤⎦ + ( −2 − 4 ) 2
=
2
2 2 [1 + 3] + ( −2 − 4 )
=
2 2 [ 4] + ( −6 )
= 16 + 36 = 52 Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c2
(
26
) +( 2
26
) =( 2
26 + 26 = 52 52 = 52
ANSWER:
52
)
2
Isosceles triangles have two legs of equal length. Show that the following points form an isosceles triangle. 75.
PROBLEM: (1, 6), (−1, 1) and (3, 1) SOLUTION: Points: (1,6) and (−1, 1) a= =
( −1 − 1) + (1 − 6 ) 2
( −2 ) + ( −5) 2
2
2
= 4 + 25 = 29
Points: (−1, 1) and (3,1)
b = ⎡⎣3 − ( −1) ⎤⎦ + (1 − 1) 2
= =
2
2 2 [3 + 1] + (1 − 1) 2 2 [ 4] + ( 0 )
= 16 =4 Points: (3,1) and (1,6) c= = =
(1 − 3) + ( 6 − 1) 2
( −2 ) + ( 5) 2
2
2
4 + 25
= 29
ANSWER: The edges a and c are equal. Hence the following points form an isosceles triangle.
76.
PROBLEM: (−6, −2), (−3, −5) and (−9, −5) SOLUTION: Points: (−6, −2) and (−3, −5)
a = ⎡⎣ −3 − ( −6 ) ⎤⎦ + ⎡⎣ −5 − ( −2 ) ⎤⎦ 2
=
[ −3 + 6] + [ −5 + 2]
=
[3] + [ −3]
2
2
2
2
2
= 9+9 = 18 Points: (−3, −5 ) and (-9,-5) b = ⎡⎣ −9 − ( −3) ⎤⎦ + ⎡⎣ −5 − ( −5) ⎤⎦ 2
=
[ −9 + 3] + [ −5 + 5]
=
[ −6] + [0]
2
= 36 =6
2
2
2
2
Points: (−9, −5 ) and (−6, −2) c = ⎡⎣ −6 − ( −9 ) ⎤⎦ + ⎡⎣ −2 − ( −5) ⎤⎦ 2
=
[ −6 + 9] + [ −2 + 5]
=
[ −3] + [3]
2
2
2
2
2
= 9+9 = 18 ANSWER: The edges a and c are equal. Hence the following points form an isosceles triangle.
77.
PROBLEM: (−3, 0), (0, 3) and (3, 0) SOLUTION: Points: (−3, 0) and (0, 3)
a = ⎡⎣0 − ( −3) ⎤⎦ + ( 3 − 0 ) 2
=
2
2 2 [3] + ( 3)
= 9+9 = 18 Points: (0, 3) and (3, 0) b= =
( 3 − 0 ) + ( 0 − 3) 2
( 3) + ( −3) 2
=
9+9
=
18
2
2
Points: (3, 0) and (−3, 0) c= =
( −3 − 3) + ( 0 − 0 ) 2
( −6 ) + ( 0 ) 2
2
2
= 36 =6 ANSWER: The edges a and b are equal. Hence the following points form an isosceles triangle.
78.
PROBLEM: (0, −1), (0, 1) and (1, 0) SOLUTION: Points: (0, −1) and (0, 1)
a=
( 0 − 0)
2
+ ⎡⎣1 − ( −1) ⎦⎤
=
( 0 − 0 ) + [1 + 1]
=
[ 2]
2
2
2
2
= 4 =2 Points: (0, 1) and (1, 0) b= =
(1 − 0 ) + ( 0 − 1) 2
(1) + (1) 2
2
2
= 1+1 = 2
Points: (1, 0) and (0, −1) c= =
( 0 − 1) + ( −1 − 0 ) 2
( −1) + ( −1) 2
2
2
= 1+1 = 2
ANSWER: The edges b and c are equal. Hence the following points form an isosceles triangle.
Calculate the area and the perimeter of the triangles formed by the following set of vertices.
79.
PROBLEM: {(−4, −5), (−4, 3), (2, 3)} SOLUTION: Points: (−4, −5) and (−4, 3)
a = ⎡⎣ −4 − ( −4 ) ⎤⎦ + ⎡⎣3 − ( −5) ⎤⎦ 2
=
[ −4 + 4] + [3 + 5]
=
[0] + [8]
2
2
2
2
2
= 64 =8 Points: (−4, 3) and (2, 3) b = ⎡⎣ 2 − ( −4 ) ⎤⎦ + ( 3 − 3) 2
= =
2
2 2 [ 2 + 4] + ( 0 )
[ 6]
2
= 36 = 6 Points: (2, 3) and (−4, −5) c= = =
( −4 − 2 ) + ( −5 − 3) 2
( −6 ) + ( −8) 2
2
2
36 + 64
= 100 = 10 ANSWER:
Perimeter = a + b + c = 8 + 6 + 10 = 24 units Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c2
(8) + ( 6 ) 2
2
= (10 )
2
64 + 36 = 100 100 = 100
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge c = 10 is the hypotenuse. 1 Area = × base × height 2 1 = × a×b 2 1 = ×8× 6 2 = 24 sq.units 80.
PROBLEM: {(−1, 1), (3, 1), (3, −2)} SOLUTION: Points: (−1, 1) and (3, 1)
a = ⎡⎣3 − ( −1) ⎤⎦ + (1 − 1) 2
=
2 2 [3 + 1] + (1 − 1)
=
2 2 [ 4] + ( 0 )
2
= 16 = 4 Points: (3, 1) and (3, −2) b= =
( 3 − 3) + ( −2 − 1) 2
( 0 ) + ( −3) 2
2
2
= 9 = 3 Points: (3, −2) and (−1, 1)
( −1 − 3)
c=
2
+ ⎡⎣1 − ( −2 ) ⎤⎦
2
( −1 − 3) + [1 + 2] 2
=
2
( −4 ) + [3] 2
=
2
= 16 + 9 = 25 = 5 Perimeter = a + b + c = 3 + 4 + 5 = 12 units Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c 2
( 3) + ( 4 ) 2
2
= ( 5)
2
9 + 16 = 25 25 = 25
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge c = 5 is the hypotenuse. 1 Area = × base × height 2 1 = × a×b 2 1 = × 4×3 2 = 6 sq.units ANSWER: 6 sq.units , 12 units
81.
PROBLEM: {(−3, 1), (−3, 5), (1, 5)} SOLUTION: Points: (−3, 1) and (−3, 5)
a = ⎡⎣ −3 − ( −3) ⎤⎦ + ( 5 − 1) 2
=
2 2 [ −3 + 3] + ( 5 − 1)
=
2 2 [ 0] + ( 4 )
2
= 16 =4 Points: (−3, 5) and (1, 5) b = ⎡⎣1 − ( −3) ⎤⎦ + ( 5 − 5) 2
2
2 2 [1 + 3] + ( 5 − 5)
=
2 2 [ 4] + ( 0 )
=
= 16 = 4 Points: (1, 5) and (−3, 1)
( −3 − 1) + (1 − 5) 2
c= =
( −4 ) + ( −4 )
=
16 + 16
2
2
2
= 32 = 2 ×16 =4 2 ANSWER: Perimeter = a + b + c = 4 + 4 + 4 2 = 8+ 4 2 units
Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c2
( 4) + ( 4) 2
2
(
= 4 2
16 + 16 = 32 32 = 32
)
2
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge c = 4 2 is the hypotenuse. 1 Area = × base × height 2 1 = × a×b 2 1 = × 4× 4 2 = 8 sq.units 82.
PROBLEM: {(−3, −1), (−3, 7), (1, −1)} SOLUTION: Points: (−3, −1) and (−3, 7)
a = ⎡⎣ −3 − ( −3) ⎤⎦ + ⎡⎣7 − ( −1) ⎤⎦ 2
=
[ −3 + 3] + [7 + 1]
=
[ 0 ] + [ 8]
2
2
2
2
= 64 =8 Points: (−3, 7) and (1, −1) b = ⎡⎣1 − ( −3) ⎦⎤ + ( −1 − 7 ) 2
=
2
2 2 [1 + 3] + ( −1 − 7 )
=
2 2 [ 4] + ( −8)
=
16 + 64
= 80 = 16 × 5 =4 5 Points: (1, −1) and (−3, −1)
2
( −3 − 1)
c=
2
+ ⎡⎣ −1 − ( −1) ⎤⎦
2
( −3 − 1) + [ −1 + 1] 2
=
2
( −4 ) + [0] 2
=
2
= 16 =4 Perimeter = a + b + c = 8 + 4 5 + 4 = 12 + 4 5 units Now we check to see if a 2 + c 2 = b 2 . a 2 + c2 = b2
(8) + ( 4 ) 2
2
(
= 4 5
)
2
64 + 16 = 80 80 = 80
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge b = 4 5 is the hypotenuse. 1 Area = × base × height 2 1 = ×a×c 2 1 = ×8× 4 2 = 16 sq.units ANSWER: 16 sq.units , 12 + 4 5 units
Part C: Midpoint Formula Find the midpoint between the given two points.
83.
PROBLEM: (−1,6) and (−7, −2) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−1, 6) (−7, −2) x1 + x2 −1 − 7 −8 = = = −4 2 2 2 y1 + y2 6 − 2 4 = = =2 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( −4, 2 ) ANSWER: ( −4, 2 )
84.
PROBLEM: (8, 0) and (4,−3) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(8, 0) (4, −3) x1 + x2 8 + 4 12 = = =6 2 2 2 y1 + y2 0 − 3 −3 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 3⎞ ⎛ ⎜ 6, − ⎟ 2⎠ ⎝
3⎞ ⎛ ANSWER: ⎜ 6, − ⎟ 2⎠ ⎝
85.
PROBLEM: (−10,0) and (10,0) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−10, 0) (10, 0) x1 + x2 −10 + 10 0 = = =0 2 2 2 y1 + y2 0 + 0 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 0, 0 ) ANSWER: ( 0, 0 )
86.
PROBLEM: (−3, −6) and (−3,6) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−3, −6) (−3, 6) x1 + x2 −3 − 3 −6 = = = −3 2 2 2 y1 + y2 −6 + 6 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( −3, 0 ) ANSWER: ( −3, 0 )
87.
PROBLEM: (−10,5) and (14,−5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−10, 5) (14, −5) x1 + x2 −10 + 14 4 = = =2 2 2 2 y1 + y2 5 − 5 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 2, 0 ) ANSWER: ( 2, 0 )
88.
PROBLEM: (0, 1) and (2, 2) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(0,1) (2, 2) x1 + x2 0 + 2 2 = = =1 2 2 2 y1 + y2 1 + 2 3 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛ 3⎞ ⎜ 1, ⎟ ⎝ 2⎠ ⎛ 3⎞ ANSWER: ⎜1, ⎟ ⎝ 2⎠
89.
PROBLEM: (5,−3) and (4,−5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( 5, −3) ( 4, −5)
x1 + x2 5 + 4 9 = = 2 2 2 y1 + y2 −3 − 5 −8 = = = −4 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
⎛9 ⎞ ⎜ ,− 4⎟ ⎝2 ⎠ ANSWER: ⎛9 ⎞ ⎜ ,− 4⎟ ⎝2 ⎠
90.
PROBLEM: (0, 0) and (1, 1) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(0, 0) (1,1) x1 + x2 0 + 1 1 = = 2 2 2 y1 + y2 0 + 1 1 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛1 1⎞ ⎜ , ⎟ ⎝2 2⎠ ⎛1 1⎞ ANSWER: ⎜ , ⎟ ⎝2 2⎠
91.
PROBLEM: (−1,−1) and (4,4) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( −1, −1) ( 4, 4 )
x1 + x2 −1 + 4 3 = = 2 2 2 y1 + y2 −1 + 4 3 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
⎛3 3⎞ ⎜ , ⎟ ⎝2 2⎠ ANSWER: ⎛3 3⎞ ⎜ , ⎟ ⎝2 2⎠
92.
PROBLEM: (3, −5) and (3, 5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(3, −5) (3,5) x1 + x2 3 + 3 6 = = =3 2 2 2 y1 + y2 −5 + 5 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 3, 0 )
ANSWER: ( 3, 0 )
93.
PROBLEM: ⎛ 1 1⎞ ⎜ − , − ⎟ and ⎝ 2 3⎠
⎛3 7⎞ ⎜ , ⎟ ⎝2 3⎠
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
⎛ 1 1⎞ ⎛ 3 7⎞ ⎜− ,− ⎟ ⎜ , ⎟ ⎝ 2 3⎠ ⎝ 2 3⎠ 1 3 2 − + x1 + x2 2 1 = 2 2= 2= = 2 2 2 4 2 1 7 6 − + y1 + y2 6 = 3 3 = 3 = =1 2 2 2 6 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛1 ⎞ ⎜ , 1⎟ ⎝2 ⎠ ANSWER: ⎛1 ⎞ ⎜ , 1⎟ ⎝2 ⎠
94.
PROBLEM: ⎛3 2⎞ ⎜ , − ⎟ and ⎝4 3⎠
⎛1 1⎞ ⎜ ,− ⎟ ⎝8 2⎠
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
⎛3 2⎞ ⎛1 1⎞ ⎜ ,− ⎟ ⎜ ,− ⎟ ⎝4 3⎠ ⎝8 2⎠ 3 1 3⋅ 2 +1 6 +1 7 + x1 + x2 4 8 7 = = 8 = 8 =8 = 2 2 2 2 2 16 2 1 −2 ⋅ 2 − 1 ⋅ 3 − 4 − 3 7 − − − y1 + y2 7 6 = 3 2= = 6 = 6 =− 2 2 2 2 2 12 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 7⎞ ⎛7 ⎜ ,− ⎟ ⎝ 16 12 ⎠
7⎞ ⎛7 ANSWER: ⎜ , − ⎟ ⎝ 16 12 ⎠ 95.
PROBLEM: ⎛5 1⎞ ⎛ 1 3⎞ ⎜ , ⎟ and ⎜ − , − ⎟ ⎝ 6 2⎠ ⎝3 4⎠ SOLUTION: ( x1 , y1 ) ( x2 , y2 )
⎛5 1⎞ ⎛ 1 3⎞ ⎜ , ⎟ ⎜− ,− ⎟ ⎝3 4⎠ ⎝ 6 2⎠ 5 1 5 ⋅ 6 − 1 ⋅ 3 30 − 3 27 − x1 + x2 3 6 27 3 18 = = = 18 = 18 = = 2 2 2 2 2 36 4 5 1 3 1− 3⋅ 2 1− 6 − − y1 + y2 4 2 5 = = 4 = 4 = 4 =− 2 2 2 2 2 8 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 5⎞ ⎛3 ⎜ ,− ⎟ 8⎠ ⎝4 ANSWER: 5⎞ ⎛3 ⎜ ,− ⎟ 8⎠ ⎝4
96.
PROBLEM: 1⎞ ⎛ 1 5⎞ ⎛ 7 ⎜ − , − ⎟ and ⎜ , − ⎟ ⎝ 5 2⎠ ⎝ 10 4 ⎠ SOLUTION: ( x1 , y1 ) ( x2 , y2 )
1⎞ ⎛ 1 5⎞ ⎛ 7 ⎜− ,− ⎟ ⎜ ,− ⎟ ⎝ 5 2 ⎠ ⎝ 10 4 ⎠ 1 7 5 − 1 ⋅ 2 + 7 −2 + 7 − + x1 + x2 5 1 10 = 5 10 = = 10 = 10 = = 2 2 2 2 2 20 4 5 1 −5 ⋅ 2 − 1 −10 − 1 11 − − − y1 + y2 11 4 4 = 2 4= = = 4 =− 2 2 2 2 2 8
⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛ 1 11 ⎞ ⎜ ,− ⎟ 8⎠ ⎝4 ⎛ 1 11 ⎞ ANSWER: ⎜ , − ⎟ 8⎠ ⎝4 97.
PROBLEM: Given the right triangle formed by the vertices (0, 0), (0, 6), and (6, 8), show that the midpoints of the sides form a right triangle. SOLUTION: Points: (0,0) and (0,6) ( x1 , y1 ) ( x2 , y2 )
(0, 0) (0, 6) x1 + x2 0 + 0 0 = = =0 2 2 2 y1 + y2 0 + 6 6 = = =3 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 0, 3) Points: (0,6) and (6,8) ( x1 , y1 ) ( x2 , y2 )
( 0, 6 ) ( 6,8)
x1 + x2 0 + 6 6 = = =3 2 2 2 y1 + y2 6 + 8 14 = = =7 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 3, 7 )
Points: (6,8) and (0,0) ( x1 , y1 ) ( x2 , y2 )
( 6,8) ( 0, 0 )
x1 + x2 6 + 0 6 = = =3 2 2 2 y1 + y2 8 + 0 8 = = =4 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 3, 4 )
Considering the midpoints as the vertices, (0,3), (3,7), and (3,4). Points: (0, 3) and (3, 7) a= =
( 3 − 0 ) + ( 7 − 3) 2
( 3) + ( 4 ) 2
2
2
= 9 + 16 = 25 =5 Points: (3, 7) and (3, 4) b= =
( 3 − 3) + ( 4 − 7 ) 2
( 0 ) + ( −3) 2
2
2
= 9 = 3 Points: (3, 4) and (0, 3) c= = =
( 0 − 3) + ( 3 − 4 ) 2
( −3) + ( −1) 2
2
2
9 +1
= 10
Now we check to see if b 2 + c 2 = a 2 . b2 + c2 = a 2
( 10 )
2
+ ( 3) = ( 5 ) 2
2
10 + 9 = 25 19 ≠ 25 ANSWER: The mid points of the vertices do not form a right triangle.
98.
PROBLEM: Given the isosceles triangle formed by the vertices (−10, −12), (0, 12), and (10, −12), show that the midpoints of the sides also forms an isosceles triangle. SOLUTION: Points: (−10, −12) and (0,12) ( x1 , y1 ) ( x2 , y2 )
(−10, −12) (0,12) x1 + x2 −10 + 0 −10 = = = −5 2 2 2 y1 + y2 −12 + 12 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( −5, 0 ) Points: (0, 12) and (10, −12) ( x1 , y1 ) ( x2 , y2 )
( 0, 12 ) (10,
− 12 )
x1 + x2 0 + 10 10 = = =5 2 2 2 y1 + y2 12 − 12 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 5, 0 )
Points: (10, −12) and (−10, −12) ( x1 , y1 ) ( x2 , y2 )
(10,
− 12 )
( −10,
− 12 )
x1 + x2 10 − 10 0 = = =0 2 2 2 y1 + y2 −12 − 12 −24 = = = −12 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 0, − 12 )
Considering the midpoints as the vertices, (−5,0), (5,0), and (0, −12). Points: (−5,0) and (5,0)
a = ⎡⎣5 − ( −5 ) ⎤⎦ + ( 0 − 0 ) 2
=
2 2 [5 + 5] + ( 0 − 0 )
=
[10]
2
2
= 100 = 10 Points: (5,0) and (0, −12)
b= = =
( 0 − 5) + ( −12 − 0 ) 2
( −5) + ( −12 ) 2
2
2
25 + 144
= 169 = 13 Points: (0, −12) and (−5,0) c= = =
( −5 − 0 )
2
+ ⎡⎣0 − ( −12 ) ⎤⎦
2
( −5) + [ −12] 2
2
25 + 144
= 169 = 13
ANSWER: The sides b and c are equal. Hence the midpoints of the sides also form an isosceles triangle.
99.
PROBLEM: Calculate the area of the triangle formed by the vertices (−4, −3), (−1, 1), and (2, −3). (Hint: the vertices form an isosceles triangle.) SOLUTION: Points: (−4, −3) and (−1, 1)
a = ⎣⎡ −1 − ( −4 ) ⎦⎤ + ⎡⎣1 − ( −3) ⎤⎦ 2
=
[ −1 + 4] + [1 + 3]
=
[3] + [ 4]
2
2
2
2
2
= 9 + 16 = 25 =5 Points: (−1, 1) and (2, −3)
b = ⎡⎣ 2 − ( −1) ⎤⎦ + ( −3 − 1) 2
=
2
2 2 [ 2 + 1] + ( −3 − 1)
=
2 2 [3] + ( −4 )
=
9 + 16
= 25 =5 Points: (2, −3) and (−4, −3)
c= = =
( −4 − 2 )
2
+ ⎡⎣ −3 − ( −3) ⎤⎦
2
( −4 − 2 ) + [ −3 + 3] 2
2
( −6 ) + [ −0] 2
2
= 36 =6 The edges a and b are equal. The unequal side of the triangle is the base of the triangle. The height of the triangle is the distance between the mid-point of the base vertices, i.e. (2, −3) and (−4, −3), and the common end of the equal sides, i.e. (−1, 1). Mid-point of (2, −3) and (−4, −3):
Points: (2, −3) and (−4, −3) ( x1 , y1 ) ( x2 , y2 )
( 2,
− 3)
( −4,
− 3)
x1 + x2 2 − 4 −2 = = = −1 2 2 2 y1 + y2 −3 − 3 −6 = = = −3 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( −1, − 3)
Distance between the (−1, −3) and (−1, 1): Points: (−1, −3) and (−1, 1) Height = ⎡⎣ −1 − ( −1) ⎦⎤ + ⎡⎣1 − ( −3) ⎤⎦ 2
=
[ −1 + 1] + [1 + 3]
=
[ 0] + [ 4 ]
2
2
2
2
2
= 16 =4 ANSWER: 1 Area = × base × height 2 1 = × 6× 4 2 = 12 sq.units
100.
PROBLEM: Calculate the area of the triangle formed by the vertices (−2, 1), (4, 1), and (1, −5). SOLUTION: Points: (−2, 1) and (4, 1)
a = ⎡⎣ 4 − ( −2 ) ⎦⎤ + (1 − 1) 2
=
2 2 [ 4 + 2] + (1 − 1)
=
2 2 [ 6] + ( 0 )
= 36 =6
2
Points: (4, 1) and (1, −5) b= =
(1 − 4 ) + ( −5 − 1) 2
( −3) + ( −6 ) 2
=
9 + 36
=
45
2
2
Points: (1, −5) and (−2, 1)
c= = =
( −2 − 1)
2
+ ⎡⎣1 − ( −5) ⎤⎦
2
( −2 − 1) + [1 + 5] 2
2
( −3) + [ −6] 2
2
= 9 + 36 = 45 The edges b and c are equal. The unequal side of the triangle is the base of the triangle. The height of the triangle is the distance between the mid-point of the base vertices, i.e. (−2, 1) and (4, 1), and the common end of the equal sides, i.e. (1, −5). Mid-point of (−2, 1) and (4, 1): ( x1 , y1 ) ( x2 , y2 )
( −2, 1) ( 4, 1)
x1 + x2 −2 + 4 2 = = =1 2 2 2 y1 + y2 1 + 1 2 = = =1 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
(1, 1)
Distance between the (1, 1) and (1, −5): Height = =
(1 − 1) + ( −5 − 1) 2
( 0 ) + ( −6 )
= 36 =6
2
2
2
ANSWER: 1 Area = × base × height 2 1 = × 6× 6 2 = 18 sq.units
Part D: Discussion Board Topics
101.
PROBLEM: Research and discuss the life and contributions to mathematics of Rene Descartes. ANSWER: Rene Descartes (31 March 1596 – 11 February 1650), was a French philosopher, mathematician, physicist. He has been dubbed the “Father of Modern Philosophy”, and much subsequent Western philosophy is a response to his writings, which are studied closely to this day. Descartes created analytic geometry, and discovered an early form of the law of conservation of momentum (the term momentum refers to the momentum of a force). He outlined his views on the universe in his Principles of Philosophy. One of Descartes most enduring legacies was his development of Cartesian geometry, which uses algebra to describe geometry. He also “invented”, the notation that uses superscripts to show the powers or exponents, for example the 4 used in x4 to indicate squaring of squaring. (Source: http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes)
102.
PROBLEM: Research and discuss the history of the right triangle and the Pythagorean Theorem. ANSWER: A right triangle is triangle with an angle of 90o(π/2 radians). The sides a, b, and c of such a triangle satisfy the Pythagorean theorem. a2 + b2 = c2 where the largest side is conventionally denoted c and is called hypotenuse. The other two sides of lengths a and b are called legs, or sometimes catheti.
103.
PROBLEM: What is a Pythagorean Triple? Provide some examples. ANSWER: A Pythagorean triple consists of three positive integers a, b, and c such that a2 + b2 = c2. The name is derived from the Pythagorean Theorem, satisfying the formula a2 + b2 = c2; thus, Pythagorean triples describe the three integer side lengths of a right triangle. A Pythagorean triple is commonly written as (a, b, c). Some examples are (3, 4, 5), (5, 12, 13), (7, 24, 25) etc.
104.
PROBLEM: Explain why we cannot use a ruler to calculate distance on a graph. ANSWER: The graph is often scaled down to obtain the representation of large objects. The physical distance on a graph measured using a ruler would not be accurate because of this scaling down process. Hence, it is not possible to measure the distance of an object from its graphical representation using a ruler. However, the physical distance measured using a ruler can be multiplied by the scaling factor to obtain the actual distance.
105.
PROBLEM: How can we use a compass and a straight edge to bisect a line segment? ANSWER: The radius of the compass should be greater than half the length of the line segment. From one end of the line segment draw arcs above and below the line segment. From the other end draw arcs intersecting the previous arcs. Connect the intersecting points of the two arcs using a straight edge. The intersecting point of the line drawn using the straight edge and the line segment is the mid-point.
3.2 Graph by Plotting Points Part A: Solutions to Linear Systems Determine whether or not the given point is a solution.
1.
PROBLEM: 5x – 2y = 4; (−1, 1) SOLUTION: 5x – 2y = 4 5(– 1) – 2(1) = 4 –5– 2 = 4 –7 ≠ 4 ANSWER: The given point (−1, 1) is not a solution of 5x – 2y = 4.
2.
PROBLEM: 3x – 4y = 10; (2, −1) SOLUTION: 3(2) – 4(–1) = 10 6 + 4 = 10 10 = 10 ANSWER: The given point (2, −1) is a solution of 3x – 4y = 10.
3.
PROBLEM: –3x + y = –6 ; (4, 6) SOLUTION: –3(4) + 6 = –6 –12 + 6 = –6 –6 = –6 ANSWER: The given point (4, 6) is a solution of –3x + y = –6.
4.
PROBLEM: –8x – y = 24 ; (−2, −3) SOLUTION: –8(–2) – (–3) = 24 16 + 3 = 24 19 ≠ 24 ANSWER: The given point (−2, −3) is not a solution of –8x – y = 24.
5.
PROBLEM: –x + y = –7; (5, −2) SOLUTION: –(5) + (–2) = –7 –7 = –7 ANSWER: The given point (5, −2) is a solution of –x + y = –7.
6.
PROBLEM: 9x – 3y = 6; (0, −2) SOLUTION: 9x – 3y = 6 9(0) – 3(–2) = 6 0+6=6 6=6 ANSWER: The given point (0, −2) is a solution of 9x – 3y = 6.
7.
PROBLEM: 1 1 1 x + y = − ; (1, −2) 2 3 6 SOLUTION: 1 1 1 x+ y =− 2 3 6 1 1 1 (1) + ( −2 ) = − 2 3 6 1 2 1 − =− 2 3 6 3 ⋅1 − 2 ⋅ 2 1 =− 6 6 3− 4 1 =− 6 6 1 1 − =− 6 6 ANSWER: The given point (1, –2) is a solution of
8.
1 1 1 x+ y =− . 2 3 6
PROBLEM: 3 1 x − y = −1 ; (2, 1) 4 2
SOLUTION: 3 1 ( 2 ) − (1) = −1 4 2 6 1 − = −1 4 2 6 − 1⋅ 2 = −1 4 6−2 = −1 4 4 = −1 4 1 ≠ −1 ANSWER: The given point (2, 1) is not a solution of
3 1 x − y = −1 . 4 2
9.
PROBLEM: ⎛1 1⎞ 4x – 3y = 1; ⎜⎝ 2 , 3 ⎟⎠ SOLUTION: 4x – 3y = 1
⎛1⎞ ⎛1⎞ 4 ⎜ ⎟ − 3⎜ ⎟ = 1 ⎝ 2⎠ ⎝3⎠ 2 −1 = 1 1=1 ⎛1 1⎞ ANSWER: The given point ⎜ , ⎟ is a solution of 4x – 3y = 1. ⎝ 2 3⎠
10.
PROBLEM: 9 −10 x + 2 y = − ; 5
( 15 , 101 )
SOLUTION: 9 5 9 ⎛1⎞ ⎛ 1⎞ −10 ⎜ ⎟ + 2 ⎜ ⎟ = − 5 ⎝5⎠ ⎝ 10 ⎠ 10 2 9 − + =− 5 10 5 −10 ⋅ 2 + 2 9 =− 10 5 −20 + 2 9 = − 10 5 18 9 − =− 10 5 9 9 − = − 5 5 −10 x + 2 y = −
9 ⎛1 1 ⎞ ANSWER: The given point ⎜ , ⎟ is a solution of −10 x + 2 y = − . 5 ⎝ 5 10 ⎠
11.
PROBLEM: 1 y = x + 3 ; (6, 3) 3 SOLUTION: 1 y = x+3 3 1 ( 3) = ( 6 ) + 3 3 3= 2+3 3≠5 ANSWER: The given point (6, 3) is a solution of y =
12.
1 x + 3. 3
PROBLEM: y = −4x + 1 ; (−2, 9) SOLUTION: 9 = −4(−2) + 1 9=8+1 9=9 ANSWER: The given point (−2, 9) is a solution of y = −4x + 1.
13.
PROBLEM: 2 y = x − 3 ; (0, −3) 3 SOLUTION: 2 y = x−3 3 2 −3 = ( 0 ) − 3 3 −3 = 0 − 3 −3 = −3 ANSWER: The given point (0, −3) is a solution of y =
2 x − 3. 3
14.
PROBLEM: 5 y = − x + 1 ; (8, −5) 8 SOLUTION: 5 y = − x +1 8 5 −5 = − ( 8 ) + 1 8 − 5 = −5 + 1 − 5 ≠ −4 5 ANSWER: The given point (8, −5) is not a solution of y = − x + 1. 8
15.
PROBLEM: 1 3 y = − x + ; ( − 12 ,1) 2 4 SOLUTION: 1 3 y =− x+ 2 4 1 1 3 (1) = − ⎛⎜ − ⎞⎟ + 2⎝ 2⎠ 4 1 3 1= + 4 4 1+ 3 1= 4 4 1= 4 1=1 1 3 ANSWER: The given point ( − 12 ,1) is a solution of y = − x + . 2 4
16.
PROBLEM: 1 1 y =− x− ; 3 2
( 12 , − 23 )
SOLUTION: 1 1 y =− x− 3 2 1⎛1⎞ 1 ⎛ 2⎞ ⎜− ⎟ = − ⎜ ⎟− 3⎝ 2⎠ 2 ⎝ 3⎠ 2 1 1 − =− − 6 2 3 2 −1 − 1⋅ 3 − = 3 6 2 −1 − 3 − = 3 6 2 −4 − = 3 6 2 2 − =− 3 3 ANSWER: The given point
17.
( 12 , − 23 ) is a solution of
1 1 y =− x− . 3 2
PROBLEM: y = 2; (−3, 2) SOLUTION: y = 2 2=2 ANSWER: The given point (−3, 2) is a solution of y = 2.
18.
PROBLEM:
y = 4; (4, −4) SOLUTION: y = 4 −4 ≠ 4 ANSWER: The given point (4, −4) is not a solution of y = 4.
19.
PROBLEM: x = 3; (3, −3) SOLUTION: x = 3 3=3 ANSWER: The given point (3, −3) is a solution of x = 3.
20.
PROBLEM: x = 0; (1,0) SOLUTION: x = 0 1≠0 ANSWER: The given point (1, 0) is not a solution of x = 0.
Find the ordered pair solutions given the set of x‐values.
21.
PROBLEM: y = −2x + 4; {−2, 0, 2} SOLUTION:
x-value
y-value
x = −2
y = −2 ( − 2 ) + 4 = 4+4 =8
x=0
y = −2 ( 0 ) + 4 = 0+4
Solution (−2, 8)
(0, 4)
=4
x=2
ANSWER: (−2, 8), (0, 4), (2, 0)
y = −2 ( 2 ) + 4 = −4 + 4 =0
(2, 0)
22.
PROBLEM: 1 y = x − 3 ; {−4, 0, 4} 2 SOLUTION:
x-value
23.
y-value 1 x −3 2 1 = ( −4 ) − 3 2 = −2−3 = −5
x = −4
y=
x=0
y=
x=4
y=
1 x −3 2 1 = (0) − 3 2 = 0−3 = −3
1 x −3 2 1 = ( 4) − 3 2 = 2−3 = −1
Answer (−4, −5)
(0, −3)
(4, −1)
PROBLEM: 3 1 y = − x + ; {−2, 0, 2} 4 2 SOLUTION:
x-value
x = −2
y-value 3 1 3 1 y = − x + = − ( −2 ) + 4 2 4 2 3 1 4 = + = =2 2 2 2
Solution (−2,2)
3 1 3 1 1 y = − x + = − (0) + = 0 + 4 2 4 2 2 1 = 2
x=0
3 1 3 1 3 1 y = − x + = − ( 2) + = − + 4 2 4 2 2 2 2 = − = −1 2
x=2
⎛ 1⎞ ⎜ 0, ⎟ ⎝ 2⎠
(2, −1)
ANSWER: ⎛ 1⎞ (−2,2), ⎜ 0, ⎟ , (2, −1) ⎝ 2⎠
24.
PROBLEM: y = −3 x + 1 ; {−1/2, 0, 1/2}
SOLUTION:
x-value
y-value
Answer
1 x=− 2
3 3 + 1⋅ 2 3 + 2 ⎛ 1⎞ y = −3 x + 1 = − 3 ⎜ − ⎟ + 1 = + 1 = = 2 2 2 ⎝ 2⎠ 5 = 2
⎛ 1 5⎞ ⎜− , ⎟ ⎝ 2 2⎠
x=0
y = −3 x + 1 = − 3 ( 0 ) + 1 = 0 + 1 = 1
(0, 1)
1 x= 2
3 −3 + 1 ⋅ 2 −3 + 2 ⎛1⎞ y = −3 x + 1 = − 3 ⎜ ⎟ + 1 = − + 1 = = 2 2 2 ⎝2⎠ 1 =− 2
⎛1 1⎞ ⎜ ,− ⎟ ⎝2 2⎠
25.
PROBLEM: y = − 4 ; {−3, 0, 3} SOLUTION:
x-value
y-value
Solution
x = −3
y = −4
(−3, −4)
x=0
y = −4
(0, −4 )
x=3
y = −4
(3, −4)
ANSWER: (−3, −4), (0, −4 ), (3, −4)
26.
PROBLEM: 1 3 y = x + ; {−1/4, 0, 1/4} 2 4 ANSWER:
x-value
x=−
1 4
x=0
1 x= 4
y-value 1⎛ 1⎞ 3 1 3 −1 + 3 ⋅ 2 −1 + 6 y = ⎜− ⎟+ =− + = = 2⎝ 4⎠ 4 8 4 8 8 5 = 8 y=
1 3 3 3 (0) + = 0 + = 2 4 4 4
1⎛1⎞ 3 1 3 1+ 3⋅ 2 1+ 6 = y = ⎜ ⎟+ = + = 2⎝4⎠ 4 8 4 8 8 7 = 8
Answer ⎛ 1 5⎞ ⎜− , ⎟ ⎝ 4 8⎠
⎛ 3⎞ ⎜ 0, ⎟ ⎝ 4⎠ ⎛1 7⎞ ⎜ , ⎟ ⎝4 8⎠
27.
PROBLEM: 2 x − 3 y = 1 ; {0, 1, 2} SOLUTION: 2x – 3y = 1 2x – 2x – 3y = 1 – 2x – 3y = 1– 2x 3y = 2x – 1 3y 2x – 1 = 3 3 2 1 y = x− 3 3 x-value
x=0
x =1
x=2
y-value
y=
2 1 2 1 1 x − = (0) − = − 3 3 3 3 3
⎛ 1⎞ ⎜1, ⎟ ⎝ 3⎠
2 1 2 1 4 1 3 x − = ( 2) − = − = 3 3 3 3 3 3 3 =1
(2, 1)
y=
PROBLEM: 3 x − 5 y = − 15 ; {−5, 0, 5}
SOLUTION: 3x – 5y = – 15 3x – 3x – 5y = –15 – 3x – 5y = –15 – 3x −5 y –15 3 = − x −5 −5 −5 3 y = 3+ x 5 ANSWER:
1⎞ ⎛ ⎜ 0, − ⎟ 3⎠ ⎝
2 1 2 1 2 1 x − = (1) − = − 3 3 3 3 3 3 1 = 3
y=
ANSWER: 1⎞ ⎛ 1⎞ ⎛ ⎜ 0, − ⎟ , ⎜1, ⎟ , (2, 1) 3⎠ ⎝ 3⎠ ⎝
28.
Solution
x-value
29.
y-value
Answer
x = −5
3 3 y = 3 + x = 3 + ( −5 ) = 3 − 3 = 0 5 5
(−5, 0)
x=0
3 3 y = 3 + x = 3 + ( 0) = 3 − 0 = 3 5 5
(0, 3)
x=5
3 3 y = 3 + x = 3 + (5) = 3 + 3 = 6 5 5
(5,6)
PROBLEM: – x + y = 3 ; {−5, −1, 0} SOLUTION: –x + y = 3 –x + x + y = 3 + x y=3+x
x-value
y-value
Solution
x = −5
y = 3 + x = 3 + (–5) = 3 – 5 = –2
(–5, –2)
x = −1
y = 3 + x = 3 + (–1) =3–1=2
(– 1, 2)
x=0
y = 3 + x = 3 + (0) =3+0=3
(0, 3)
ANSWER: (–5, –2), (– 1, 2), (0, 3)
30.
PROBLEM: 1 1 x − y = −4 ; {−4, −2, 0} 2 3 SOLUTION: 1 1 x − y = −4 2 3 1 1 1 1 x − x − y = −4 − x 2 2 3 2 1 1 − y = −4 − x 3 2 1 1 ( −3) ⋅ ⎛⎜ − y ⎞⎟ = ( −3) ⋅ ⎛⎜ −4 − x ⎞⎟ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 y = 12 + x 2
ANSWER:
x-value
x = −4
x = −2
y-value 3 3 12 12 ⋅ 2 − 12 24 − 12 y = 12 + x = 12 + ( −4 ) = 12 − = = 2 2 2 2 2 12 = =6 2 y = 12 + =
x=0
3 3 6 12 ⋅ 2 − 6 24 − 6 = x = 12 + ( −2 ) = 12 − = 2 2 2 2 2
Answer (−4, 6)
(−2, 9)
18 =9 2
y = 12 +
3 3 x = 12 + ( 0 ) = 12 + 0 = 12 2 2
(0,12)
31.
PROBLEM: 3 1 x + y = 2 ; {−15, −10, −5} 5 10 SOLUTION: 3 1 x+ y = 2 5 10 3 3 1 3 x− x+ y = 2− x 5 5 10 5 1 3 y = 2− x 10 5 1 3 ⎞ ⎛ 10 ⋅ y = 10 ⋅ ⎜ 2 − x ⎟ 10 5 ⎠ ⎝ y = 20 − 6 x
x-value
y-value
Solution
x = −15
y = 20 – 6x = 20 – 6(–15) = 20 + 90 = 110
(–15, 110)
x = −10
y = 20 – 6x = 20 – 6(–10) = 20 + 60 = 80
(– 10, 80)
x = −5
y = 20 – 6x = 20 – 6(–5) = 20 + 30 = 50
(–5, 50)
ANSWER: (–15, 110), (– 10, 80), (–5, 50)
32.
PROBLEM: x − y = 0 ; {10, 20, 30} SOLUTION: x – y = 0 –y = –x y=x ANSWER:
x-value
y-value
x = 10
y = x = 10
(10, 10)
x = 20
y = x = 20
(20, 20)
x = 30
y = x = 30
(30,30)
Find the ordered pair solutions given the set of y‐values.
33.
Solution
PROBLEM: 1 y = x − 1 ; {−5, 0, 5} 2 SOLUTION: 1 y = x −1 2 1 y +1 = x −1+1 2 1 x = y +1 2 1 2 ⋅ x = 2 ⋅ ( y + 1) 2 x = 2y + 2
y-value
x-value
Solution
y = −5
x = 2y + 2 = 2(–5) + 2 = –10 + 2 = –8
(–8, –5)
y=0
x = 2y + 2 = 2(0) + 2 =0+2 =2
(2, 0)
y =5
x = 2y + 2 = 2(5) + 2 = 10 + 2 = 12
(12, 5)
ANSWER: (–8, –5), (2, 0), (12, 5)
34.
PROBLEM: 3 y = − x + 2 ; {0, 2, 4} 4 SOLUTION: 3 y =− x+2 4 3 y−2 = − x+2−2 4 3 − x = y−2 4 ⎛ 4 ⎞⎛ 3 ⎞ ⎛ 4 ⎞ ⎜ − ⎟ ⎜ − x ⎟ = ⎜ − ⎟ ( y − 2) ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 3 ⎠ 8 4 x= − y 3 3 ANSWER:
y-value
35.
x-value
Solution
y=0
8 4 8 4 8 x = − y = − (0) = 3 3 3 3 3
⎛8 ⎞ ⎜ , 0⎟ ⎝3 ⎠
y=2
8 4 8 4 8 8 x = − y = − ( 2) = − = 0 3 3 3 3 3 3
(0, 2)
y=4
8 4 8 4 8 16 8 x = − y = − ( 4) = − = − 3 3 3 3 3 3 3
⎛ 8 ⎞ ⎜ − , 0⎟ ⎝ 3 ⎠
PROBLEM: 3 x − 2 y = 6 ; {−3, −1, 0} SOLUTION: 3x − 2 y = 6 3x − 2 y + 2 y = 6 + 2 y 3x = 6 + 2 y 3x 6 2 y = + 3 3 3 2 x = 2+ y 3 y-value x-value 2 2 2 2 x = + y = + ( −3) = 2 − 2 = 0 y = −3 3 3
y = −1
x = 2+ =
y=0
ANSWER:
2 2 2 2⋅3 − 2 6 − 2 = y = 2 + ( −1) = 2 − = 3 3 3 3 3
4 3
x = 2+
2 2 y = 2 + ( 0) = 2 + 0 = 2 3 3
Solution (0, –3) ⎛4 ⎞ ⎜ , −1⎟ ⎝3 ⎠
(2, 0)
⎛4 ⎞ (0, –3), ⎜ , −1⎟ , (2, 0) ⎝3 ⎠ 36.
PROBLEM: − x + 3 y = 4 ; {−4, −2, 0} SOLUTION: −x + 3y = 4 −x + 3y − 3y = 4 − 3y −x = 4 − 3y x = 3y − 4 ANSWER:
y-value
x-value
Solution
y = −4
x = 3y – 4 = 3(–4) – 4 = –12 – 4 = –16
(–16, –4)
y = −2
x = 3y – 4 = 3(–2) – 4 = –6 – 4 = –10
(–10, –2)
y=0
x = 3y – 4 = 3(0) – 4 = 0 – 4 = –4
(–4, 0)
37.
PROBLEM: 1 1 x − y = −4 ; {−1, 0, 1} 3 2
SOLUTION: 1 1 x − y = −4 3 2 1 1 1 1 x − y + y = −4 + y 3 2 2 2 1 1 x = −4 + y 3 2 1 1 ⎞ ⎛ 3 ⋅ x = 3 ⋅ ⎜ −4 + y ⎟ 3 2 ⎠ ⎝ 3 x = −12 + y 2 3 x = y − 12 2 y-value x-value
y = −1
y=0
y =1
Solution
x=
3 3 3 −3 − 12 ⋅ 2 −3 − 24 −27 = = y − 12 = ( −1) − 12 = − − 12 = 2 2 2 2 2 2
x=
3 3 y − 12 = ( 0 ) − 12 = 0 − 12 = −12 2 2
x=
3 3 3 3 − 12 ⋅ 2 3 − 24 21 = =− y − 12 = (1) − 12 = − 12 = 2 2 2 2 2 2
ANSWER: ⎛ 27 ⎞ ⎜ − , −1⎟ , (-12, 0), ⎝ 2 ⎠
⎛ 21 ⎜− , ⎝ 2
⎞ 1⎟ ⎠
⎛ 27 ⎞ ⎜ − , −1⎟ ⎝ 2 ⎠
(-12, 0)
⎛ 21 ⎞ ⎜ − , 1⎟ ⎝ 2 ⎠
38.
PROBLEM: 3 1 x + y = 2 ; {−20, −10, −5} 5 10 SOLUTION: 3 1 x+ y = 2 5 10 3 1 1 1 x+ y− y = 2− y 5 10 10 10 3 1 x = 2− y 5 10 5 3 5 ⎛ 1 ⎞ ⋅ x = ⋅⎜ 2 − y ⎟ 3 5 3 ⎝ 10 ⎠ 10 1 x= − y 3 6 ANSWER:
y-value
y = −20 y = −10
y = −5
x-value 10 1 10 1 10 10 20 x = − y = − ( −20 ) = + = 3 6 3 6 3 3 3
x=
10 1 10 1 10 5 15 − y = − ( −10 ) = + = = 5 3 6 3 6 3 3 3
10 1 10 1 10 5 10 ⋅ 2 + 5 20 + 5 − y = − ( −5 ) = + = = 3 6 3 6 3 6 6 6 25 = 6
x=
Solution ⎛ 20 ⎞ ⎜ , −20 ⎟ ⎝ 3 ⎠ (5, −10)
⎛ 25 ⎞ ⎜ , −5 ⎟ ⎝ 6 ⎠
Part B: Graphing Lines Given the set of x‐values {−2, −1, 0, 1, 2} find the corresponding y‐values and graph.
39.
PROBLEM: y=x+1 ANSWER:
x y −2 −1 −1 0 0 1 1 2 2 3
40.
y = x +1 y = −2 + 1 = −1 y = −1 + 1 = 0 y = 0 +1 = 1 y = 1+1 = 2 y = 2 +1 = 3
PROBLEM: y = −x + 1 ANSWER:
x −2
y 3
y = −x + 1
−1
2
y = − ( −1) + 1 = 1 + 1 = 2
0 1
1 0
y = 0 +1 = 1 y = − (1) + 1 = −1 + 1 = 0
2
−1
y = − ( 2 ) + 1 = −2 + 1 = −1
y = − ( −2 ) + 1 = 2 + 1 = 3
41.
PROBLEM: y = 2x − 1 ANSWER:
x y −2 −5
y = 2x − 1
−1 −3
y = 2 ( −1) − 1 = −2 − 1 = −3
0
−1
y = 2 ( 0 ) − 1 = 0 − 1 = −1
1
1
y = 2 (1) − 1 = 2 − 1 = −1
2
3
y = 2 ( 2) −1 = 4 −1 = 3
y = 2 ( −2 ) − 1 = −4 − 1 = −5
42.
PROBLEM: y = −3x + 2 ANSWER:
y = −3x + 2
x −2
y 8
−1
5
y = −3 ( −1) + 2 = 3 + 2 = 5
0
2
y = −3 ( 0 ) + 2 = 0 + 2 = 2
1
−1
y = −3 (1) + 2 = −3 + 2 = −1
2
−4
y = −3 ( 2 ) + 2 = −6 + 2 = −4
43.
y = −3 ( −2 ) + 2 = 6 + 2 = 8
PROBLEM: y = 5x – 10 ANSWER:
x y −2 −20
y = 5x – 10
y = 5 ( −2 ) − 10 = −10 − 10 = −20
−1 −10
y = 5 ( −1) − 10 = −5 − 10 = −15
0
−10
y = 5 ( 0 ) − 10 = 0 − 10 = −10
1
−5
y = 5 (1) − 10 = 5 − 10 = −5
2
0
y = 5 ( 2 ) − 10 = 10 − 10 = 0
44.
PROBLEM: 5 x + y = 15
SOLUTION: 5x + y = 15 5x – 5x + y = 15 – 5x y = 15 – 5x ANSWER:
x y −2 25
y = 15 – 5x
y = 15 − 5 ( −2 ) = 15 + 10 = 25
−1 20
y = 15 − 5 ( −1) = 15 + 5 = 20
0
15
y = 15 − 5 ( 0 ) = 15 − 0 = 15
1
10
y = 15 − 5 (1) = 15 − 5 = 10
2
5
y = 15 − 5 ( 2 ) = 15 − 10 = 5
45.
PROBLEM: 3x – y = 9 SOLUTION: 3x – 3x – y = 9 – 3x y = 3x – 9 ANSWER:
x y −2 −15
y = 3x – 9
y = 3 ( −2 ) – 9 = − 6 − 9 = −15
−1 −12
y = 3 ( −1) – 9 = − 3 − 9 = −12
0
−9
y = 3 ( 0 ) – 9 = 0 − 9 = −9
1
−6
y = 3 (1) – 9 = 3 − 9 = −6
2
−3
y = 3 ( 2 ) – 9 = 6 − 9 = −3
46.
PROBLEM: 6x – 3y = 9 SOLUTION: 6x – 6x – 3y = 9 – 6x – 3y = 9 – 6x 3y = 6x – 9 3y 6x 9 = − 3 3 3 y = 2x – 3
ANSWER:
x y −2 −7
y = 2x – 3
y = 2 ( −2 ) – 3 = −4 − 3 = −7
−1 −5
y = 2 ( −1) – 3 = −2 − 3 = −5
0
−3
y = 2 ( 0 ) – 3 = 0 − 3 = −3
1
−1
y = 2 (1) – 3 = 2 − 3 = −1
2
1
y = 2 ( 2) – 3 = 4 − 3 = 1
47.
PROBLEM: y=–5 ANSWER:
x −2 −1 0 1 2
y −5 −5 −5 −5 −5
48.
PROBLEM: y=3 ANSWER:
x −2 −1 0 1 2
y 3 3 3 3 3
Find at least five ordered pair solutions and graph.
49.
PROBLEM: y = 2x – 1 ANSWER:
y = 2x – 1
x −2
y −5
−1
−3
y = 2 ( −1) − 1 = −2 − 1 = −3
0
−1
y = 2 ( 0 ) − 1 = 0 − 1 = −1
1
1
y = 2 (1) − 1 = 2 − 1 = 1
2
3
y = 2 ( 2) −1 = 4 −1 = 3
y = 2 ( −2 ) − 1 = −4 − 1 = −5
50.
PROBLEM: y = –5x + 3 ANSWER:
y = –5x + 3
x 0
y 3
y = –5 ( 0 ) + 3 = 0 + 3 = 3
0.6
0
y = –5 ( 0.6 ) + 3 = − 3 + 3 = 0
1
−2
y = –5 (1) + 3 = − 5 + 3 = − 2
1.4
−4
y = –5 (1.4 ) + 3 = − 7 + 3 = − 4
1.6
−5
y = –5 (1.6 ) + 3 = − 8 + 3 = − 5
51.
PROBLEM: y = –4x + 2 ANSWER:
y = –4x + 2
x −0.75
y 5
−0.5
4
y = –4 ( −0.5) + 2 = 2 + 2 = 4
0
2
y = –4 ( 0 ) + 2 = 0 + 2 = 2
1
−2
y = –4 (1) + 2 = −4 + 2 = −2
2
−6
y = –4 ( 2 ) + 2 = −8 + 2 = −6
52.
y = –4 ( −0.75) + 2 = 3 + 2 = 5
PROBLEM: y = 10 x − 20
ANSWER:
x y −1 −30
y = 10 x − 20 y = 10 ( − 1) − 20 = − 10 − 20 = − 30
0
−20
y = 10 ( 0 ) − 20 = 0 − 20 = − 20
2
0
y = 10 ( 2 ) − 20 = 20 − 20 = 0
3
10
y = 10 ( 3 ) − 20 = 30 − 20 = 10
4
20
y = 10 ( 4 ) − 20 = 40 − 20 = 20
53.
PROBLEM: 1 y =− x+2 2 ANSWER:
x
y
−4
4
−2
3
0
2
2
1
4
0
1 y =− x+2 2 1 y = − ( −4 ) + 2 = 2 + 2 = 4 2 1 y = − ( −2 ) + 2 = 1 + 2 = 3 2 1 y = − ( 0) + 2 = 0 + 2 = 2 2 1 y = − ( 2 ) + 2 = −1 + 2 = 1 2 1 y = − ( 4 ) + 2 = −2 + 2 = 0 2
54.
PROBLEM: 1 y = x −1 3 ANSWER:
x
y
−6
−3
−3
−2
0
−1
3
0
6
1
55.
1 x −1 3 1 y = ( − 6 ) − 1 = − 2 − 1 = −3 3 1 y = ( −3 ) − 1 = −1 − 1 = − 2 3 1 y = ( 0 ) − 1 = 0 − 1 = −1 3 1 y = ( 3) − 1 = 1 − 1 = 0 3 1 y = (6) −1 = 2 −1 = 1 3 y=
PROBLEM: 2 y = x−6 3 ANSWER:
x
y
−9
−12
−6
−10
2 x−6 3 2 y = ( −9 ) − 6 = −6 − 6 = −12 3 2 y = ( −6 ) − 6 = −4 − 6 = −10 3
y=
−3
−8
0
−6
3
−4
56.
2 ( −3) − 6 = −2 − 6 = −8 3 2 y = ( 0 ) − 6 = 0 − 6 = −6 3 2 y = ( 3) − 6 = 2 − 6 = −4 3 y=
PROBLEM: 2 y =− x+2 3 ANSWER:
x
y
−3
4
0
2
3
0
6
−2
9
−4
2 y =− x+2 3 2 y = − ( −3 ) + 2 = 2 + 2 = 4 3 2 y = − (0) + 2 = 0 + 2 = 2 3 2 y = − ( 3 ) + 2 = −2 + 2 = 0 3 2 y = − ( 6 ) + 2 = −4 + 2 = − 2 3 2 y = − ( 9 ) + 2 = −6 + 2 = − 4 3
57.
PROBLEM: y=x ANSWER:
x −5 −4 −2 0 1
y −5 −4 −2 0 1
58.
PROBLEM: y = −x ANSWER:
x −5 −4 −2 0 1
y 5 4 2 0 −1
59.
PROBLEM:
− 2 x + 5 y = − 15
SOLUTION:
− 2 x + 5 y = − 15
− 2 x + 2 x + 5 y = − 15 + 2 x 5 y = 2 x − 15 5 2 15 y = x− 5 5 5 2 y = x−3 5
ANSWER:
x
y
−5
−5
0
−3
5
−1
10
1
15
3
y= y= y= y= y= y=
2 x −3 5 2 ( − 5 ) − 3 = − 2 − 3 = −5 5 2 ( 0 ) − 3 = 0 − 3 = −3 5 2 ( 5 ) − 3 = 2 − 3 = −1 5 2 (10 ) − 3 = 4 − 3 = 1 5 2 (15 ) − 3 = 6 − 3 = 3 5
60.
PROBLEM: x + 5y = 5
SOLUTION: x + 5y = 5
x − x + 5y = 5 − x 5y = 5 − x 5 5 1 y= − x 5 5 5 1 y = 1− x 5
ANSWER:
x
y
−5
2
0
1
10
−1
15
−2
20
−3
1 y = 1− x 5 1 y = 1 − ( −5 ) = 1 + 1 = 2 5 1 y = 1 − (0) = 1 − 0 = 1 5 1 y = 1 − (10 ) = 1 − 2 = −1 5 1 y = 1 − (15 ) = 1 − 3 = −2 5 1 y = 1 − ( 20 ) = 1 − 4 = −3 5
61.
PROBLEM: 6x − y = 2
SOLUTION:
6x − y = 2 6x − 6x − y = 2 − 6x − y = 2 − 6x y = 6x − 2
x −1
ANSWER: y = 6x − 2 y −8 y = 6 ( −1) − 2 = −6 − 2 = −8
0
−2
y = 6 ( 0 ) − 2 = 0 − 2 = −2
1
4
y = 6 (1) − 2 = 6 − 2 = 4
2
10
y = 6 ( 2 ) − 2 = 12 − 2 = 10
3
16
y = 6 ( 3) − 2 = 18 − 2 = 16
62.
PROBLEM: 4 x + y = 12
SOLUTION:
4 x + y = 12 4 x − 4 x + y = 12 − 4 x y = 12 − 4 x
ANSWER:
y = 12 − 4 x
x −1
y 16
y = 12 − 4 ( −1) = 12 + 4 = 16
0
12
y = 12 − 4 ( 0 ) = 12 − 0 = 12
1
8
y = 12 − 4 (1) = 12 − 4 = 8
2
4
y = 12 − 4 ( 2 ) = 12 − 8 = 4
3
0
y = 12 − 4 ( 3) = 12 − 12 = 0
63.
PROBLEM: −x + 5y = 0
SOLUTION:
−x + 5y = 0 −x + x + 5y = 0 + x 5y = x 5 1 y= x 5 5 1 y= x 5
ANSWER:
x
y
−10
−2
1 x 5 1 y = ( −10 ) = −2 5
y=
−5
−1
0
0
5
1
10
2
64.
1 ( −5) = −1 5 1 y = (0) = 0 5 1 y = ( 5) = 1 5 1 y = (10 ) = 2 5
y=
PROBLEM: x + 2y = 0 SOLUTION: x + 2y = 0 x − x + 2 y = −x 2 y = −x 2 1 y=− x 2 2 1 y=− x 2 ANSWER:
x
y
−4
2
−2
1
0
0
1 y=− x 2 1 y = − ( −4 ) = 2 2 1 y = − ( −2 ) = 1 2 1 y = − ( 0) = 1 2
2
−1
4
−2
65.
1 ( 2 ) = −1 2 1 y = − ( 4 ) = −2 2
y=−
PROBLEM: 1 x− y =3 10 SOLUTION: 1 x− y =3 10 1 1 1 x − x − y = 3− x 10 10 10 1 −y = 3− x 10 1 y = x −3 10
x −20 −10 0 10 20
ANSWER: y 1 y = x−3 10 −5 1 y = ( −20 ) − 3 = −2 − 3 = −5 10 −4 1 y = ( −10 ) − 3 = −1 − 3 = −4 10 −3 1 y = ( 0 ) − 3 = 0 − 3 = −3 10 −2 1 y = (10 ) − 3 = 1 − 3 = −2 10 −1 1 y = ( 20 ) − 3 = 2 − 3 = −1 10
66.
PROBLEM: 3 x + 5 y = 30 2 SOLUTION: 3 x + 5 y = 30 2 3 3 3 x − x + 5 y = 30 − x 2 2 2 3 5 y = 30 − x 2 5 30 3 y= − x 5 5 2⋅5 3 y = 6− x 10 ANSWER:
x
y
−20
12
−10
9
0
6
10
3
20
0
3 x 10 3 y = 6 − ( −20 ) = 6 + 6 = 12 10 3 y = 6 − ( −10 ) = 6 + 3 = 9 10 3 y = 6 − (0) = 6 − 0 = 6 10 3 y = 6 − (10 ) = 6 − 3 = 3 10 3 y = 6 − ( 20 ) = 6 − 6 = 0 10
y = 6−
Part C: Horizontal and Vertical Lines Find at least five ordered pair solutions and graph. 67.
PROBLEM: y =4 ANSWER:
x −5 −4 0 1 2
y 4 4 4 4 4
68.
PROBLEM: y = − 10
ANSWER:
x −5 −4 0 1 2
69.
y −10 −10 −10 −10 −10
PROBLEM: x=4 ANSWER:
x 4 4 4 4 4
y 1 2 3 4 5
70.
PROBLEM: x = −1 ANSWER:
x −1 −1 −1 −1 −1
y 1 2 3 4 5
71.
PROBLEM: y=0
x 1 2 3 4 5
ANSWER: y 0 0 0 0 0
72.
PROBLEM: x=0 ANSWER:
x 0 0 0 0 0
y 1 2 3 4 5
73.
PROBLEM: 3 y= 4 ANSWER:
x −1 −2 0 1 2
y 3 4 3 4 3 4 3 4 3 4
74.
PROBLEM: 5 x=− 4 ANSWER:
x 5 − 4 5 − 4 5 − 4
y −1 −2 0
5 4 5 − 4
1
75.
PROBLEM: Graph the lines y = − 4 and x = 2 on the same set of axes. Where do they intersect?
−
2
ANSWER:
x −1 −2 0 1 2 x=2 2 2 2 2 2
y = −4 −4 −4 −4 −4 −4 y −4 −3 0 3 4
The lines intersect at (2, −4). 76.
PROBLEM: Graph the lines y = 5 and x = −5 on the same set of axes. Where do they intersect? SOLUTION:
x −5 −4 0 4 5
y =5 5 5 5 5 5
x = −5 −5 −5 −5 −5 −5
y =5 −5 −4 0 4 5
ANSWER: The lines intersect at (−5,5).
77.
PROBLEM: What is the equation that describes the x-axis? SOLUTION: If the line is considered to be the x-axis or the line lies completely on the x-axis, the value of y for any value of x is zero. Hence the equation is y = 0. ANSWER: y=0
78.
PROBLEM: What is the equation that describes the y-axis? SOLUTION: If the line is considered to be the y-axis or the line lies completely on the y-axis, the value of x for any value of y is zero. ANSWER: Hence the equation is x = 0.
Part D: Mixed Practice Graph by plotting points
79.
PROBLEM: 3 y = − x+6 5 SOLUTION:
x
y
−5
9
0
6
5
3
10
0
15
−3
3 y = − x+6 5 3 y = − ( −5) + 6 = 3 + 6 = 9 5 3 y = − ( 0) + 6 = 0 + 6 = 6 5 3 y = − ( 5) + 6 = −3 + 6 = 3 5 3 y = − (10 ) + 6 = −6 + 6 = 0 5 3 y = − (15 ) + 6 = −9 + 6 = −3 5
ANSWER:
80.
PROBLEM: 3 y = x −3 5
x
y
−5
−6
0
−3
5
0
10
3
15
6
SOLUTION: 3 y = x −3 5 3 y = ( −5) − 3 = −3 − 3 = −6 5 3 y = ( 0 ) − 3 = 0 − 3 = −3 5 3 y = ( 5) − 3 = 3 − 3 = 0 5 3 y = (10 ) − 3 = 6 − 3 = 3 5 3 y = (15) − 3 = 9 − 3 = 6 5 ANSWER:
81.
PROBLEM: y = −3 SOLUTION:
x −10 −5 0 5 10
y −3 −3 −3 −3 −3 ANSWER:
82.
PROBLEM: x = −5 SOLUTION:
x −5 −5 −5 −5 −5
y −8 −4 0 4 8
ANSWER:
83.
x −4 −2 0 2 4
PROBLEM: 3x − 2 y = 6 SOLUTION: 3x − 2 y = 6 3x − 3x − 2 y = 6 − 3x −2 y = 6 − 3 x 2 y = 3x − 6 2 3 6 y = x− 2 2 2 3 y = x −3 2 y 3 y = x −3 2 −9 3 y = ( −4 ) − 3 = −6 − 3 = −9 2 −6 3 y = ( −2 ) − 3 = −3 − 3 = −6 2 −3 3 y = ( 0 ) − 3 = 0 − 3 = −3 2 0 3 y = ( 2) − 3 = 3 − 3 = 0 2 3 3 y = ( 4) − 3 = 6 − 3 = 3 2
ANSWER:
84.
PROBLEM: −2 x + 3 y = −12
x
y
−6
−8
−3
−6
0
−4
3
−2
6
0
SOLUTION: −2 x + 3 y = −12 −2 x + 2 x + 3 y = −12 + 2 x 3 y = 2 x − 12 3 2 12 y = x− 3 3 3 2 y = x−4 3 2 y = x−4 3 2 y = ( −6 ) − 4 = −4 − 4 = −8 3 2 y = ( −3) − 4 = −2 − 4 = −6 3 2 y = ( 0 ) − 4 = 0 − 4 = −4 3 2 y = ( 3) − 4 = 2 − 4 = −2 3 2 y = ( 6) − 4 = 4 − 4 = 0 3
ANSWER:
Part E: Discussion Board Topics
85.
PROBLEM: Discuss the significance of the relationship between algebra and geometry in describing lines. ANSWER: Many concepts in geometry have their counterpart in algebra. For example, a point in geometry corresponds to an ordered pair (x, y) of numbers in algebra, a line corresponds to a set of ordered pairs satisfying an equation of the form ax + by = c (a, b, c ∈R), the intersection of two lines to the set of ordered pairs that satisfy the corresponding equations.
86.
PROBLEM: Give real world examples relating two unknowns. ANSWER: The cost of a flower and one box of chocolates are $1 and $10, respectively. If John spends $100 on flowers and chocolates for his girlfriend on Valentine’s day, how many flowers and boxes of chocolates did he buy? Let the number of flowers bought be x. Let the number of boxes of chocolates bought be y. Amount spent on flowers = 1⋅x = x Amount spent on chocolates = 10⋅y = 10y Total amount spent = x + 10y x + 10y = 100
The cost of a kilo of fruits and a liter of vanilla ice-cream is $15 and $20, respectively. If Chris, an ice-cream vendor, spends $300 per day on fruits and icecream, to prepare fruit salads with vanilla ice-cram in them, what is the quantity of fruits and ice-cream bought by Chris? Let the number of kilos of fruits bought be x. Let the number of liters of ice-cream bought be y. Amount spent on fruits = 15⋅x = 15x
Amount spent on ice-cream = 20⋅y Total amount spent = 15x + 20y 15x + 20y = 300 15 20 300 x+ y= 5 5 5 3 x + 4 y = 60
3.3 Graph using Intercepts
Part A: Intercepts Given the graph find the x‐ and y‐intercepts.
1.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = −3. Hence the y-intercept is (0, −3). The line intersects the x-axis at x = 4. Hence the x-intercept is (4, 0). ANSWER: x-intercept is (4, 0), y-intercept is (0, −3).
2.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = 1. Hence the y-intercept is (0, 1). The line intersects the x-axis at x = 2. Hence the x-intercept is (2, 0). ANSWER: x-intercept is (2, 0), y-intercept is (0, 1).
3.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = −3. Hence the y-intercept is (0,
−3). The line does not intersect the x-axis. Hence there is no x-intercept. ANSWER: no x-intercept, y-intercept is (0, −3).
4.
PROBLEM:
SOLUTION: The line does not intersect the y-axis. Hence there is no y-intercept. The line intersects the x-axis at x = 6. Hence the x-intercept is (6, 0). ANSWER: x-intercept is (6, 0), no y-intercept.
5.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = 0. Hence the y-intercept is (0,
0). The line intersects the x-axis at x = 0. Hence the x-intercept is (0, 0). ANSWER: x-intercept is (0, 0), y-intercept is (0, 0).
6.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = −1. Hence the y-intercept is (0, −1). The line intersects the x-axis at x = −2. Hence the x-intercept is (−2, 0). ANSWER: x-intercept is (−2, 0), y-intercept is (0, −1). Find the x‐ and y‐intercepts.
7.
PROBLEM: 5 x − 4 y = 20 SOLUTION: Set y = 0 5 x − 4 y = 20
5 x − 4 ( 0 ) = 20 5 x = 20 5 20 x= 5 5 x=4 x-intercept: (4, 0) Set x = 0 5 x − 4 y = 20 5 ( 0 ) − 4 y = 20 −4 y = 20 −4 20 y=− −4 4 y = −5 y-intercept: (0, −5)
ANSWER: (4, 0), (0, −5)
8.
PROBLEM: −2 x + 7 y = −28 SOLUTION: Set y = 0 −2 x + 7 y = −28
−2 x + 7 ( 0 ) = −28 −2 x = −28 −2 −28 x= −2 −2 x = 14 x-intercept: (14, 0) Set x = 0 −2 x + 7 y = −28 −2 ( 0 ) + 7 y = −28 7 y = −28 −28 7 y= 7 7 y = −4 y-intercept: (0, −4)
ANSWER: (14, 0), (0, −4)
9.
PROBLEM: x− y =3 SOLUTION: Set y = 0 x− y =3
x − (0) = 3 x=3 x-intercept: (3, 0) Set x = 0 x− y =3
( 0) − y = 3 y = −3 y-intercept: (0, −3) ANSWER: (3, 0), (0, −3)
10.
PROBLEM: −x + y = 0 SOLUTION: Set y = 0 −x + y = 0
− x + ( 0) = 0 x=0 x-intercept: (0, 0) Set x = 0 − ( 0) + y = 0 y=0 y-intercept: (0, 0)
ANSWER: (0, 0), (0, 0)
11.
PROBLEM: 3x − 4 y = 1 SOLUTION: Set y = 0 3x − 4 y = 1 3x − 4 ( 0 ) = 1 3x = 1 3 1 x= 3 3 1 x= 3
⎛1 ⎞ x-intercept: ⎜ , 0 ⎟ ⎝3 ⎠ Set x = 0 3x − 4 y = 1 3( 0) − 4 y = 1 −4 y = 1 −4 1 y=− −4 4 1 y=− 4
1⎞ ⎛ y-intercept: ⎜ 0, − ⎟ 4⎠ ⎝
1⎞ ⎛1 ⎞ ⎛ ANSWER: ⎜ , 0 ⎟ , ⎜ 0, − ⎟ 4⎠ ⎝3 ⎠ ⎝ 12.
PROBLEM: −2 x + 5 y = 3 SOLUTION: Set y = 0 −2 x + 5 y = 3 −2 x + 5 ( 0 ) = 3 −2 x = 3 3 −2 x=− −2 2 3 x=− 2
⎛ 3 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 2 ⎠ Set x = 0 −2 x + 5 y = 3 −2 ( 0 ) + 5 y = 3 5y = 3 5 3 y= 5 5 3 y= 5
⎛ 3⎞ y-intercept: ⎜ 0, ⎟ ⎝ 5⎠ ⎛ 3 ⎞ ⎛ 3⎞ ANSWER: ⎜ − , 0 ⎟ , ⎜ 0, ⎟ ⎝ 2 ⎠ ⎝ 5⎠
13.
PROBLEM: 1 1 x − y =1 4 3 SOLUTION: Set y = 0 1 1 x − y =1 4 3 1 1 x − (0) = 1 4 3 1 x =1 4 x = 1⋅ 4 x=4 x-intercept: (4, 0)
Set x = 0 1 1 x − y =1 4 3 1 1 ( 0) − y = 1 4 3 1 − y =1 3 y = −1 ⋅ 3 y = −3 y-intercept: (0, – 3) ANSWER: (4, 0), (0, – 3)
14.
PROBLEM: 2 3 − x+ y = 2 5 4 SOLUTION: Set y = 0 2 3 − x+ y =2 5 4 2 3 − x + (0) = 2 5 4 2 − x=2 5 ⎛ 5⎞ ⎛ 2 ⎞ ⎛ 5⎞ ⎜ − ⎟⋅⎜ − x ⎟ = 2⋅⎜ − ⎟ ⎝ 2⎠ ⎝ 5 ⎠ ⎝ 2⎠ x = −5 x-intercept: (– 5, 0)
Set x = 0 2 3 − x+ y = 2 5 4 2 3 − (0) + y = 2 5 4 4 3 4 ⋅ y = 2⋅ 3 4 3 8 y= 3 ⎛ 8⎞ y-intercept: ⎜ 0, ⎟ ⎝ 3⎠ ⎛ 8⎞ ANSWER: (– 5, 0), ⎜ 0, ⎟ ⎝ 3⎠ 15.
PROBLEM: y=6 SOLUTION: x-intercept: none
y=6 y-intercept: (0, 6) ANSWER: none, (0, 6)
16.
PROBLEM: y = −3 SOLUTION: x-intercept: none
y = –3 y-intercept: (0, –3) ANSWER: x-intercept: none, y-intercept: (0, –3)
17.
PROBLEM: x=2 SOLUTION: x = 2 x-intercept: (2, 0)
y-intercept: none ANSWER: (2, 0), none
18.
PROBLEM: x = −1 SOLUTION: x = –1 x-intercept: (–1, 0)
y-intercept: none ANSWER: x-intercept: (–1, 0), y-intercept: none
19.
PROBLEM: y = mx + b SOLUTION: Set y = 0 y = mx + b
( 0 ) = mx + b mx + b = 0 mx + b − b = −b mx = −b m b x=− m m b x=− m
⎛ b ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ m ⎠ Set x = 0 y = mx + b y = m ( 0) + b y =b y-intercept: (0, b)
⎛ b ⎞ ANSWER: ⎜ − , 0 ⎟ , (0, b) ⎝ m ⎠ 20.
PROBLEM: ax + by = c SOLUTION: Set y = 0 ax + by = c ax + b ( 0 ) = c ax = c a c x= a a c x= a
⎛c ⎞ x-intercept: ⎜ , 0 ⎟ ⎝a ⎠ Set x = 0 ax + by = c a ( 0 ) + by = c by = c b c y= b b c y= b
⎛ c⎞ y-intercept: ⎜ 0, ⎟ ⎝ b⎠ ⎛c ⎞ ANSWER: ⎜ , 0 ⎟ , ⎝a ⎠
⎛ c⎞ ⎜ 0, ⎟ ⎝ b⎠
Part B: Graph using Intercepts Find the intercepts and graph.
21.
PROBLEM: 3x + 4 y = 12 SOLUTION: Set y = 0 3x + 4 y = 12
3x + 4 ( 0 ) = 12 3x = 12 3 12 x= 3 3 x=4 x-intercept: (4, 0) Set x = 0 3 x + 4 y = 12 3 ( 0 ) + 4 y = 12 4 y = 12 4 12 y= 4 4 y=3 y-intercept: (0, 3)
ANSWER: (4, 0), (0, 3)
22.
PROBLEM: −2 x + 3 y = 6 SOLUTION: Set y = 0 −2 x + 3 y = 6
−2 x + 3 ( 0 ) = 6 −2 x = 6 −2 6 x=− −2 2 x = −3 x-intercept: (–3 , 0) Set x = 0 −2 x + 3 y = 6 −2 ( 0 ) + 3 y = 6 3y = 6 3 6 y= 3 3 y=2 y-intercept: (0, 2)
ANSWER: (–3 , 0), (0, 2)
23.
PROBLEM: 5 x − 2 y = 10 SOLUTION: Set y = 0 5 x − 2 y = 10
5 x − 2 ( 0 ) = 10 5 x = 10 5 10 x= 5 5 x=2 x-intercept: (2, 0) Set x = 0 5 x − 2 y = 10 5 ( 0 ) − 2 y = 10 −2 y = 10 10 −2 y=− 2 −2 y = −5 y-intercept: (0, –5)
ANSWER: (2, 0), (0, –5)
24.
PROBLEM: −4 x − 8 y = 16 SOLUTION: Set y = 0 −4 x − 8 y = 16
−4 x − 8 ( 0 ) = 16 −4 16 x=− −4 4 x = −4 x-intercept: (–4 , 0) Set x = 0 −4 x − 8 y = 16 −4 ( 0 ) − 8 y = 16 −8 y = 16 16 −8 y=− 8 −8 y = −2 y-intercept: (0, –2 )
ANSWER: (–4 , 0), (0, –2 )
25.
PROBLEM: 1 1 − x + y =1 2 3 SOLUTION: Set y = 0 1 1 − x + y =1 2 3 1 1 − x + ( 0) = 1 2 3 1 − x =1 2 x = −2 x-intercept: (–2 , 0)
Set x = 0 1 1 − x + y =1 2 3 1 1 − ( 0) + y = 1 2 3 1 y =1 3 y =3 y-intercept: (0, 3) ANSWER: (–2 , 0), (0, 3)
26.
PROBLEM: 3 1 x − y = −3 4 2 SOLUTION: Set y = 0 3 1 x − y = −3 4 2 3 1 x − ( 0 ) = −3 4 2 3 x = −3 4 4 3 4 ⋅ x = −3 ⋅ 3 4 3 x = −4 x-intercept: (–4 , 0)
Set x = 0 3 1 x − y = −3 4 2 3 1 ( 0 ) − y = −3 4 2 1 0 − y = −3 2 1 − y = −3 2 y=6 y-intercept: (0, 6) ANSWER: (–4 , 0), (0, 6)
27.
PROBLEM: 5 2 x − y = 10 2 SOLUTION: Set y = 0 5 2 x − y = 10 2 5 2 x − ( 0 ) = 10 2 2 x = 10
2 10 x= 2 2 x=5 x-intercept: (5, 0) Set x = 0 5 2 x − y = 10 2 5 2 ( 0 ) − y = 10 2 ⎛ 2 ⎞⎛ 5 ⎞ ⎛ 2⎞ ⎜ − ⎟⎜ − ⎟ y = 10 ⋅ ⎜ − ⎟ ⎝ 5 ⎠⎝ 2 ⎠ ⎝ 5⎠ y = −4 y-intercept: (0, –4) ANSWER: (5, 0), (0, –4)
28.
PROBLEM: 7 2 x − y = −14 3 SOLUTION: Set y = 0 7 2 x − y = −14 3 7 2 x − ( 0 ) = −14 3 2 14 x=− 2 2 x = −7 x-intercept: (–7 , 0)
Set x = 0 7 2 x − y = −14 3 7 2 ( 0 ) − y = −14 3 ⎛ 3⎞ ⎛ 7⎞ ⎛ 3⎞ ⎜ − ⎟ ⋅ ⎜ − ⎟ y = −14 ⋅ ⎜ − ⎟ ⎝ 7⎠ ⎝ 3⎠ ⎝ 7⎠ y=6 y-intercept: (0, 6) ANSWER: (–7 , 0), (0, 6)
29.
PROBLEM: 4 x − y = −8 SOLUTION: Set y = 0 4 x − y = −8 4 x − 0 = −8
4 8 x=− 4 4 x = −2 x-intercept: (–2, 0) Set x = 0 4 x − y = −8 4 ( 0 ) − y = −8 − y = −8 y =8 y-intercept: (0, 8) ANSWER: (–2, 0), (0, 8)
30.
PROBLEM: 6x − y = 6 SOLUTION: Set y = 0 6x − y = 6 6x − 0 = 6 6x = 6 6 6 x= 6 6 x =1 x-intercept: (1 , 0)
Set x = 0 6x − y = 6
6 ( 0) − y = 6 −y = 6 y = −6 y-intercept: (0, –6) ANSWER: (1 , 0), (0, –6)
31.
PROBLEM: –x + 2y =1 SOLUTION: Set y = 0 –x + 2y =1
− x + 2 ( 0) = 1 −x = 1 x = −1 x-intercept: (–1, 0) Set x = 0 –x + 2y =1 0 + 2y =1 2y =1 2 1 y= 2 2 1 y= 2 ⎛ 1⎞ y-intercept: ⎜ 0, ⎟ ⎝ 2⎠ ⎛ 1⎞ ANSWER: (–1, 0), ⎜ 0, ⎟ ⎝ 2⎠
32.
PROBLEM: 3x + 4 y = 6 SOLUTION: Set y = 0 3x + 4 y = 6
3x + 4 ( 0 ) = 6 3x = 6 3 6 x= 3 3 x=2 x-intercept: (2, 0) Set x = 0 3x + 4 y = 6 3 ( 0) + 4 y = 6 4y = 6 4 6 y= 4 4 3 y= 2
⎛ 3⎞ y-intercept: ⎜ 0, ⎟ ⎝ 2⎠ ⎛ 3⎞ ANSWER: (2, 0), ⎜ 0, ⎟ ⎝ 2⎠
33.
PROBLEM: 2 x + y = −1 SOLUTION: Set y = 0 2 x + y = −1 2 x + 0 = −1 2 x = −1 2 1 x=− 2 2 1 x=− 2 ⎛ 1 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 2 ⎠
Set x = 0 2 x + y = −1 2 ( 0 ) + y = −1 y = −1 y-intercept: (0, –1) ⎛ 1 ⎞ ANSWER: ⎜ − , 0 ⎟ , (0, –1) ⎝ 2 ⎠
34.
PROBLEM: −2 x + 6 y = 3 SOLUTION: Set y = 0 −2 x + 6 y = 3 −2 x + 6 ( 0 ) = 3 −2 x = 3 3 −2 x=− −2 2 3 x=− 2
⎛ 3 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 2 ⎠ Set x = 0 −2 x + 6 y = 3 −2 ( 0 ) + 6 y = 3 0 + 6y = 3 6y = 3 6 3 y= 6 6 1 y= 2 ⎛ 1⎞ y-intercept: ⎜ 0, ⎟ ⎝ 2⎠ ⎛ 3 ⎞ ANSWER: ⎜ − , 0 ⎟ , ⎝ 2 ⎠
⎛ 1⎞ ⎜ 0, ⎟ ⎝ 2⎠
35.
PROBLEM: 15 x + 4 y = −60 SOLUTION: Set y = 0 15 x + 4 y = −60
15 x + 4 ( 0 ) = −60 15 x = −60 15 60 x=− 15 15 x = −4 x-intercept: (–4, 0) Set x = 0 15 x + 4 y = −60 15 ( 0 ) + 4 y = −60 4 y = −60 4 60 y=− 4 4 y = −15 y-intercept: (0, –15)
ANSWER: (–4, 0), (0, –15)
36.
PROBLEM: −25 x + 3 y = 75 SOLUTION: Set y = 0 −25 x + 3 y = 75
−25 x + 3 ( 0 ) = 75 −25 x = 75 75 −25 x=− 25 −25 x = −3 x-intercept: (–3, 0) Set x = 0 −25 x + 3 y = 75 −25 ( 0 ) + 3 y = 75 3 y = 75 3 75 y= 3 3 y = 25 y-intercept: (0, 25)
ANSWER: (–3, 0), (0, 25)
37.
PROBLEM: 4x + 2 y = 0 SOLUTION: Set y = 0 4x + 2 y = 0
4x + 2 ( 0) = 0 x=0 x-intercept: (0, 0) Set x = 0 4x + 2 y = 0 4 ( 0) + 2 y = 0 2y = 0 y=0 y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
38.
PROBLEM: 3x − y = 0 SOLUTION: Set y = 0 3x − y = 0
3x − ( 0 ) = 0 x=0 x-intercept: (0, 0) Set x = 0 3x − y = 0 3( 0) − y = 0 y=0
y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
39.
PROBLEM: −12 x + 6 y = −4 SOLUTION: Set y = 0 −12 x + 6 y = −4 −12 x + 6 ( 0 ) = −4 −12 x = −4 −12 −4 x= −12 −12 1 x= 3 ⎛1 ⎞ x-intercept: ⎜ , 0 ⎟ ⎝3 ⎠
Set x = 0 −12 x + 6 y = −4 −12 ( 0 ) + 6 y = −4 6 y = −4 6 4 y=− 6 6 2 y=− 3
2⎞ ⎛ y-intercept: ⎜ 0, − ⎟ 3⎠ ⎝ 2⎞ ⎛1 ⎞ ⎛ ANSWER: ⎜ , 0 ⎟ , ⎜ 0, − ⎟ 3⎠ ⎝3 ⎠ ⎝
40.
PROBLEM: 3x + 12 y = −4 SOLUTION: Set y = 0 3x + 12 y = −4 3x + 12 ( 0 ) = −4 3x = −4 3 4 x=− 3 3 4 x=− 3
⎛ 4 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 3 ⎠ Set x = 0 3 x + 12 y = −4 3 ( 0 ) + 12 y = −4 12 y = −4 12 4 y=− 12 12 1 y=− 3
1⎞ ⎛ y-intercept: ⎜ 0, − ⎟ 3⎠ ⎝ ⎛ 4 ⎞ ANSWER: ⎜ − , 0 ⎟ , ⎝ 3 ⎠
1⎞ ⎛ ⎜ 0, − ⎟ 3⎠ ⎝
41.
PROBLEM: y = 2x + 4 SOLUTION: Set y = 0 y = 2x + 4
0 = 2x + 4 2x + 4 − 4 = 0 − 4 2 x = −4 2 4 x=− 2 2 x = −2 x-intercept: (–2, 0) Set x = 0 y = 2x + 4 y = 2 ( 0) + 4 y=4 y-intercept: (0, 4) ANSWER: (–2, 0), (0, 4)
42.
PROBLEM: y = −x + 3 SOLUTION: Set y = 0 y = −x + 3 0 = −x + 3 − x + 3 − 3 = −3 − x = −3 x=3 x-intercept: (3, 0)
Set x = 0 y = −x + 3 y = 0+3 y =3 y-intercept: (0, 3) ANSWER: (3, 0), (0, 3)
43.
PROBLEM: 1 y = x +1 2 SOLUTION: Set y = 0 1 y = x +1 2 1 0 = x +1 2 1 0 −1 = x + 1 −1 2 1 −1 = x 2 x = −2 x-intercept: (–2, 0)
Set x = 0 1 y = x +1 2 1 y = ( 0) + 1 2 y =1 y-intercept: (0, 1) ANSWER: (–2, 0), (0, 1)
44.
PROBLEM: 2 y = x −3 3 SOLUTION: Set y = 0 2 y = x−3 3 2 0 = x −3 3 2 0+3 = x −3+3 3 2 x=3 3 3 2 3 ⋅ x = 3⋅ 2 3 2 9 x= 2 ⎛ 1 ⎞ x-intercept: ⎜ 4 , 0 ⎟ ⎝ 2 ⎠ Set x = 0 2 y = x−3 3 2 y = ( 0) − 3 3 y = −3 y-intercept: (0, –3)
⎛ 1 ⎞ ANSWER: ⎜ 4 , 0 ⎟ , (0, –3) ⎝ 2 ⎠
45.
PROBLEM: 2 y = − x +1 5 SOLUTION: Set y = 0 2 y = − x +1 5 2 0 = − x +1 5 2 0 −1 = − x + 1 −1 5 2 −1 = − x 5 ⎛ 5⎞ ⎛ 2⎞ ⎛ 5⎞ ⎜ − ⎟ ⋅ ⎜ − ⎟ x = −1 ⋅ ⎜ − ⎟ ⎝ 2⎠ ⎝ 5⎠ ⎝ 2⎠ 5 x= 2 ⎛ 1 ⎞ x-intercept: ⎜ 2 , 0 ⎟ ⎝ 2 ⎠
Set x = 0 2 y = − x +1 5 2 y = − ( 0) + 1 5 y =1 y-intercept: (0, 1) ⎛ 1 ⎞ ANSWER: ⎜ 2 , 0 ⎟ , (0, 1) ⎝ 2 ⎠
46.
PROBLEM: 5 5 y =− x− 8 4 SOLUTION: Set y = 0 5 5 y =− x− 8 4 5 5 0=− x− 8 4 5 5 5 5 0+ = − x− + 4 8 4 4 5 ⎛ 8⎞ ⎛ 8⎞ ⎛ 5⎞ ⎜ − ⎟⋅⎜ − ⎟ x = ⋅⎜ − ⎟ 4 ⎝ 5⎠ ⎝ 5⎠ ⎝ 8⎠ x = −2 x-intercept: (–2, 0)
Set x = 0 5 5 y =− x− 8 4 5 5 y = − ( 0) − 8 4 5 y=− 4 1⎞ ⎛ y-intercept: ⎜ 0, −1 ⎟ 4⎠ ⎝ 1⎞ ⎛ ANSWER: (–2, 0), ⎜ 0, −1 ⎟ 4⎠ ⎝
47.
PROBLEM: 7 7 y =− x− 8 2 SOLUTION: Set y = 0 7 7 y =− x− 8 2 7 7 0=− x− 8 2 7 7 7 7 0+ = − x− + 2 8 2 2 7 7 − x= 8 2 ⎛ 8⎞ ⎛ 7⎞ ⎛7⎞ ⎛ 8⎞ ⎜ − ⎟⋅⎜ − ⎟ x = ⎜ ⎟⋅⎜ − ⎟ ⎝ 7⎠ ⎝ 8⎠ ⎝2⎠ ⎝ 7⎠ x = −4 x-intercept: (–4, 0)
Set x = 0 7 7 y =− x− 8 2 7 7 y = − ( 0) − 8 2 7 y=− 2 1⎞ ⎛ y-intercept: ⎜ 0, −3 ⎟ 2⎠ ⎝ 1⎞ ⎛ ANSWER: (–4, 0), ⎜ 0, −3 ⎟ 2⎠ ⎝
48.
PROBLEM: 3 y = −x + 2 SOLUTION: Set y = 0 3 y = −x + 2 3 0 = −x + 2 3 3 3 −x + − = 0 − 2 2 2 3 −x = − 2 3 x= 2 ⎛ 1 ⎞ x-intercept: ⎜1 , 0 ⎟ ⎝ 2 ⎠
Set x = 0
y = −x + y = 0+ y=
3 2
3 2
3 2
⎛ 1⎞ y-intercept: ⎜ 0,1 ⎟ 2⎠ ⎝ ⎛ 1 ⎞ ANSWER: ⎜1 , 0 ⎟ , ⎝ 2 ⎠
⎛ 1⎞ ⎜ 0,1 ⎟ 2⎠ ⎝
49.
PROBLEM: y=3 SOLUTION: No x-intercept
y =3 y-intercept: (0, 3) ANSWER: No x-intercept, (0, 3)
50.
PROBLEM: 3 y= 2 SOLUTION: No x-intercept
y=
3 2
⎛ 1⎞ y-intercept: ⎜ 0,1 ⎟ 2⎠ ⎝ ⎛ 1⎞ ANSWER: No x-intercept, ⎜ 0,1 ⎟ 2⎠ ⎝
51.
PROBLEM: x=5 SOLUTION: x = 5 x-intercept: (5, 0)
No y-intercept ANSWER: (5, 0), No y-intercept.
52.
PROBLEM: x = −2 SOLUTION: x = −2 x-intercept: (–2, 0)
No y-intercept ANSWER: (–2, 0), No y-intercept.
53.
PROBLEM: y = 5x SOLUTION: Set y = 0 y = 5x 0 = 5x x=0 x-intercept: (0, 0)
Set x = 0 y = 5x
y = 5 ( 0) y=0 y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
54.
PROBLEM: y = −x SOLUTION: Set y = 0 y = −x y=0 x-intercept: (0, 0)
Set x = 0 y = −x
0 = −x x=0 y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
Part C: Intercepts of Nonlinear Graphs. Given the graph find the x‐ and y‐intercepts.
55.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (–3, 0) and (3, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −3).
56.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (–3, 0) and (1, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −1).
57.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (–4, 0) and (0, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, 0).
58.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in one place. Therefore, this graph has only one x-intercept, namely (2, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only y-intercept is (0, −2).
59.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (−2, 0) and (2, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −1).
60.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (−1, 0) and (3, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, 3).
61.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in three places. Therefore, this graph has three x-intercepts, namely (−3, 0), (0, 0), and (2, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, 0).
62.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (−4, 0) and (−1, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −2).
63.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has three x-intercepts, namely (−4, 0) and (4, 0). Furthermore, the graph intersects the y-axis in two places, hence, the y-intercept are (0, −4) and (0, 4).
64.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in one place. Therefore, this graph has one x-intercept, namely (0, 0). Furthermore, the graph intersects the y-axis in two places, hence, the y-intercept are (0, 0) and (0, −6).
Part D: Discussion Board Topics
65.
PROBLEM: What are the x-intercepts of the line y = 0 ? ANSWER: The line y = 0 does not have any x-intercepts.
66.
PROBLEM: What are the y-intercepts of the line x = 0 ? ANSWER: The line x = 0 does not have any y-intercepts.
67.
PROBLEM: Do all lines have intercepts? ANSWER: No. An example is a horizontal line like y = 3. This line never intersects the x-axis, therefore it does not have an x-intercept. (The value of y may vary)
68.
PROBLEM: How many intercepts can a circle have? Draw circles showing all possible numbers of intercepts. ANSWER: A circle can have a maximum of 4 intercepts.
From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has three x-intercepts, namely (−4, 0) and (4, 0). Furthermore, the graph intersects the y-axis in two places, hence, the y-intercept are (0, 4) and (0, −4). (The x- and y-intercepts may vary.)
69.
PROBLEM: Research and post the definitions of line segment, ray and line. Why are the arrows important? ANSWER: Line segment: A line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. (Source: http://en.wikipedia.org/wiki/Line_segment)
Ray: A ray is a part of a line that begins at a particular point (called the endpoint) and extends endlessly in one direction. A ray is also called half-line. (Source: http://www.icoachmath.com/SiteMap/Ray.html) Line: A straight line is the shortest distance between any two points on a plane. (Source: http://www.mathopenref.com/line.html) The arrows at the end of a line indicate the line can extend beyond the endpoints in either direction. 70.
PROBLEM: Can a function have more than one y-intercept? Explain ANSWER: Functions associate x values to no more than one y value as part of their definition, they can have at most one y-intercept. Some 2-dimensional mathematical relationships such as circles, ellipses, and hyperbolas can have more than one y-intercept. (Source: http://en.wikipedia.org/wiki/Y-intercept)
3.4 Graph using the yintercept and Slope
Part A: Slope Determine the slope and the y‐intercept of the given graph.
1.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, 3). ( x1 , y1 ) ( x2 , y2 )
(4, 0) (0,3) y − y 3−0 3 m= 2 1 = =− x2 − x1 0 − 4 4 3 ANSWER: − , (0, 3). 4 2.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, –2). ( x1 , y1 ) ( x2 , y2 )
(3, 0) (0, − 2) y − y −2 − 0 − 2 2 = = m= 2 1 = x2 − x1 0 − 3 −3 3 ANSWER:
2 , y-intercept is (0, –2). 3
3.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, 2). ( x1 , y1 ) ( x2 , y2 )
(1, 2) (0, 2) y − y 2−2 0 = =0 m= 2 1 = x2 − x1 1 − 0 1 ANSWER: 0, (0, 2).
4.
PROBLEM:
SOLUTION: From the graph, we can see that it does not intersect the y-axis. Hence, there is no y-intercept. ( x1 , y1 ) ( x2 , y2 )
(2, 0) (2, 2) y − y 2−0 2 = m= 2 1 = x2 − x1 2 − 2 0 ANSWER: The slope is not defined, no y-intercept.
5.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, 0). ( x1 , y1 ) ( x2 , y2 )
(0, 0) (1, 2) y − y 2−0 2 m= 2 1 = = =2 x2 − x1 1 − 0 1 ANSWER: 2, (0, 0).
6.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis. Hence, the y-intercept is (0, –1). ( x1 , y1 ) ( x2 , y2 )
(−2, 0) (0, − 1) y −y −1 − 0 1 m= 2 1 = =− x2 − x1 0 − ( −2 ) 2 1 ANSWER: − , y-intercept is (0, –1). 2
Determine the slope given two points.
7.
PROBLEM: (3, 2) and (5, 1) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(3, 2) (5,1) y − y 1− 2 1 m= 2 1 = =− x2 − x1 5 − 3 2 ANSWER: −
8.
1 2
PROBLEM: (7, 8) and (−3, 5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(7,8) (−3,5) y −y 5−8 3 −3 m= 2 1 = = = x2 − x1 −3 − 7 −10 10 ANSWER:
9.
3 10
PROBLEM: (2, −3) and (−3, 2) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(2, −3) (−3, 2) y − y 2 − ( −3 ) 5 m= 2 1 = = − = −1 x2 − x1 −3 − 2 5 ANSWER: −1
10.
PROBLEM: (−3, 5) and (7, −5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−3, 5) (7, −5) y −y −5 − 5 −10 m= 2 1 = = = −1 x2 − x1 7 − ( −3) 10 ANSWER: −1
11.
PROBLEM: (−1, −6) and (3, 2) SOLUTION: ( x1 , y1 ) ( x2 , y2 ) (−1, −6) m=
( 3, 2 )
y2 − y1 2 − ( −6 ) 8 = = =2 x2 − x1 3 − ( −1) 4
ANSWER: 2
12.
PROBLEM: (5, 3) and (4, 12) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( 5, 3) m=
( 4, 12 )
y2 − y1 12 − 3 9 = = = −9 x2 − x1 4 − 5 −1
ANSWER: −9
13.
PROBLEM: (−9, 3) and (−6, −5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−9, 3) (−6, −5) y −y −5 − 3 −8 8 = m= 2 1 = =− x2 − x1 −6 − ( −9 ) 3 3 ANSWER: −
14.
8 3
PROBLEM: (−22, 4) and (−8, −12) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−22, 4) (−8, −12) y −y −12 − 4 −16 8 m= 2 1 = = =− x2 − x1 −8 − ( −22 ) 14 7 ANSWER: −
15.
8 7
PROBLEM: ( 12 , − 13 ) and ( − 12 ,
2 3
)
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( 12 , − 13 ) ( − 12 , 23 )
2 ⎛ 1⎞ 2 1 3 − − + y2 − y1 3 ⎜⎝ 3 ⎟⎠ m= = = 3 3 = 3 = −1 1 1 1 1 2 x2 − x1 − − − − − 2 2 2 2 2 ANSWER: −1
16.
PROBLEM: ( − 34 , 32 ) and ( 14 , − 12 ) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( − 34 , 32 ) ( 14 , − 12 )
1 3 4 − − − y −y 2 2 = 2 = −2 m= 2 1 = 4 x2 − x1 1 ⎛ 3 ⎞ −⎜− ⎟ 4 4 ⎝ 4⎠ ANSWER: −2
17.
PROBLEM: ( − 13 , 85 ) and ( 12 , − 34 ) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( − 13 , 85 ) ( 12 , − 43 )
3 5 11 −6 − 5 − − − y2 − y1 4 8 = 8 = 8 = − 11 ⋅ 6 = − 33 m= = 3+ 2 5 8⋅5 20 x2 − x1 1 ⎛ 1 ⎞ −⎜− ⎟ 6 6 2 ⎝ 3⎠ 33 ANSWER: − 20
18.
PROBLEM: ( − 53 , − 23 ) and
( 101 , 54 )
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( − 53 , − 32 ) ( 101 , 54 ) 4 ⎛ 3 ⎞ 8 + 15 23 −⎜− ⎟ y2 − y1 23 5 ⎝ 2⎠ m= = = 10 = 10 = 7 1 ⎛ 3 ⎞ 1+ 6 x2 − x1 7 −⎜− ⎟ 10 10 10 ⎝ 5 ⎠
ANSWER:
23 7
19.
PROBLEM: (3, −5) and (5, −5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(3, −5)
(5, −5) y − y − 5 − ( −5 ) 0 m= 2 1 = = =0 x2 − x1 5−3 2
ANSWER: 0
20.
PROBLEM: (−3, 1) and (−14, 1) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−3, 1) (−14, 1) y −y 1−1 0 m= 2 1 = = =0 x2 − x1 −14 − ( −3 ) −11 ANSWER: 0
21.
PROBLEM: (−2, 3) and (−2, −4) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−2, 3) (−2, −4) y −y −4 − 3 −7 m= 2 1 = = x2 − x1 −2 − ( −2 ) 0 ANSWER: The slope is not defined.
22.
PROBLEM: (−4, −4) and (5, 5) SOLUTION: ( x1 , y1 ) ( x2 , y2 ) (−4, −4) m=
( 5, 5)
y2 − y1 5 − ( −4 ) 9 = = =1 x2 − x1 5 − ( −4 ) 9
ANSWER: 1
23.
PROBLEM: A roof drops 4 ft for every 12 ft forward. Determine the slope of the roof. SOLUTION: If the roof moves forward, the x-intercept is 12, i.e. (12, 0). If the roof drops down, the y-intercept is 4, i.e. (0, 4). ( x1 , y1 ) ( x2 , y2 )
(0, 4) (12, 0) y −y 0−4 4 1 =− =− m= 2 1 = x2 − x1 12 − 0 12 3 ANSWER: −
24.
1 3
PROBLEM: A road drops 300 ft for every 5,280 ft forward. Determine the slope of the road. SOLUTION: If the road moves forward, the x-intercept is 5280, i.e. (5280, 0). If the road drops down, the y-intercept is 300, i.e. (0, 300). ( x1 , y1 ) ( x2 , y2 )
( 5280, 0 ) ( 0, 300 ) m=
y2 − y1 300 − 0 300 5 = =− =− x2 − x1 0 − 5280 5280 88
ANSWER: −
5 88
25.
PROBLEM: The following graph gives the US population of persons 65 years old and over (Source: U.S. Census Bureau). At what rate was this population increasing from the year 2000 to 2008?
SOLUTION: The rate of population increasing from the year 2000 to 2008 can be found out by calculating the slope between the points corresponding to 2000 and 2008 on the x-axis and 33 and 37 corresponding to the y-axis. 37 − 33 4 1 m= = = 2008 − 2000 8 2 1 ANSWER: The population is increasing at a rate of million per year from 2 2000 to 2008.
26.
PROBLEM: The following graph gives total consumer credit outstanding in the US (Source: U.S. Census Bureau). At what rate was consumer credit increasing from the year 2002 to 2008?
SOLUTION: The rate of increasing consumer credit from the year 2002 to 2008 can be found out by calculating the slope between the points corresponding to 2002 and 2008 on the x-axis and 2.0 and 2.6 corresponding to the y-axis. 2.6 − 2.0 0.6 m= = = 0.1 2008 − 2002 6 ANSWER: The consumer credit increases at a rate of $0.1 billion per year from 2002 to 2008.
27.
PROBLEM: A commercial van was purchased new for $20,000 and is expected to be worth $4,000 in 8 years. Determine the rate at which the van will depreciate in value. SOLUTION: The rate of depreciation can be found out by calculating the slope, which helps in determining the rate at which the van will depreciate in value. Price is plotted in the y-axis and the number of years in the x-axis. $20, 000 − $4000 $16, 000 m= = = $2, 000 8−0 8 ANSWER: The van will depreciate by $2,000 per year.
28.
PROBLEM: A commercial grade copy machine was purchased new for $4,800 and will be considered worthless in 6 years. Determine the rate at which the copy machine will depreciate in value. SOLUTION: The rate of depreciation can be found out by calculating the slope, which helps in determining the rate at which the commercial grade copy machine will depreciate in value. Price is plotted in the y-axis and the number of years in the x-axis. $4800 − 0 $4800 m= = = $800 6−0 6 ANSWER: The commercial grade copy machine will depreciate by $800 per year.
29.
PROBLEM: Find y if the slope of the line passing through (−2, 3) and (4, y) is 12. SOLUTION: m = 12 ( x1 , y1 ) ( x2 , y2 )
(−2, 3) (4, y ) y −y y −3 y −3 12 = 2 1 = = 6 x2 − x1 4 − ( −2 ) y − 3 + 3 = 72 + 3 y = 75
ANSWER: y = 75
30.
PROBLEM: Find y if the slope of the line passing through (5, y) and (6, −1) is 10. SOLUTION: m = 10 ( x1 , y1 ) ( x2 , y2 )
( 5, y )
10 =
(6, −1)
y2 − y1 −1 − y −1 − y = = 6−5 1 x2 − x1
−1 − y = 10 −1 + 1 − y = 10 + 1 − y = 11 y = −11
ANSWER: y = −11
31.
PROBLEM: Find y if the slope of the line passing through (5, y) and (−4, 2) is 0. SOLUTION: m=0 ( x1 , y1 ) ( x2 , y2 )
( 5, y ) 0=
(−4, 2)
y2 − y1 2 − y 2 − y = = x2 − x1 −4 − 5 −9
2−2− y = 0−2 − y = −2 y=2
ANSWER: y = 2
32.
PROBLEM: Find x if the slope of the line passing through (−3, 2) and (x, 5) is undefined. SOLUTION: m = undefined ( x1 , y1 ) ( x2 , y2 ) (−3, 2)
( x, 5 )
3 y2 − y1 5−2 3 = = = 0 x2 − x1 x − ( −3) x + 3 x+3= 0 x + 3 − 3 = −3 x = −3
ANSWER: x = −3
Part B: Slope‐Intercept Form Express the given linear equation in slope‐intercept form and identify the slope and y‐intercept.
33.
PROBLEM: 6 x − 5 y = 30 SOLUTION: 6 x − 5 y = 30 6 x − 6 x − 5 y = 30 − 6 x 5 y = 6 x − 30 5 6 30 y = x− 5 5 5 6 y = x−6 5
6 ANSWER: The y-intercept is (0, – 6) and the slope is m = . 5 34.
PROBLEM: −2 x + 7 y = 28 SOLUTION: −2 x + 7 y = 28 −2 x + 2 x + 7 y = 28 + 2 x 7 y = 2 x + 28 7 2 28 y = x+ 7 7 7 2 y = x+4 7 ANSWER: The y-intercept is (0, 4) and the slope is m =
35.
2 . 7
PROBLEM: 9 x − y = 17 SOLUTION: 9 x − y = 17 9 x − 9 x − y = 17 − 9 x
y = 9 x − 17 ANSWER: The y-intercept is (0, – 17) and the slope is m = 9.
36.
PROBLEM: x − 3 y = 18 SOLUTION: x − 3 y = 18 x − x − 3 y = 18 − x
3 y = x − 18 3 1 18 y = x− 3 3 3 1 y = x−6 3
1 ANSWER: The y-intercept is (0, –6) and the slope is m = . 3 37.
PROBLEM: 2x − 3 y = 0 SOLUTION: 2x − 3y = 0 2x − 2x − 3y = 0 − 2x
3 y = 2x 3 2 y= x 3 3 2 y= x 3 2 ANSWER: The y-intercept is (0, 0) and the slope is m = . 3 38.
PROBLEM: −6 x + 3 y = 0 SOLUTION: −6 x + 3 y = 0 −6 x + 6 x + 3 y = 0 + 6 x 3y = 6x 3 6 y= x 3 3 y = 2x ANSWER: The y-intercept is (0, 0) and the slope is m = 2.
39.
PROBLEM: 5 2 3 x − 4 y = 10 SOLUTION: 5 2 3 x − 4 y = 10 2 3
x − 23 x − 54 y = 10 − 32 x
4 5
⋅ 54 y = 54 ⋅ 23 x − 54 ⋅10
y = 158 x − 8 ANSWER: The y-intercept is (0, –8) and the slope is m =
40.
8 . 15
PROBLEM: − 43 x + 15 y = −5 SOLUTION: − 43 x + 15 y = −5
− 43 x + 43 x + 15 y = −5 + 34 x 1 5
y = 43 x − 5
5 1
⋅ 15 y = 15 ⋅ 34 x − 15 ⋅ 5
y=
20 3
x − 25
ANSWER: The y-intercept is (0, –25) and the slope is m =
20 . 3
Graph the line given the slope and the y‐intercept.
41.
PROBLEM: 1 m = and (0, −2) 3 SOLUTION: 1 rise m= = 3 run Starting from the point (0, −2) we will use the slope to mark another point up 1 unit and to the right 3 units.
ANSWER:
42.
PROBLEM: 2 m = − and (0, 4) 3 SOLUTION:
2 rise = 3 run Starting from the point (0, 4) we will use the slope to mark another point down 2 units and to the right 3 units. m=−
ANSWER:
43.
PROBLEM: m = 3 and (0, 1) SOLUTION: 3 rise m= = 1 run Starting from the point (0, 1) we will use the slope to mark another point up 3 units and to the right 1 unit.
ANSWER:
44.
PROBLEM: m = −2 and (0, −1) SOLUTION: 2 rise m=− = 1 run Starting from the point (0, −1) we will use the slope to mark another point down 2 units and to the right 1 unit. ANSWER:
45.
PROBLEM: m = 0 and (0, 5) SOLUTION: rise m=0= run Starting from the point (0, 5) the line is a horizontal line with no slope.
ANSWER:
46.
PROBLEM: m undefined and (0, 0) SOLUTION:
rise run Starting from the point (0, 0) the line is along the y-axis since the slope is undefined. m = undefined =
ANSWER:
47.
PROBLEM: m = 1 and (0, 0) SOLUTION: 1 rise m= = 1 run Starting from the point (0, 0) we will use the slope to mark another point up 1 unit and to the right 1 unit.
ANSWER:
48.
PROBLEM: m = −1 and (0, 0) SOLUTION: 1 rise m=− = 1 run Starting from the point (0, 0) we will use the slope to mark another point down 1 unit and to the right 1 unit. ANSWER:
49.
PROBLEM: 15 m=− and (0, 20) 3 SOLUTION: 15 rise m=− = 3 run Starting from the point (0, 20) we will use the slope to mark another point down 15 units and to the right 3 units.
ANSWER:
50.
PROBLEM: m = −10 and (0, −5) SOLUTION: 10 rise m=− = 1 run Starting from the point (0, −5) we will use the slope to mark another point down 10 units and to the right 1 unit. ANSWER:
Graph using the slope and y‐intercept.
51.
PROBLEM: 2 y = x−2 3 SOLUTION:
2 The y-intercept is (0, –2) and the slope is m = . 3
2 rise = 3 run Starting from the point (0, –2) we will use the slope to mark another point up 2 units and to the right 3 units. m=
ANSWER:
52.
PROBLEM: 1 y = − x +1 3 SOLUTION:
1 The y-intercept is (0, 1) and the slope is m = − . 3 1 rise m=− = 3 run Starting from the point (0, 1) we will use the slope to mark another point down 1 unit and to the right 3 units. ANSWER:
53.
PROBLEM: y = −3 x + 6 SOLUTION: The y-intercept is (0, 6) and the slope is m = –3. 3 rise m=− = 1 run Starting from the point (0, 6) we will use the slope to mark another point down 3 units and to the right 1 unit. ANSWER:
54.
PROBLEM: y = 3x + 1 SOLUTION: The y-intercept is (0, 1) and the slope is m = 3. 3 rise m= = 1 run Starting from the point (0, 1) we will use the slope to mark another point down 3 units and to the right 1 unit. ANSWER:
55.
PROBLEM: 3 y= x 5 SOLUTION:
3 The y-intercept is (0, 0) and the slope is m = . 5 3 rise m= = 5 run Starting from the point (0, 0) we will use the slope to mark another point up 3 units and to the right 5 units. ANSWER:
56.
PROBLEM: 3 y=− x 7 SOLUTION:
3 The y-intercept is (0, 0) and the slope is m = − . 7 3 rise m=− = 7 run Starting from the point (0, 0) we will use the slope to mark another point down 3 units and to the right 7 units.
ANSWER:
57.
PROBLEM: y = −8 SOLUTION: The y-intercept is (0, –8) and the slope is m = 0. rise m=0= run Starting from the point (0, –8) the line is a horizontal line with no slope. ANSWER:
58.
PROBLEM: y=7 SOLUTION: The y-intercept is (0, 7) and the slope is m = 0. rise m=0= run Starting from the point (0, 7) the line is a horizontal line with no slope.
ANSWER:
59.
PROBLEM: y = −x + 2 SOLUTION: The y-intercept is (0, 2) and the slope is m = –1. 1 rise m=− = 1 run Starting from the point (0, 2) we will use the slope to mark another point down 1 unit and to the right 1 unit. ANSWER:
60.
PROBLEM: y = x +1 SOLUTION: The y-intercept is (0, 1) and the slope is m = 1. 1 rise m= = 1 run Starting from the point (0, 1) we will use the slope to mark another point up 1 unit and to the right 1 unit.
ANSWER:
61.
PROBLEM: 1 3 y = x+ 2 2 SOLUTION:
1 ⎛ 1⎞ The y-intercept is ⎜ 0,1 ⎟ and the slope is m = . 2 2⎠ ⎝ 1 rise m= = 2 run ⎛ 1⎞ Starting from the point ⎜ 0,1 ⎟ we will use the slope to mark another point up 1 2⎠ ⎝ unit and to the right 2 units. ANSWER:
62.
PROBLEM: 3 5 y =− x+ 4 2 SOLUTION:
1⎞ 3 ⎛ The y-intercept is ⎜ 0, 2 ⎟ and the slope is m = − . 2⎠ 4 ⎝ 3 rise m=− = 4 run 1⎞ ⎛ Starting from the point ⎜ 0, 2 ⎟ we will use the slope to mark another point down 2⎠ ⎝ 3 units and to the right 4 units. ANSWER:
63.
PROBLEM: 4x + y = 7 SOLUTION: 4x + y = 7 4x − 4x + y = 7 − 4x
y = −4 x + 7 The y-intercept is (0, 7) and the slope is m = –4. 4 rise m=− = 1 run Starting from the point (0, 7) we will use the slope to mark another point down 4 units and to the right 1 unit.
ANSWER:
64.
PROBLEM: 3x − y = 5 SOLUTION:
3x − y = 5 3x − 3x − y = 5 − 3x
y = 3x − 5 The y-intercept is (0, –5) and the slope is m = 3. 3 rise m= = 1 run Starting from the point (0, –5) we will use the slope to mark another point up 3 units and to the right 1 unit. ANSWER:
65.
PROBLEM: 5 x − 2 y = 10 SOLUTION: 5 x − 2 y = 10 5 x − 5 x − 2 y = 10 − 5 x 2 y = 5 x − 10 2 5 10 y = x− 2 2 2 5 y = x −5 2
5 The y-intercept is (0, –5) and the slope is m = . 2 5 rise m= = 2 run Starting from the point (0, –5) we will use the slope to mark another point up 5 units and to the right 2 units. ANSWER:
66.
PROBLEM: −2 x + 3 y = 18 SOLUTION: −2 x + 3 y = 18 −2 x + 2 x + 3 y = 18 + 2 x 3 y = 2 x + 18 3 2 18 y = x+ 3 3 3 2 y = x+6 3
2 The y-intercept is (0, 6) and the slope is m = . 3 2 rise m= = 3 run Starting from the point (0, 6) we will use the slope to mark another point up 2 units and to the right 3 units. ANSWER:
67.
PROBLEM: x− y =0 SOLUTION: x− y =0 x− x− y = 0− x
y=x The y-intercept is (0, 0) and the slope is m = 1. 1 rise m= = 1 run Starting from the point (0, 0) we will use the slope to mark another point up 1 unit and to the right 1 unit.
ANSWER:
68.
PROBLEM: x+ y =0 SOLUTION:
x+ y =0 x− x+ y = 0− x
y = −x The y-intercept is (0, 0) and the slope is m = –1. 1 rise m=− = 1 run Starting from the point (0, 0) we will use the slope to mark another point down 1 unit and to the right 1 unit. ANSWER:
69.
PROBLEM: 1 1 x − y =1 2 3 SOLUTION: 1 1 x − y =1 2 3 1 1 1 1 x − x − y = 1− x 2 2 3 2 1 1 y = x −1 3 2 1 1 3 ⋅ y = 3 ⋅ x − 3 ⋅1 3 2 3 y = x−3 2
3 The y-intercept is (0, –3) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, –3) we will use the slope to mark another point up 3 units and to the right 2 units. ANSWER:
70.
PROBLEM: 2 1 − x+ y =2 3 2 SOLUTION:
2 1 − x+ y =2 3 2 2 2 1 2 − x+ x+ y = 2+ x 3 3 2 3 1 2 y = x+2 2 3 1 2 2⋅ y = 2⋅ x + 2⋅2 2 3 4 y = x+4 3 4 The y-intercept is (0, 4) and the slope is m = . 3 4 rise m= = 3 run Starting from the point (0, 4) we will use the slope to mark another point up 4 units and to the right 3 units. ANSWER:
71.
PROBLEM: 3x + 2 y = 1 SOLUTION: 3x + 2 y = 1 3x − 3x + 2 y = 1 − 3x 2 y = −3 x + 1 2 3 1 y =− x+ 2 2 2 3 1 y =− x+ 2 2 3 ⎛ 1⎞ The y-intercept is ⎜ 0, ⎟ and the slope is m = − . 2 ⎝ 2⎠ 3 rise m=− = 2 run ⎛ 1⎞ Starting from the point ⎜ 0, ⎟ we will use the slope to mark another point down 3 ⎝ 2⎠ units and to the right 2 units. ANSWER:
72.
PROBLEM: −5 x + 3 y = −1 SOLUTION; −5 x + 3 y = −1 −5 x + 5 x + 3 y = −1 + 5 x 3 y = 5x −1 3 5 1 y = x− 3 3 3 5 1 y = x− 3 3 1⎞ 5 ⎛ The y-intercept is ⎜ 0, − ⎟ and the slope is m = . 3 3⎠ ⎝ 5 rise m= = 3 run Starting from the point (0, 4) we will use the slope to mark another point up 5 units and to the right 3 units. ANSWER:
73.
PROBLEM: On the same set of axes graph the three lines where y = 32 x + b where b = {−2, 0, 2}. SOLUTION: y = 32 x + b When b = −2 y = 32 x − 2
3 The y-intercept is (0, –2) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, –2) we will use the slope to mark another point up 3 units and to the right 2 units. When b = 0 y = 32 x + 0 y = 32 x 3 The y-intercept is (0, 0) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, 0) we will use the slope to mark another point up 3 units and to the right 2 units. When b = 2 y = 32 x + 2 3 The y-intercept is (0, 2) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, 2) we will use the slope to mark another point up 3 units and to the right 2 units.
ANSWER:
74.
PROBLEM: On the same set of axes graph the three lines where y = mx + 1 where m = {−1/2, 0, 1/2}. SOLUTION:
y = mx + 1
When m = −1/2
1 y = − x +1 2 1 The y-intercept is (0, 1) and the slope is m = − . 2 1 rise m=− = 2 run Starting from the point (0, 1) we will use the slope to mark another point down 1 unit and to the right 2 units. When m = 0
y = ( 0) x + 1
y =1 The y-intercept is (0, 1) and the slope is m = 0. rise m=0= run Starting from the point (0, 1) the line will be a horizontal line with no slope.
When m = 1/2
y=
1 x +1 2
1 The y-intercept is (0, 1) and the slope is m = . 2
1 rise = 2 run Starting from the point (0, 1) we will use the slope to mark another point up 1 unit and to the right 2 units. m=
ANSWER:
Part C: Discussion Board Topics
75.
PROBLEM: Name three methods for graphing lines. Discuss the pros and cons of each method? ANSWER: The three methods of graphing lines are by using the x-intercepts and y-intercepts, slope intercept form, and point slope form. Pros of the x-intercept and y-intercept method Identifies the y intercept Identifies x intercept Cons of the x-intercept and y-intercept method Does not involve understanding of gradient or rate of change. Pros of the slope intercept form Identifies y intercept. Shows understanding of gradient or rate of change. Cons of the slope intercept form Rule must be in y = mx + b form. Does not find x intercept
Pros of the point-slope form The beauty of the slope formula is that we can obtain the slope, given two points, using only algebra. Cons of the point-slope formula When using the slope formula, take care to be consistent since order does matter. We must subtract the coordinates of the first point from the second point for both the numerator and denominator in the same order. When considering the slope as a rate of change it is important to include the correct units. (Source: henchmaths.wikispaces.com/.../Sketching+Linear+Graphs+4+ways.ppt)
76.
PROBLEM: Choose a linear equation and graph it 3 different ways. Scan the work and share it on the discussion board. ANSWER: Let the equation be x – y = 0 x-intercept and the y-intercept form Set x = 0 y=0 y-intercept: (0, 0) Set y = 0 x=0 x-intercept: (0, 0) Slope intercept form x–y–x=0–x y=x m = 1; y intercept: (0, 0) Point slope form Let the two coordinates be (0, 0) and (1, 1). y −y y − y1 m= 2 1 = x2 − x1 x − x1
1− 0 y − 0 = 1 −1 x − 0 y=x
m=
m = 1; y intercept: (0, 0) (Answers may vary)
77.
PROBLEM: Why do we use the letter m for slope? ANSWER: A third theory distinguishes between variables in the beginning of the alphabet--a, b, c--from variables in the middle of the alphabet--l, m, n--from variables at the end of the alphabet--x, y, z. Variables in the beginning of the alphabet represent constants, such as Avogadro's number. Variables in the middle of the alphabet represent parameters. Variables at the end of the alphabet are, well, variable.
78.
PROBLEM: How are equivalent fractions useful when working with slopes? ANSWER: Equivalent fractions help in simplifying the graphing. The rise and the run of the slope are equal which simplifies the graphing method.
79.
PROBLEM: Can we graph a line knowing only its slope? ANSWER: Yes. If the coordinates of two points in a line are known then the line can be graphed.
80.
PROBLEM:
Research and discuss the alternative notation m =
Δy . Δx
ANSWER: Δy y2 − y1 m= = Δx x2 − x1
Δy shows the change in the y-coordinate or y value. Δx shows the change in the x-coordinate or the x value. 81.
PROBLEM: What strategies for graphing lines should be brought to an exam? Explain. ANSWER: A clear straight edge, a mechanical pencil (preferably with a lead of 0.5 mm thickness), and graph paper. The three methods to graph lines should be studied thoroughly. Mathematical calculations should be done accurately.
3.5 Finding Linear Equations Part A: Slope‐Intercept Form Determine the slope and y‐intercept
1.
PROBLEM: 5x − 3 y = 18 SOLUTION: 5x − 3 y = 18
5x − 5x − 3 y = 18 − 5 x 3 y = 5 x − 18 3 5 18 y = x− 3 3 3 5 y = x−6 3 ANSWER: The y-intercept is (0, –6) and the slope is m =
2.
5 . 3
PROBLEM: −6 x + 2 y = 12 SOLUTION: −6 x + 2 y = 12
−6 x + 6 x + 2 y = 12 + 6 x 2 y = 6 x + 12 2 6 12 y = x+ 2 2 2 y = 3x + 6 ANSWER: The y-intercept is (0, 6) and the slope is m = 3 .
3.
PROBLEM: x− y =5 SOLUTION: x− y =5
x + y − x = 5− x y = x −5 ANSWER: The y-intercept is (0, –5) and the slope is m = 1 .
4.
PROBLEM: −x + y = 0 SOLUTION: −x + y = 0
−x + x + y = 0 + x y=x ANSWER: The y-intercept is (0, 0) and the slope is m = 1 .
5.
PROBLEM: 4 x − 5 y = 15 SOLUTION: 4 x − 5 y = 15
4 x − 4 x − 5 y = 15 − 4 x 5 y = 4 x − 15 5 4 15 y = x− 5 5 5 4 y = x −3 5 ANSWER: The y-intercept is (0, –3) and the slope is m =
4 . 5
6.
PROBLEM: −7 x + 2 y = 3 SOLUTION: −7 x + 2 y = 3
−7 x + 7 x + 2 y = 3 + 7 x 2 y = 7x + 3 2 7 3 y = x+ 2 2 2 7 3 y = x+ 2 2 7 ⎛ 3⎞ ANSWER: The y-intercept is ⎜ 0, ⎟ and the slope is m = . 2 ⎝ 2⎠ 7.
PROBLEM: y =3 SOLUTION: y =3 ANSWER: The y-intercept is (0,3) and the slope is m = 0 .
8.
PROBLEM: 3 y=− 4 SOLUTION: 3 y=− 4
3⎞ ⎛ ANSWER: The y-intercept is ⎜ 0, − ⎟ and the slope is m = 0 . 4⎠ ⎝
9.
PROBLEM: 1 1 x − y = −1 5 3 SOLUTION: 1 1 x − y = −1 5 3 1 1 1 x − x − y = −1 − x 5 3 5 1 1 y = x +1 3 5 1 1 3 ⋅ y = 3 ⋅ x + 3 ⋅1 3 5 3 y = x+3 5 ANSWER: The y-intercept is (0, 3) and the slope is m =
10.
3 . 5
PROBLEM: 5 3 x+ y =9 16 8 SOLUTION: 5 3 x+ y =9 16 8 5 5 3 5 x− x+ y =9− x 16 16 8 16 3 5 y = − x+9 8 16 8 3 8 ⎛ 5 ⎞ 8 ⋅ y = ⋅⎜ − x⎟ + ⋅9 3 8 3 ⎝ 16 ⎠ 3 5 y = − x + 24 6
5 ANSWER: The y-intercept is (0, 24) and the slope is m = − . 6
11.
PROBLEM: 2 5 5 − x+ y = 3 2 4 SOLUTION: 2 5 5 − x+ y = 3 2 4 2 2 5 5 2 − x+ x+ y = + x 3 3 2 4 3 5 2 5 y = x+ 2 3 4 2 5 2 2 2 5 ⋅ y = ⋅ x+ ⋅ 5 2 5 3 5 4 4 1 y = x+ 15 2
4 ⎛ 1⎞ ANSWER: The y-intercept is ⎜ 0, ⎟ and the slope is m = . 15 ⎝ 2⎠ 12.
PROBLEM: 1 3 1 x− y = − 2 4 2 SOLUTION: 1 3 1 x− y = − 2 4 2 1 1 3 1 1 x− x− y =− − x 2 2 4 2 2 3 1 1 y = x+ 4 2 2 4 3 4 1 4 1 ⋅ y = ⋅ x+ ⋅ 3 4 3 2 3 2 2 2 y = x+ 3 3
2 ⎛ 2⎞ ANSWER: The y-intercept is ⎜ 0, ⎟ and the slope is m = . 3 ⎝ 3⎠
Part B: Finding Equations in Slope‐Intercept Form Given the slope and y‐intercept determine the equation of the line.
13.
PROBLEM: m = 1/2; (0, 5) SOLUTION: y = mx + b 1 b = 5; m = 2 ANSWER: y =
14.
1 x+5 2
PROBLEM: m = 4; (0, −1) SOLUTION: y = mx + b b = −1; m = 4 ANSWER: y = 4x – 1
15.
PROBLEM: m = −2/3; (0, −4) SOLUTION: y = mx + b
b=–4;m= −
2 3
2 ANSWER: y = − x – 4 3 16.
PROBLEM: m = −3; (0, 9) SOLUTION: y = mx + b b = 9; m = −3 ANSWER: y = −3x + 9
17.
PROBLEM: m = 0; (0, −1) SOLUTION: y = mx + b b=–1;m=0 y = (0)x – 1 ANSWER: y = –1
18.
PROBLEM: m = 5; (0, 0) SOLUTION: y = mx + b b = 0; m = 5 y = 5x + 0 ANSWER: y = 5x
Given the graph find the equation in slope‐intercept form.
19.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 3). Hence the b = 3.
rise 2 = − = −1 run 2 y = mx + b
m=
ANSWER: y = –x + 3
20.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, –2). Hence the b = –2.
rise 4 = run 3 y = mx + b
m=
ANSWER: y =
21.
4 x−2 3
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 2). Hence the b = 2.
rise 0 = =0 run 1 y = mx + b y = (0)x + 2
m=
ANSWER: y = 2
22.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 2). Hence the b = 2.
rise −4 = =1 run −4 y = mx + b m=
ANSWER: y = x + 2
23.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 0). Hence the b = 0.
rise 4 1 = = run 8 2 y = mx + b 1 y= x+0 2 1 ANSWER: y = x 2 m=
24.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 0). Hence the b = 0.
rise −2 2 = =− run 5 5 y = mx + b
m=
ANSWER: 2 y = − x+0 5 2 y=− x 5 Find the equation given the slope and a point.
25.
PROBLEM: m = 2/3; (−9, 2) SOLUTION: y = mx + b 2 y = x+b 3 Substituting the point (−9, 2) in the equation, 2 y = x+b 3 2 2 = ( −9 ) + b 3 2 = −6 + b 8=b Completing the equation y = mx + b,
y=
2 x+b 3
ANSWER: y =
26.
2 x +8 3
PROBLEM: m = −1/5; (5, −5) SOLUTION: y = mx + b 1 y = − x+b 5 Substituting the point (5, −5) in the equation, 1 y = − x+b 5 1 ( −5 ) = − ( 5 ) + b 5 −5 = −1 + b −4 = b Completing the equation y = mx + b, 1 y = − x+b 5 1 ANSWER: y = − x − 4 5
27.
PROBLEM: m = 0; (−4,3) SOLUTION: y = mx + b y = (0) x + b
y=b Substituting the point (−4,3) in the equation, y=b 3=b Completing the equation y = mx + b, y=b ANSWER: y = 3
28.
PROBLEM: m = 3; (−2,1) SOLUTION: y = mx + b y = 3x + b Substituting the point (−2, 1) in the equation, y = 3x + b 1 = 3 ( −2 ) + b 1 = −6 + b 7=b Completing the equation y = mx + b, y = 3x + b
ANSWER: y = 3 x + 7
29.
PROBLEM: m = −5; (−2, 8) SOLUTION: y = mx + b y = ( −5 ) x + b
y = −5x + b Substituting the point (−2, 8) in the equation, y = −5 x + b 8 = −5 ( −2 ) + b 8 = 10 + b –2 = b Completing the equation y = mx + b, ANSWER: y = –5x – 2
30.
PROBLEM: m = −4; (1/2, −3/2) SOLUTION: y = mx + b y = −4 x + b Substituting the point (1/2, −3/2) in the equation, y = −4 x + b 3 ⎛1⎞ = −4 ⎜ ⎟ + b 2 ⎝2⎠ 3 4 − = − +b 2 2 1 =b 2 Completing the equation y = mx + b, 1 ANSWER: y = −4 x + 2 −
31.
PROBLEM: m = −1/2; (3, 2) SOLUTION: y = mx + b 1 y = − x+b 2 Substituting the point (3, 2) in the equation, 1 y = − x+b 2 1 2 = − ( 3) + b 2 3 2 = − +b 2 7 =b 2 Completing the equation y = mx + b, 1 7 ANSWER: y = − x + 2 2
32.
PROBLEM: m = 3/4; (1/3, 5/4) SOLUTION: y = mx + b 3 y = x+b 4 Substituting the point (1/3, 5/4) in the equation, 3 y = x+b 4 5 3 1 = ⋅ +b 4 4 3 5 1 = +b 4 4 1= b Completing the equation y = mx + b, 3 ANSWER: y = x + 1 4
33.
PROBLEM: m = 0; (3,0) SOLUTION: y = mx + b
y = ( 0) x + b y =b Substituting the point (3,0) in the equation, y=b 0=b Completing the equation y = mx + b, ANSWER: y = 0
34.
PROBLEM: m undefined; (3,0) ANSWER: If the slope is undefined, the line is vertical line x = 3.
Given two points find the equation of the line.
35.
PROBLEM: (−6, 6), (2, 2) SOLUTION: y −y 2−6 −4 1 m= 2 1 = = =− x2 − x1 2 − ( −6 ) 8 2 y = mx + b 1 y = − x+b 2 Substitute the point (2, 2) in the equation, ⎛ 1⎞ y = ⎜− ⎟x +b ⎝ 3⎠ ⎛ 1⎞ 2 = ⎜− ⎟⋅2 + b ⎝ 2⎠ 3=b Completing the equation y = mx + b, 1 ANSWER: y = − x + 3 2
36.
PROBLEM: (−10, −3), (5, 0) SOLUTION: 0 − ( −3 ) y −y 3 1 m= 2 1 = = = x2 − x1 5 − ( −10 ) 15 5 y = mx + b 1 y = x+b 5 Substitute the point (5, 0) in the equation, 1 y = x+b 5 1 0 = ( 5) + b 5 −1 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 1 5
37.
PROBLEM: (0, 1/2), (1/2, −1) SOLUTION: 1 3 −1 − − y2 − y1 2 = 2 = −3 = m= 1 1 x2 − x1 −0 2 2 y = mx + b
y = −3 x + b Substitute the point (1/2, −1) in the equation, y = −3 x + b ⎛1⎞ −1 = −3 ⎜ ⎟ + b ⎝2⎠ 3 −1 = − + b 2 1 =b 2 Completing the equation y = mx + b, 1 ANSWER: y = −3 x + 2
38.
PROBLEM: (1/3, 1/3), (2/3, 1) SOLUTION: 1 2 y −y 3 = 3 =2 m= 2 1 = 2 1 1 x2 − x1 − 3 3 3 y = mx + b 1−
y = 2x + b Substitute the point (2/3, 1) in the equation, y = 2x + b ⎛2⎞ 1 = 2⎜ ⎟ + b ⎝3⎠ 4 1− = b 3 1 − =b 3 Completing the equation y = mx + b, 1 ANSWER: y = 2 x − 3
39.
PROBLEM: (3, −4), (−6, −7) SOLUTION: y − y −7 − ( − 4 ) −3 1 m= 2 1 = = = x2 − x1 −6 − 3 −9 3 y = mx + b 1 y = x+b 3 Substitute the point (−6, −7) in the equation, 1 y = x+b 3 1 −7 = ( −6 ) + b 3 −5 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 5 3
40.
PROBLEM: (−5, 2), (3, 2) SOLUTION: y −y 2−2 0 m= 2 1 = = =0 x2 − x1 3 − ( −5 ) 8 y = mx + b
y = ( 0) x + b y =b Substitute the point (3, 2) in the equation, y=b 2=b Completing the equation y = mx + b, ANSWER: y = 2
41.
PROBLEM: (−6, 4), (−6,−3) SOLUTION: y −y −3 − 4 −7 m= 2 1 = = = undefined x2 − x1 −6 − ( −6 ) 0 ANSWER: Since the slope is undefined, the line is a vertical line x = –6.
42.
PROBLEM: (−4, −4), (−1, −1) SOLUTION: y − y −1 − ( −4 ) 3 m= 2 1 = = =1 x2 − x1 −1 − ( −4 ) 3 y = mx + b y = x+b Substitute the point (−1, −1) in the equation, y = x+b − 1 = −1 + b 0=b Completing the equation y = mx + b,
ANSWER: y = x
43.
PROBLEM: (3, −3), (−5, 5) SOLUTION: y − y 5 − ( −3 ) 8 m= 2 1 = = = −1 x2 − x1 −5 − 3 −8 y = mx + b
y = −x + b Substitute the point (−5, 5) in the equation, y = −x + b
5 = − ( −5 ) + b 0=b Completing the equation y = mx + b, ANSWER: y = − x
44.
PROBLEM: (0, 8), (−4, 0) SOLUTION: y −y 0−8 m= 2 1 = =2 x2 − x1 −4 − 0 y = mx + b
y = 2x + b Substitute the point (0, 8) in the equation, y = 2x + b
8 = 2 ( 0) + b 8=b Completing the equation y = mx + b, ANSWER: y = 2 x + 8 Part C: Equations using Point‐Slope Form Find the equation given the slope and a point.
45.
PROBLEM: m = 1/2; (4, 3) SOLUTION: y = mx + b 1 y = x+b 2 Substituting the point (4, 3) in the equation, 1 y = x+b 2 1 3 = ( 4) + b 2 1= b Completing the equation y = mx + b, 1 ANSWER: y = x + 1 2
46.
PROBLEM: m = −1/3; (9, −2) SOLUTION: y = mx + b
1 y = − x+b 3 Substituting the point (9, −2) in the equation, 1 y = − x+b 3 1 −2 = − ( 9 ) + b 3 1= b Completing the equation y = mx + b, 1 ANSWER: y = − x + 1 3 47.
PROBLEM: m = 6; (1, −5) SOLUTION: y = mx + b
y = 6x + b Substituting the point (1, −5) in the equation, y = 6x + b
−5 = 6 (1) + b −11 = b Completing the equation y = mx + b, ANSWER: y = 6 x − 11
48.
PROBLEM: m = −10; (1, −20) SOLUTION: y = mx + b y = −10 x + b Substituting the point (1, −20) in the equation, −20 = −10 (1) + b
−10 = b Completing the equation y = mx + b, ANSWER: y = −10 x − 10
49.
PROBLEM: m = −3; (2, 3) SOLUTION: y = mx + b y = −3 x + b Substituting the point (2, 3) in the equation, y = −3x + b
3 = −3 ( 2 ) + b 9=b Completing the equation y = mx + b, ANSWER: y = −3 x + 9
50.
PROBLEM: m = 2/3; (−3, −5) SOLUTION: y = mx + b 2 y = x+b 3 Substituting the point (−3, −5) in the equation, 2 y = x+b 3 2 −5 = ( −3) + b 3 −3 = b Completing the equation y = mx + b, 2 ANSWER: y = x − 3 3
51.
PROBLEM: m = −3/4; (−8,3) SOLUTION: y = mx + b 3 y = − x+b 4 Substituting the point (−8,3) in the equation, 3 y = − x+b 4 3 3 = − ( −8 ) + b 4 −3 = b Completing the equation y = mx + b, 3 ANSWER: y = − x − 3 4
52.
PROBLEM: m = 5; (1/5, −3) SOLUTION: y = mx + b
y = 5x + b Substituting the point (1/5, −3) in the equation, y = 5x + b
⎛1⎞ −3 = 5 ⎜ ⎟ + b ⎝5⎠ −4 = b Completing the equation y = mx + b, ANSWER: y = 5 x − 4
53.
PROBLEM: m = −3; (−1/9, 2) SOLUTION: y = mx + b
y = −3 x + b Substituting the point (−1/9, 2) in the equation, y = −3x + b ⎛ 1⎞ 2 = −3 ⎜ − ⎟ + b ⎝ 9⎠ 5 =b 3 Completing the equation y = mx + b, 5 ANSWER: y = −3 x + 3 54.
PROBLEM: m = 0; (4, −6) SOLUTION: y = mx + b
y = ( 0) x + b y =b Substituting the point (4, −6) in the equation, y=b −6 = b Completing the equation y = mx + b, ANSWER: y = −6
55.
PROBLEM: m = 0; (−5, 10) SOLUTION: y = mx + b
y = ( 0) x + b y =b Substituting the point (−5, 10) in the equation, y=b 10 = b Completing the equation y = mx + b, ANSWER: y = 10
56.
PROBLEM: m = 5/8; (4,3) SOLUTION: y = mx + b 5 y = x+b 8 Substituting the point (4,3) in the equation, 5 y = x+b 8 5 3 = ( 4) + b 8 1 =b 2 Completing the equation y = mx + b, 5 1 ANSWER: y = x + 8 2
57.
PROBLEM: m = −3/5; (−2, −1) SOLUTION: y = mx + b 3 y = − x+b 5 Substituting the point (−2, −1) in the equation, 3 y = − x+b 5 ⎛ 3⎞ −1 = ⎜ − ⎟ ( −2 ) + b ⎝ 5⎠ 11 − =b 5 Completing the equation y = mx + b, 3 11 ANSWER: y = − x − 5 5
58.
PROBLEM: m = 1/4; (12, −2) SOLUTION: y = mx + b 1 y = x+b 4 Substituting the point (12, –2) in the equation, 1 y = x+b 4 1 −2 = (12 ) + b 4 −5 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 5 4
59.
PROBLEM: m = 1; (0, 0) SOLUTION: y = mx + b
y = x+b Substituting the point (0, 0) in the equation, y = x+b 0 = 0+b 0=b Completing the equation y = mx + b, ANSWER: y = x
60.
PROBLEM: m = −3/4; (0, 0) SOLUTION: y = mx + b 3 y = − x+b 4 Substituting the point (0, 0) in the equation, 3 y = − x+b 4 3 0 = − ( 0) + b 4 0=b Completing the equation y = mx + b, 3 ANSWER: y = − x 4
Use the point‐slope formula to find the equation given the Graph.
61.
PROBLEM:
SOLUTION: The two points seen in the graph are (2, 1) and (1, 3) y − y 3 −1 2 m= 2 1 = = − = −2 x2 − x1 1 − 2 1 y = mx + b y = −2 x + b Substitute the point (2, 1) in the equation, y = −2 x + b
1 = −2 ( 2 ) + b 5=b Completing the equation y = mx + b, ANSWER: y = −2 x + 5
62.
PROBLEM:
SOLUTION: The two points seen in the graph are (–1, –2) and (4, 1) y − y 1 − ( −2 ) 3 = m= 2 1 = x2 − x1 4 − ( −1) 5 y = mx + b 3 y = x+b 5 Substitute the point (4, 1) in the equation, 3 y = x+b 5 3 1 = ( 4) + b 5 7 − =b 5 Completing the equation y = mx + b, 3 7 ANSWER: y = x − 5 5
63.
PROBLEM:
SOLUTION: The two points seen in the graph are (–7, 1) and (–1, 5) y −y 5 −1 4 2 m= 2 1 = = = x2 − x1 −1 − ( −7 ) 6 3 y = mx + b 2 y = x+b 3 Substitute the point (–1, 5) in the equation, 2 y = x+b 3 2 5 = ( −1) + b 3 17 =b 3 Completing the equation y = mx + b, 2 17 ANSWER: y = x + 3 3
64.
PROBLEM:
SOLUTION: The two points seen in the graph are (–6, 5) and (–1, 5) y −y 5−5 0 m= 2 1 = = =0 x2 − x1 1 − ( −6 ) 7 y = mx + b
y = ( 0) x + b y =b Substitute the point (–1, 5) in the equation, y=b 5=b Completing the equation y = mx + b, ANSWER: y = 5
65.
PROBLEM:
SOLUTION: The two points seen in the graph are (1, –1) and (–9, 5) y − y 5 − ( −1) 6 3 m= 2 1 = =− =− x2 − x1 −9 − 1 10 5 y = mx + b 3 y = − x+b 5 Substitute the point (1, –1) in the equation, 3 y = − x+b 5 3 −1 = − (1) + b 5 2 − =b 5 Completing the equation y = mx + b, 3 2 ANSWER: y = − x − 5 5
66.
PROBLEM:
SOLUTION: The two points seen in the graph are (–7, 3) and (2, 2) y −y 2−3 1 m= 2 1 = =− x2 − x1 2 − ( −7 ) 9 y = mx + b 1 y = − x+b 9 Substitute the point (2, 2) in the equation, 1 y = − x+b 9 1 2 = − ( 2) + b 9 20 =b 9 Completing the equation y = mx + b, 1 20 ANSWER: y = − x + 9 9
Use the point‐slope formula to find the equation passing through the two points.
67.
PROBLEM: (−4, 0), (0, 5) SOLUTION: y −y 5−0 5 m= 2 1 = = x2 − x1 0 − ( −4 ) 4 y = mx + b 5 y = x+b 4 Substitute the point (0, 5) in the equation, 5 y = x+b 4 5 5 = ( 0) + b 4 5=b Completing the equation y = mx + b, 5 ANSWER: y = x + 5 4
68.
PROBLEM: (−1, 2), (0, 3) SOLUTION: y −y 3− 2 1 m= 2 1 = = =1 x2 − x1 0 − ( −1) 1 y = mx + b
y = x+b Substitute the point (0, 3) in the equation, y = x+b 3= 0+b 3=b Completing the equation y = mx + b, ANSWER: y = x + 3
69.
PROBLEM: (−3, −2), (3, 2) SOLUTION: y − y 2 − ( −2 ) 4 2 = = m= 2 1 = x2 − x1 3 − ( −3) 6 3 y = mx + b 2 y = x+b 3 Substitute the point (3, 2) in the equation, 2 y = x+b 3 2 2 = ( 3) + b 3 0=b Completing the equation y = mx + b, 2 ANSWER: y = x 3
70.
PROBLEM: (3, −1), (2, −3) SOLUTION: y − y −3 − ( −1) −2 m= 2 1 = = =2 x2 − x1 2−3 −1 y = mx + b
y = 2x + b Substitute the point (2, −3) in the equation, y = 2x + b −3 = 2 ( 2 ) + b −7 = b Completing the equation y = mx + b, ANSWER: y = 2 x − 7
71.
PROBLEM: (−2, 4), (2, −4) SOLUTION: y −y −4 − 4 −8 m= 2 1 = = = −2 x2 − x1 2 − ( −2 ) 4 y = mx + b
y = −2 x + b Substitute the point (2, −4) in the equation, y = −2 x + b − 4 = −2 ( 2 ) + b 0=b Completing the equation y = mx + b, ANSWER: y = −2 x 72.
PROBLEM: (−5, −2), (5, 2) SOLUTION: y − y 2 − ( −2 ) 4 2 = = m= 2 1 = x2 − x1 5 − ( −5 ) 10 5 y = mx + b 2 y = x+b 5 Substitute the point (5, 2) in the equation, 4 y = x+b 5 2 2 = ( 5) + b 5 0=b Completing the equation y = mx + b, 2 ANSWER: y = x 5
73.
PROBLEM: (−3, −1), (3, 3) SOLUTION: y − y 3 − ( −1) 4 2 = = m= 2 1 = x2 − x1 3 − ( −3) 6 3 y = mx + b 2 y = x+b 3 Substitute the point (3, 3) in the equation, 2 y = x+b 3 2 3 = ( 3) + b 3 1= b Completing the equation y = mx + b, 2 ANSWER: y = x + 1 3
74.
PROBLEM: (1, 5), (0, 5) SOLUTION: y − y 5−5 0 =0 m= 2 1 = = x2 − x1 0 − 1 −1 y = mx + b
y = ( 0) x + b y =b Substitute the point (0, 5) in the equation, y=b 5=b Completing the equation y = mx + b, ANSWER: y = 5
75.
PROBLEM: (1, 2), (2, 4) SOLUTION: y − y 4−2 2 m= 2 1 = = =2 x2 − x1 2 − 1 1 y = mx + b
y = 2x + b Substitute the point (2, 4) in the equation, y = 2x + b 4 = 2 ( 2) + b 0=b Completing the equation y = mx + b, ANSWER: y = 2 x
76.
PROBLEM: (6, 3), (2, −3) SOLUTION: y − y −3 − 3 −6 3 m= 2 1 = = = x2 − x1 2 − 6 −4 2 y = mx + b 3 y = x+b 2 Substitute the point (2, −3) in the equation, 3 y = x+b 2 3 −3 = ( 2 ) + b 2 −6 = b Completing the equation y = mx + b, 3 ANSWER: y = x − 6 2
77.
PROBLEM: (10, −3), (5, −4) SOLUTION: y − y −4 − ( −3) −1 1 m= 2 1 = = = x2 − x1 5 − 10 −5 5 y = mx + b 1 y = x+b 5 Substitute the point (5, −4) in the equation, 1 −4 = ( 5 ) + b 5 −5 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 5 5
78.
PROBLEM: (−3, 3), (−1,12) SOLUTION: y −y 12 − 3 9 m= 2 1 = = x2 − x1 −1 − ( −3) 2 y = mx + b 9 y = x+b 2 Substitute the point (−1,12) in the equation, 9 y = x+b 2 9 12 = ( −1) + b 2 33 =b 2 Completing the equation y = mx + b, 9 33 ANSWER: y = x + 2 2
79.
PROBLEM: (4/5, −1/3), (−1/5, 2/3) SOLUTION: 2 ⎛ 1⎞ 3 −⎜− ⎟ y −y 3 ⎝ 3⎠ m= 2 1 = = 3 = −1 1 4 5 x2 − x1 − − − 5 5 5 y = mx + b
y = −x + b Substitute the point (−1/5, 2/3) in the equation, y = −x + b 2 ⎛ 1⎞ = −⎜− ⎟ + b 3 ⎝ 5⎠ 7 =b 15 Completing the equation y = mx + b, 7 ANSWER: y = − x + 15 80.
PROBLEM: (5/3, 1/3), (−10/3, −5/3) SOLUTION: 5 1 6 − − − y2 − y1 6 2 m= = 3 3 = 3 = = x2 − x1 − 10 − 5 − 15 15 5 3 3 3 y = mx + b
2 x+b 5 Substitute the point (−10/3, −5/3) in the equation, 2 y = x+b 5 5 2 ⎛ 10 ⎞ − = ⎜− ⎟+b 3 5⎝ 3 ⎠ 1 − =b 3 Completing the equation y = mx + b, 2 1 ANSWER: y = x − 5 3 y=
81.
PROBLEM: (3, −1/4), (4, −1/2) SOLUTION: 1 ⎛ 1⎞ − −⎜− ⎟ − 1 y −y 1 2 ⎝ 4⎠ m= 2 1 = = 4 =− x2 − x1 4−3 1 4 y = mx + b
1 y = − x+b 4 Substitute the point (4, −1/2) in the equation, 1 y = − x+b 4 1 1 − = − ( 4) + b 2 4 1 =b 2 Completing the equation y = mx + b, 1 1 ANSWER: y = − x + 4 2 82.
PROBLEM: (0, 0), (−5, 1) SOLUTION: y −y 1− 0 1 =− m= 2 1 = x2 − x1 −5 − 0 5 y = mx + b
1 y = − x+b 5 Substitute the point (−5, 1) in the equation, 1 y = − x+b 5 1 1 = − ( −5 ) + b 5 0=b Completing the equation y = mx + b, 1 ANSWER: y = − x 5
83.
PROBLEM: (2, −4), (0, 0) SOLUTION: y − y 0 − ( −4 ) 4 m= 2 1 = = = −2 x2 − x1 0−2 −2 y = mx + b y = −2 x + b Substitute the point (0, 0) in the equation, y = −2 x + b
0 = −2 ( 0 ) + b 0=b Completing the equation y = mx + b, ANSWER: y = −2 x
84.
PROBLEM: (3, 5), (3, −2) SOLUTION: y − y − 2 − 5 −7 m= 2 1 = = = undefined x2 − x1 3−3 0 ANSWER: Since slope is not defined, the line is a vertical line x = 3.
85.
PROBLEM: (−4, 7), (−1, 7) SOLUTION: y −y 7−7 0 = =0 m= 2 1 = x2 − x1 −1 − ( −4 ) 3 y = mx + b
y = ( 0) x + b y =b Substitute the point (−1, 7) in the equation, y=b 7=b Completing the equation y = mx + b, ANSWER: y = 7
86.
PROBLEM: (−8, 0), (6, 0) SOLUTION: y −y 0−0 0 m= 2 1 = = =0 x2 − x1 6 − ( −8 ) 14 y = mx + b
y = ( 0) x + b y =b Substitute the point (6, 0) in the equation, y=b 0=b Completing the equation y = mx + b, ANSWER: y = 0 Part D: Applications
87.
PROBLEM: Joe has been keeping track of his cellular phone bills for the last two months. The bill for the first month was $38 for 100 minutes of usage and the bill for the second month was $45.50 for 150 minutes of usage. Find a linear equation that gives the total monthly bill based on the minutes of usage. SOLUTION: Bill amount is plotted on the y-axis; minutes of usage on the x-axis. Bill for the first month is $38 for 100 minutes of usage: (100, 38) Bill for the first month is $45.50 for 150 minutes of usage: (150, 45.50) y − y 45.50 − 38 3 Slope, m = 2 1 = = x2 − x1 150 − 100 20 y = mx + b 3 x+b y= 20 Substitute the point (100, 38) in the equation, 3 y= x+b 20 3 38 = (100 ) + b 20 23 = b Completing the equation y = mx + b,
3 x + 23 20 y = 0.15 x + 23 y=
ANSWER: y = 0.15 x + 23
88.
PROBLEM: A company in its first year of business produced 150 training manuals for a total cost of $2,350. The following year the company produced 50 more manuals at a cost of $1,450. Use this information to find a linear equation that gives the total cost of producing training manuals. SOLUTION: Cost of the training manuals produced is plotted on the y-axis; number of training manuals is plotted on the x-axis. 150 training manuals for a total cost of $2,350: (150, 2350) 50 training manuals for a total cost of $1,450: (50, 1450) y − y 1450 − 2350 Slope, m = 2 1 = =9 x2 − x1 50 − 150 y = mx + b y = 9x + b Substitute the point (50, 1450) in the equation, y = 9x + b
1450 = 9 ( 50 ) + b 1000 = b Completing the equation y = mx + b, ANSWER: y = 9 x + 1000
89.
PROBLEM: A corn farmer in California was able to produce 154 bushels of corn per acre 2 years after starting his operation. Currently, after 7 years of operation, he has increased his yield to 164 bushels per acre. Use this information to write a linear equation that gives the total yield per acre based on number of years of operation and use it to predict the yield for next year. SOLUTION: Number of bushels produced in a year is plotted on the y-axis; number of years of operation is plotted on the x-axis. 154 bushels of corn per acre for 3 years: (2, 154) 164 bushels of corn per acre for 7 years: (7, 164) y − y 164 − 154 10 Slope, m = 2 1 = = =2 x2 − x1 7−2 5
y = mx + b y = 2x + b Substitute the point (7, 164) in the equation, y = 2x + b 164 = 2 ( 7 ) + b 150 = b Completing the equation y = mx + b, y = 2 x + 150 To find the yield for the next year; put x = 8 in the above equation. y = 2 x + 150 y = 2 ( 8 ) + 150 y = 166 ANSWER: y = 2 x + 150 . Yield for the 8th year is 166 bushels.
90.
PROBLEM: A webmaster has noticed that the number of registered users has been steadily increasing since beginning an advertising campaign. Before starting to advertise he had 1,200 registered users and after 3 months of advertising he now has 1,590 registered users. Use this data to write a linear equation that gives the total number of registered users given the number of months after starting to advertise. Use the equation to predict the number of users 7 months into the advertising campaign. SOLUTION: Number of registered users is plotted on the y-axis; number of months of advertising is plotted on the x-axis. Number of registered users before advertising 1,200: (0, 1200) Number of registered users after 3 months of advertising: (3, 1590) y − y 1590 − 1200 Slope, m = 2 1 = = 130 x2 − x1 3−0 y = mx + b y = 130 x + b Substitute the point (0, 1200) in the equation, y = 130 x + b
1200 = 130 ( 0 ) + b 1200 = b Completing the equation y = mx + b, y = 130 x + 1200
To find number of registered users after 7 months of advertising, put x = 7 in the equation above. y = 130x + 1200 y = 130 ( 7 ) + 1200 y = 2110 ANSWER: Number of registered users after 7 months of advertising is 2,110.
91.
PROBLEM: A car purchased new cost $22,000 and was sold 10 years later for $7,000. Write a linear equation that gives the value of the car in terms of its age in years. SOLUTION: Value of the car is plotted on the y-axis; number of years is plotted on the x-axis. Initial cost of the car is $22,000: (0, 22000) Cost of the car after 10 years is $7,000: (10, 7000) y − y 7000 − 22000 −15000 Slope, m = 2 1 = = = −1500 x2 − x1 10 − 0 10 y = mx + b y = 1500 x + b Substitute the point (0, 22000) in the equation, y = −1500 x + b
22000 = −1500 ( 0 ) + b 22000 = b Completing the equation y = mx + b, ANSWER: y = −1500 x + 22000 92.
PROBLEM: An antique clock was purchased in 1985 for $1,500 and sold at auction in 1997 for $5,700. Determine a linear equation that models the value of the clock in terms of years since 1985. SOLUTION: Value of the clock is plotted on the y-axis; years are plotted on the x-axis. Assume the year 1985 represents the origin. Value of the clock in the year 1985 is $1,500: (0, 1500) Value of the clock in the year 1997 is $5,700: (12, 5700) y − y 5700 − 1500 4200 Slope, m = 2 1 = = = 350 x2 − x1 12 − 0 12 y = mx + b y = 350 x + b Substitute the point (0, 1500) in the equation,
y = 350 x + b 1500 = 350 ( 0 ) + b 1500 = b Completing the equation y = mx + b, ANSWER: y = 350 x + 1500 Part E: Discussion Board Topics
93.
PROBLEM: Discuss the merits and drawbacks of point-slope form and y-intercept form. ANSWER: Merits of the point-slope form The beauty of the slope formula is that we can obtain the slope, given two points, using only algebra. Drawbacks of the point-slope formula When using the slope formula, take care to be consistent since order does matter. We must subtract the coordinates of the first point from the second point for both the numerator and denominator in the same order. When considering the slope as a rate of change it is important to include the correct units. Merits of the y-intercept form Identifies y intercept. Shows understanding of gradient or rate of change. Drawbacks of the y-intercept form Rule must be in y = mx + b form. Does not find x intercept (Source: henchmaths.wikispaces.com/.../Sketching+Linear+Graphs+4+ways.ppt)
94.
PROBLEM: Research and discuss linear depreciation. In a linear depreciation model, what does the slope and y-intercept represent? ANSWER: The simplest and most commonly used depreciation method, linear depreciation is calculated by taking the purchase or acquisition price of an asset subtracted by the salvage value divided by the total productive years the asset can be reasonably expected to benefit the company (called “useful life” in accounting jargon).
Purchase cost = (Depreciation expense/year)(Number of years) + Salvage value Slope represents the depreciation expense/year. The y-intercept represents the salvage value of the asset. (Source: http://beginnersinvest.about.com/od/incomestatementanalysis/a/straightline-depreciation.htm)
3.6 Parallel and Perpendicular Lines Part A: Parallel and Perpendicular Lines Determine the slope of a parallel and perpendicular line.
1.
PROBLEM: 3 y = − x +8 4 SOLUTION: 3 y = − x +8 4
2.
3 The slope of the line is m = − . 4 ANSWER: Slope of the parallel line m|| = m 3 m|| = − 4 1 1 4 Slope of the perpendicular line m⊥ = − = − = 3 m 3 − 4 PROBLEM: 1 y = x −3 2 SOLUTION: 1 y = x −3 2
1 . 2 ANSWER: Slope of the parallel line m|| = m 1 m|| = 2 1 1 Slope of the perpendicular line m⊥ = − = − = −2 1 m 2
The slope of the line is m =
3.
PROBLEM: y = 4x + 4 SOLUTION: y = 4x + 4
The slope of the line is m = 4 . ANSWER: Slope of the parallel line m|| = m m|| = 4 1 1 Slope of the perpendicular line m⊥ = − = − m 4 4.
PROBLEM: y = −3x + 7 SOLUTION: y = −3x + 7
The slope of the line is m = −3 . ANSWER: Slope of the parallel line m|| = m m|| = −3 1 1 1 Slope of the perpendicular line m⊥ = − = − = m −3 3
5.
PROBLEM: 5 1 y =− x− 8 2 SOLUTION: 5 1 y =− x− 8 2
5 The slope of the line is m = − . 8 ANSWER: Slope of the parallel line m|| = m 5 m|| = − 8 1 1 8 Slope of the perpendicular line m⊥ = − = − = 5 m 5 − 8
6.
PROBLEM: 7 3 y = x+ 3 2 SOLUTION: 7 3 y = x+ 3 2
7 . 3 ANSWER: Slope of the parallel line m|| = m 7 m|| = 3 1 1 3 Slope of the perpendicular line m⊥ = − = − = − 7 m 7 3 The slope of the line is m =
7.
PROBLEM: 2 y = 9x − 5 SOLUTION: 2 y = 9x − 5 The slope of the line is m = 9 . ANSWER: Slope of the parallel line m|| = m m|| = 9 1 1 Slope of the perpendicular line m⊥ = − = − m 9
8.
PROBLEM: 1 y = −10 x + 5 SOLUTION: 1 y = −10 x + 5 The slope of the line is m = −10 . ANSWER: Slope of the parallel line m|| = m m|| = −10 1 1 1 Slope of the perpendicular line m⊥ = − = − = m −10 10
9.
PROBLEM: y =5 SOLUTION: y =5
The slope of the line is m = 0 . ANSWER: Slope of the parallel line m|| = m m|| = 0 1 1 Slope of the perpendicular line m⊥ = − = − = undefined m 0 10.
PROBLEM: 1 x=− 2 SOLUTION: 1 x=− 2 The slope of the line is m = undefined ANSWER: Slope of the parallel line m|| = m m|| = undefined
Slope of the perpendicular line m⊥ = −
1 1 =− =0 m undefined
11.
PROBLEM: x− y =0 SOLUTION: x− y =0
x − x − y = −x y=x The slope of the line is m =1. ANSWER: Slope of the parallel line m|| = m m|| = 1 1 1 Slope of the perpendicular line m⊥ = − = − = −1 m 1 12.
PROBLEM: x+ y =0 SOLUTION: x+ y =0
x − x + y = −x y = −x The slope of the line is m = −1 ANSWER: Slope of the parallel line m|| = m m|| = −1 1 1 Slope of the perpendicular line m⊥ = − = − = 1 m −1
13.
PROBLEM: 4x + 3y = 0 SOLUTION: 4x + 3y = 0
4 x − 4 x + 3 y = −4 x 3 y = −4 x 3 4 y=− x 3 3 4 y=− x 3
4 The slope of the line is m = − . 3 ANSWER: Slope of the parallel line m|| = m 4 m|| = − 3 1 1 3 = Slope of the perpendicular line m⊥ = − = − 4 4 m − 3 14.
PROBLEM: 3x − 5 y = 10 SOLUTION: 3x − 5 y = 10
3x − 3x − 5 y = 10 − 3x 5 y = 3x − 10 5 3 10 y = x− 5 5 5 3 y = x−2 5 3 5 ANSWER: Slope of the parallel line m|| = m 3 m|| = 5 1 1 5 Slope of the perpendicular line m⊥ = − = − = − 3 m 3 5 The slope of the line is m =
15.
PROBLEM: −2 x + 7 y = 14 SOLUTION: −2 x + 7 y = 14 −2 x + 2 x + 7 y = 14 + 2 x
7 y = 2 x + 14 7 2 14 y = x+ 7 7 7 2 y = x+2 7 2 The slope of the line is m = . 7 ANSWER: Slope of the parallel line m|| = m 2 m|| = 7 1 1 7 Slope of the perpendicular line m⊥ = − = − = − 2 m 2 7 16.
PROBLEM: − x − y = 15 SOLUTION: − x − y = 15
− x + x − y = 15 + x y = − x − 15 The slope of the line is m = −1 ANSWER: Slope of the parallel line m|| = m m|| = −1 1 1 Slope of the perpendicular line m⊥ = − = − = 1 m −1
17.
PROBLEM: 1 1 x − y = −1 2 3 SOLUTION: 1 1 x − y = −1 2 3 1 1 1 1 x − x − y = −1 − x 2 2 3 2 1 1 y = x +1 3 2 1 1 3 ⋅ y = 3 ⋅ x + 3 ⋅1 3 2 3 y = x+3 2
3 . 2 ANSWER: Slope of the parallel line m|| = m 3 m|| = 2 1 1 2 Slope of the perpendicular line m⊥ = − = − = − 3 m 3 2 The slope of the line is m =
18.
PROBLEM: 2 4 − x+ y =8 3 5 SOLUTION: 2 4 − x+ y =8 3 5 2 2 4 2 − x + x + y = 8+ x 3 3 5 3 4 2 y = x +8 5 3 5 4 5 2 5 ⋅ y = ⋅ x + ⋅8 4 5 4 3 4 5 y = x + 10 6
5 6 ANSWER: Slope of the parallel line m|| = m 5 m|| = 6 1 1 6 Slope of the perpendicular line m⊥ = − = − = − 5 m 5 6 The slope of the line is m =
19.
PROBLEM: 1 1 2x − y = 5 10 SOLUTION: 1 1 2x − y = 5 10 1 1 2x − 2x − y = − 2x 5 10 1 1 y = 2x − 5 10 1 1 5 ⋅ y = 5 ⋅ 2x − 5 ⋅ 5 10 1 y = 10 x − 2 The slope of the line is m = 10 . ANSWER: Slope of the parallel line m|| = m m|| = 10 1 1 Slope of the perpendicular line m⊥ = − = − m 10
20.
PROBLEM: 4 − x − 2y = 7 5 SOLUTION: 4 − x − 2y = 7 5 4 4 4 − x + x − 2y = 7 + x 5 5 5 4 2y = − x − 7 5 2 4 7 y=− x− 2 5⋅ 2 2 2 7 y =− x− 5 2
2 5 ANSWER: Slope of the parallel line m|| = m 2 m|| = − 5 1 1 5 Slope of the perpendicular line m⊥ = − = − = 2 2 m − 5 The slope of the line is m = −
Determine if the lines are parallel, perpendicular, or neither.
21.
PROBLEM: 2 ⎧ ⎪⎪ y = 3 x + 3 ⎨ ⎪y = 2 x −3 ⎪⎩ 3 SOLUTION: 2 y = x+3 3 2 m= 3
2 x −3 3 2 m= 3 y=
ANSWER: Slopes of the two lines are equal. Hence lines are parallel to each other.
22.
PROBLEM: 3 ⎧ ⎪⎪ y = 4 x − 1 ⎨ ⎪y = 4 x +3 ⎪⎩ 3 SOLUTION: 3 y = x −1 4 3 m= 4
4 x+3 3 4 m= 3 y=
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
23.
PROBLEM: ⎧ y = −2 x + 1 ⎪ ⎨ 1 ⎪⎩ y = 2 x + 8 SOLUTION:
1 x +8 y = −2 x + 1 2 1 m = −2 m= 2 ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other. y=
24.
PROBLEM: 1 ⎧ ⎪ y = 3x − 2 ⎨ ⎪⎩ y = 3x + 2 SOLUTION: 1 y = 3x − 2 m=3
y = 3x + 2 m=3
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
25.
PROBLEM: ⎧ y =5 ⎨ ⎩ x = −2 SOLUTION: x = −2 y=5 m = undefined m=0 ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
26.
PROBLEM: ⎧y = 7 ⎪ 1 ⎨ ⎪⎩ y = − 7 SOLUTION: y=7 m=0
y=−
1 7
m=0 ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other. 27.
PROBLEM: ⎧3x − 5 y = 15 ⎨ ⎩ 5x + 3 y = 9 SOLUTION: 3x − 5 y = 15 3x − 3 x − 5 y = 15 − 3 x 5 y = 3 x − 15 5 3 15 y = x− 5 5 5 3 y = x−3 5 3 m= 5
5x + 3 y = 9 5x − 5x + 3 y = 9 − 5x 3 y = −5 x + 9 3 5 9 y = − x+ 3 3 3 5 y = − x+3 3 5 m=− 3
ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
28.
PROBLEM: ⎧ x− y =7 ⎨ ⎩3x + 3 y = 2 SOLUTION:
3x + 3 y = 2 3x − 3x + 3 y = 2 − 3x x− y =7 3 y = −3 x + 2 x−x− y =7−x 3 3 2 y = − x+ y = x−7 3 3 3 m =1 2 y = −x + 3 m = −1 ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other. 29.
PROBLEM: ⎧ 2x − 6 y = 4 ⎨ ⎩ − x + 3 y = −2 SOLUTION: 2x − 6 y = 4 2x − 2x − 6 y = 4 − 2x 6 y = 2x − 4 6 2 4 y = x− 6 6 6 1 2 y = x− 3 3 1 m= 3
− x + 3 y = −2 − x + x + 3 y = −2 + x 3y = x − 2 3 1 2 y = x− 3 3 3 1 2 y = x− 3 3 1 m= 3
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
30.
PROBLEM: ⎧−4 x + 2 y = 3 ⎨ ⎩ 6 x − 3 y = −3 SOLUTION: −4 x + 2 y = 3 −4 x + 4 x + 2 y = 3 + 4 x 2 y = 3 + 4x 2 4 3 y = x+ 2 2 2 3 y = 2x + 2 m=2
6 x − 3 y = −3 6 x − 6 x − 3 y = −3 − 6 x 3y = 6 y + 3 3 6 3 y = y+ 3 3 3 y = 2 y +1 m=2
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
31.
PROBLEM: ⎧ x + 3y = 9 ⎨ ⎩2 x + 3 y = 6 SOLUTION: x + 3y = 9 x − x + 3y = 9 − x 3y = −x + 9 3 1 9 y = − x+ 3 3 3 1 y = − x+3 3 1 m=− 3
2x + 3y = 6 2x − 2x + 3y = 6 − 2x 3 y = −2 x + 6 3 2 6 y =− x+ 3 3 3 2 y =− x+2 3 2 m=− 3
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
32.
PROBLEM: ⎧ y − 10 = 0 ⎨ ⎩ x − 10 = 0 SOLUTION: y − 10 = 0 m=0
x − 10 = 0 m = undefined
ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
33.
PROBLEM: ⎧ y+2=0 ⎨ ⎩2 y − 10 = 0 SOLUTION: y+2=0 m=0
2 y − 10 = 0 m=0
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
34.
PROBLEM: ⎧3x + 2 y = 6 ⎨ ⎩2 x + 3 y = 6 SOLUTION: 3x + 2 y = 6 3x − 3x + 2 y = 6 − 3x 2 y = −3x + 6 2 3 6 y =− x+ 2 2 2 3 y = − x+3 2 3 m=− 2
2x + 3y = 6 2x − 2x + 3y = 6 − 2x 3 y = −2 x + 6 3 2 6 y =− x+ 3 3 3 2 y =− x+2 3 2 m=− 3
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
35.
PROBLEM: ⎧−5 x + 4 y = 20 ⎨ ⎩ 10 x − 8 y = 16 SOLUTION: −5 x + 4 y = 20 −5 x + 5 x + 4 y = 20 + 5 x 4 y = 5 x + 20 4 5 20 y = x+ 4 4 4 5 y = x+5 4 5 m= 4
10 x − 8 y = 16 10 x − 10 x − 8 y = 16 − 10 x 8 y = 10 x − 16 8 10 16 y = x− 8 8 8 5 y = x−2 4 5 m= 4
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
36.
PROBLEM: 1 ⎧ 1 ⎪⎪ 2 x − 3 y = 1 ⎨ ⎪ 1 x + 1 y = −2 ⎪⎩ 6 4 SOLUTION: 1 1 x − y =1 2 3 1 1 1 1 x − x − y = 1− x 2 2 3 2 1 1 y = x −1 3 2 1 1 3 ⋅ y = 3 ⋅ x − 3 ⋅1 3 2 3 y = x −3 2 3 m= 2
1 1 x + y = −2 6 4 1 1 1 1 x − x + y = −2 − x 6 6 4 6 1 1 y = − x−2 4 6 1 1 4 ⋅ y = −4 ⋅ x − 4 ⋅ 2 4 6 2 y = − x −8 3 2 m=− 3
ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
Part B: Equations in Point‐Slope Form Find the equation of the line.
37.
PROBLEM:
Parallel to y =
1 x + 2 and passing through (6, −1). 2
1 1 and thus m& = . We substitute 2 2 this slope and the given point into point-slope form. Point Slope
Solution: Here the given line has slope m =
(6, −1)
m& =
y − y1 = m ( x − x1 ) y − ( −1) = y +1 = y=
1 ( x − 6) 2
1 x −3 2
1 x−4 2
ANSWER: y =
38.
1 2
1 x−4 2
PROBLEM:
3 Parallel to y = − x − 3 and passing through (−8, 2). 4 3 3 SOLUTION: Here the given line has slope m = − and thus m& = − . We 4 4 substitute this slope and the given point into point-slope form. Point
Slope
(−8, 2)
m& = −
y − y1 = m ( x − x1 ) 3 ( x − ( −8) ) 4 3 y−2 = − x−6 4 3 y = − x−4 4 y−2= −
3 4
3 ANSWER: y = − x − 4 4 39.
PROBLEM: Perpendicular to y = 3x − 1 and passing through (−3, 2). 1 Solution: Here the given line has slope m = 3 and thus m⊥ = − . We substitute 3 this slope and the given point into point-slope form. Point Slope 1 (−3, 2) m⊥ = − 3 y − y1 = m ( x − x1 )
1 ( x − ( −3) ) 3 1 y − 2 = − x −1 3 1 y = − x +1 3 1 ANSWER: y = − x + 1 3 y−2= −
40.
PROBLEM:
1 Perpendicular to y = − x + 2 and passing through (4, −3). 3 1 1 and thus m⊥ = − = 3 . We 1 3 − 3 substitute this slope and the given point into point-slope form.
SOLUTION: Here the given line has slope m = −
Point (4, −3)
Slope m⊥ = 3
y − y1 = m ( x − x1 )
y − ( −3) = 3 ( x − 4 ) y + 3 = 3x − 12 y = 3x − 15 ANSWER: y = 3x − 15
41.
PROBLEM:
Perpendicular to y = −2 and passing through (−1, 5). 1 = undefined . 0 We substitute this slope and the given point into point-slope form. Point Slope
Solution: Here the given line has slope m = 0 and thus m⊥ = −
(−1, 5)
m⊥ = undefined
y − y1 = m ( x − x1 )
y −5 = x − ( −1) undefined x = −1 ANSWER: x = −1
42.
PROBLEM: Perpendicular to x = 15 and passing through (5, −3). SOLUTION: Here the given line has slope m = undefined and thus m⊥ = −
1 = 0 . We substitute this slope and the given point into pointundefined
slope form. Point (5, −3)
Slope m⊥ = 0
y − y1 = m ( x − x1 )
y − ( −3) = 0 ( x − 5 ) y+3= 0 y = −3 ANSWER: y = −3
43.
PROBLEM: Parallel to y = 3 and passing through (2, 4). Solution: Here the given line has slope m = 0 and thus m& = 0 . We substitute this slope and the given point into point-slope form. Point Slope
( 2, 4 )
m& = 0
y − y1 = m ( x − x1 ) y − 4 = 0 ( x − 2) y=4
ANSWER: y = 4
44.
PROBLEM: Parallel to x = 2 and passing through (7, −3). SOLUTION: Here the given line has slope m = undefined and thus m& = undefined . We substitute this slope and the given point into point-slope
form. Point (7, −3)
Slope m& = undefined
y − y1 = m ( x − x1 )
y − ( −3) = undefined ( x − 7 ) y − ( −3 ) undefined x−7 = 0
= x−7
x=7
ANSWER: x = 7
45.
PROBLEM: Perpendicular to y = x and passing through (7, −13). Solution: Here the given line has slope m = 1 and thus m⊥ = −1 . We substitute this slope and the given point into point-slope form. Point Slope (7, −13) m⊥ = −1
y − y1 = m ( x − x1 )
y − ( −13) = −1( x − 7 ) y + 13 = − x + 7 y = −x − 6 ANSWER: y = − x − 6
46.
PROBLEM: Perpendicular to y = 2 x + 9 and passing through (3, −1). 1 SOLUTION: Here the given line has slope m = 2 and thus m⊥ = − . We 2 substitute this slope and the given point into point-slope form.
Point
Slope 1 (3, −1) m⊥ = − 2 y − y1 = m ( x − x1 ) 1 ( x − 3) 2 1 3 y +1 = − x + 2 2 1 1 y =− x+ 2 2 y − ( −1) = −
1 1 ANSWER: y = − x + 2 2
47.
PROBLEM:
Parallel to y =
1 x − 5 and passing through (−2, 1). 4
1 1 and thus m& = . We substitute 4 4 this slope and the given point into point-slope form. Point Slope
Solution: Here the given line has slope m =
(−2, 1)
m& =
y − y1 = m ( x − x1 )
1 4
1 ( x − ( −2 ) ) 4 1 y −1 = ( x + 2) 4 1 3 y = x+ 4 2 1 3 ANSWER: y = x + 4 2 y −1 =
48.
PROBLEM:
3 Parallel to y = − x + 1 and passing through (4, 1/4). 4 3 3 and thus m& = − . We 4 4 substitute this slope and the given point into point-slope form.
SOLUTION: Here the given line has slope m = −
Point
Slope
1⎞ 3 ⎛ m& = − ⎜ 4, ⎟ 4⎠ 4 ⎝ y − y1 = m ( x − x1 ) 1 3 = − ( x − 4) 4 4 3 13 y =− x+ 4 4 y−
3 13 ANSWER: y = − x + 4 4
49.
PROBLEM: Parallel to 2 x − 3 y = 6 and passing through (6,−2). SOLUTION: 2x − 3 y = 6 2x − 2x − 3y = 6 − 2x 3y = 2x − 6 3 2 6 y = x− 3 3 3 2 y = x−2 3 2 2 and thus m& = . We substitute this slope 3 3 and the given point into point-slope form.
Here the given line has slope m =
Point
Slope 2 (6, −2) m& = 3 y − y1 = m ( x − x1 ) y − ( −2 ) = y=
2 ( x − 6) 3
2 x−6 3
ANSWER: y =
50.
2 x−6 3
PROBLEM: Parallel to − x + y = 4 and passing through (9, 7). SOLUTION: −x + y = 4 −x + x + y = 4 + x
y = x+4 Here the given line has slope m = 1 and thus m& = 1 . We substitute this slope and the given point into point-slope form. Point
( 9, 7 )
Slope m& = 1
y − y1 = m ( x − x1 ) y − 7 = 1( x − 9 ) y = x−2 ANSWER: y = x − 2
51.
PROBLEM: Perpendicular to 5 x − 3 y = 18 and passing through (−9, 10). SOLUTION: 5 x − 3 y = 18 5 x − 5 x − 3 y = 18 − 5 x 3 y = 5 x − 18 3 5 18 y = x− 3 3 3 5 y = x−6 3 5 3 and thus m⊥ = − . We substitute this slope 3 5 and the given point into point-slope form.
Here the given line has slope m =
Point
Slope
(−9, 10)
m⊥ = −
y − y1 = m ( x − x1 )
3 5
3 ( x − ( −9 ) ) 5 3 23 y =− x+ 5 5 3 23 ANSWER: y = − x + 5 5 y − 10 = −
52.
PROBLEM: Perpendicular to x − y = 11 and passing through (6, −8). SOLUTION: x − y = 11
x − x − y = 11 − x y = x − 11 Here the given line has slope m = 1 and thus m⊥ = −1 . We substitute this slope and the given point into point-slope form. Point Slope (6, −8) m⊥ = −1 y − y1 = m ( x − x1 )
y − ( −8 ) = −1( x − 6 ) y = −x − 2 ANSWER: y = − x − 2
53.
PROBLEM: 1 1 Parallel to x − y = 2 and passing through (−15, 6). 5 3 SOLUTION: 1 1 x− y = 2 5 3 1 1 1 1 x− x− y = 2− x 5 5 3 5 1 1 y = x−2 3 5 1 1 3⋅ y = 3⋅ x − 3⋅ 2 3 5 3 y = x−6 5 3 3 and thus m& = . We substitute this slope and 5 5 the given point into point-slope form.
Here the given line has slope m =
Point
Slope 3 (−15, 6) m& = 5 y − y1 = m ( x − x1 ) y−6 = y=
3 ( x − ( −15) ) 5
3 x + 15 5
ANSWER: y =
54.
3 x + 15 5
PROBLEM:
Parallel to −10 x −
5 1 y = and passing through (−1, 1/2). 7 2
SOLUTION: 5 1 −10 x − y = 7 2 5 1 −10 x + 10 x − y = + 10 x 7 2 5 1 y = −10 x − 7 2 7 5 7 7 1 ⋅ y = ⋅ ( −10 ) x − ⋅ 5 7 5 5 2 7 y = −14 x − 10 Here the given line has slope m = −14 and thus m& = −14 . We substitute this slope and the given point into point-slope form.
Point
Slope
1⎞ ⎛ m& = −14 ⎜ −1, ⎟ 2⎠ ⎝ y − y1 = m ( x − x1 ) 1 = −14 ( x − ( −1) ) 2 27 y = −14 x − 2 y−
ANSWER: y = −14 x −
27 2
55.
PROBLEM:
Perpendicular to
1 1 x − y = 1 and passing through (−10, 3). 2 3
SOLUTION: 1 1 x − y =1 2 3 1 1 1 1 x − x − y = 1− x 2 2 3 2 1 1 y = x −1 3 2 1 1 3 ⋅ y = 3 ⋅ x − 3 ⋅1 3 2 3 y = x −3 2 3 1 2 and thus m⊥ = − = − . We substitute this 3 2 3 2 slope and the given point into point-slope form.
Here the given line has slope m =
Point
Slope
(−10, 3)
m⊥ = −
y − y1 = m ( x − x1 )
2 3
2 ( x − ( −10 ) ) 3 2 11 y =− x− 3 3 2 11 ANSWER: y = − x − 3 3 y −3 = −
56.
PROBLEM: Perpendicular to −5 x + y = −1 and passing through (−4, 0).
SOLUTION: −5 x + y = −1
−5 x + 5 x + y = −1 + 5 x y = 5x −1 1 Here the given line has slope m = 5 and thus m⊥ = − . We substitute this slope 5 and the given point into point-slope form.
Point
Slope
(−4, 0)
m⊥ = −
y − y1 = m ( x − x1 )
1 5
1 ( x − ( −4 ) ) 5 1 4 y =− x− 5 5 y−0 = −
1 4 ANSWER: y = − x − 5 5 57.
PROBLEM: Parallel to x + 4 y = 8 and passing through (−1, −2). SOLUTION: x + 4y = 8 x − x + 4y = 8 − x
4 y = −x + 8 4 1 8 y =− x+ 4 4 4 1 y =− x+2 4 1 1 and thus m& = − . We substitute this slope 4 4 and the given point into point-slope form.
Here the given line has slope m = −
Point (−1, −2)
Slope 1 m& = − 4
y − y1 = m ( x − x1 ) y − ( −2 ) = −
1 ( x − ( −1) ) 4
1 9 y =− x− 4 4 1 9 ANSWER: y = − x − 4 4 58.
PROBLEM: Parallel to 7 x − 5 y = 35 and passing through (2, −3). SOLUTION: 7 x − 5 y = 35 7 x − 7 x − 5 y = 35 − 7 x
5 y = 7 x − 35 5 7 35 y = x− 5 5 5 7 y = x −5 5 7 7 and thus m& = . We substitute this slope 5 5 and the given point into point-slope form.
Here the given line has slope m =
Point
Slope 7 (2, −3) m& = 5 y − y1 = m ( x − x1 ) 7 ( x − 2) 5 7 29 y = x− 5 5 y − ( −3) =
ANSWER: y =
7 29 x− 5 5
59.
PROBLEM: Perpendicular to 6 x + 3 y = 1 and passing through (8, −2). SOLUTION: 6x + 3 y = 1 6x − 6x + 3y = 1 − 6x
3 y = −6 x + 1 3 6 1 y =− x+ 3 3 3 1 y = −2 x + 3 Here the given line has slope m = −2 and thus m⊥ = and the given point into point-slope form. Slope 1 (8, −2) m⊥ = 2 y − y1 = m ( x − x1 ) Point
y − ( −2 ) = y=
1 ( x − 8) 2
1 x−6 2
ANSWER: y =
1 x−6 2
1 . We substitute this slope 2
60.
PROBLEM: Perpendicular to −4 x − 5 y = 1 and passing through (−1, −1). SOLUTION: −4 x − 5 y = 1 −4 x + 4 x − 5 y = 1 + 4 x
5 y = −4 x − 1 5 4 1 y =− x− 5 5 5 4 1 y =− x− 5 5 4 1 5 and thus m⊥ = − = . We substitute 4 4 5 − 5 this slope and the given point into point-slope form.
Here the given line has slope m = −
Point
Slope 5 (−1, −1) m⊥ = 4 y − y1 = m ( x − x1 ) 5 ( x − ( −1) ) 4 5 1 y = x+ 4 4 y − ( −1) =
ANSWER: y =
5 1 x+ 4 4
61.
PROBLEM: ⎛1 1⎞ Parallel to −5 x − 2 y = 4 and passing through ⎜ , − ⎟ . ⎝5 4⎠
SOLUTION: −5 x − 2 y = 4 −5 x + 5 x − 2 y = 4 + 5 x
2 y = −5 x − 4 2 5 4 y =− x− 2 2 2 5 y = − x−2 2 5 5 and thus m& = − . We substitute this slope 2 2 and the given point into point-slope form.
Here the given line has slope m = −
Point
Slope
5 ⎛1 1⎞ m& = − ⎜ ,− ⎟ 2 ⎝5 4⎠ y − y1 = m ( x − x1 ) 5⎛ 1⎞ ⎛ 1⎞ y −⎜− ⎟ = − ⎜ x − ⎟ 2⎝ 5⎠ ⎝ 4⎠ 5 1 y =− x+ 2 4 5 1 ANSWER: y = − x + 2 4
62.
PROBLEM:
Parallel to 6 x −
3 y = 9 and passing through 2
⎛1 2⎞ ⎜ , ⎟. ⎝3 3⎠
SOLUTION: 3 6x − y = 9 2 3 6x − 6x − y = 9 − 6x 2 3 y = 6x − 9 2 2 3 2 2 ⋅ y = ⋅ 6x − ⋅ 9 3 2 3 3 y = 4x − 6 Here the given line has slope m = 4 and thus m& = 4 . We substitute this slope and the given point into point-slope form.
Point
Slope
⎛1 2⎞ m& = 4 ⎜ , ⎟ ⎝3 3⎠ y − y1 = m ( x − x1 ) 2 1⎞ ⎛ = 4⎜ x − ⎟ 3 3⎠ ⎝ 2 y = 4x − 3 y−
ANSWER: y = 4 x −
2 3
63.
PROBLEM: Perpendicular to y − 3 = 0 and passing through (−6, 12). SOLUTION: y −3 = 0
Here the given line has slope m = 0 and thus m⊥ = −
1 = undefined . We substitute 0
this slope and the given point into point-slope form. Point (−6, 12)
Slope m⊥ = undefined
y − y1 = m ( x − x1 )
y − 12 = ( x − ( −6 ) ) undefined x+6 = 0 x = −6 ANSWER: x = −6
64.
PROBLEM: Perpendicular to x + 7 = 0 and passing through (5, −10). SOLUTION: x+7 =0
Here the given line has slope m = undefined and thus m⊥ = − substitute this slope and the given point into point-slope form. Point Slope (5, −10) m⊥ = 0 y − y1 = m ( x − x1 ) y − ( −10 ) = 0 ( x − 5 ) y = −10 ANSWER: y = −10
1 = 0 . We undefined
3.7 Introduction to Functions Part A: Functions Does the following correspondence represent a function?
1.
PROBLEM: Algebra students to their scores on the first exam. ANSWER: An algebra student corresponds exactly to an algebra score. Hence the above statement represents a function.
2.
PROBLEM: Family members to their ages. ANSWER: A family member corresponds exactly to an age. Hence the above statement represents a function.
3.
PROBLEM: Lab computers to their users. ANSWER: A single lab computer does not correspond exactly to one user. Hence the above statement does not represent a function.
4.
PROBLEM: Students to the schools they have attended. ANSWER: A single student does not correspond exactly to a single school attended. Hence the above statement does not represent a function.
5.
PROBLEM: People to their citizenships. ANSWER: A single person does not correspond exactly to a single citizenship. Hence the above statement does not represent a function.
6.
PROBLEM: Local businesses to their number of employees. ANSWER: A local business corresponds exactly to a number of employees. Hence the above statement represents a function.
Determine the domain and range and state whether the relation is a function or not.
7.
PROBLEM: {(3, 2), (5, 3), (7, 4)} SOLUTION:
3 5 7
2 3 4
ANSWER: Domain: {3, 5, 7} Range: {2, 3, 4} The relation is a function because each x-value corresponds to exactly one of yvalue.
8.
PROBLEM: {(−5, −3), (0, 0), (5, 0)} SOLUTION:
−5 0
−3 0
5 ANSWER: Domain:{–5, 0, 5} Range: {–3, 0} The relation is a function because each x-value corresponds to exactly one of yvalue.
9.
PROBLEM: {(−10, 2), (−8, 1), (−8, 0)} SOLUTION:
–10 −8
2 1 0
ANSWER: Domain: {−10, −8} Range: {2, 1, 0}. The relation is not a function because each x-value does not correspond to exactly one of y-value.
10.
PROBLEM: {(9, 12), (6, 6), (6, 3)} SOLUTION:
9 6
−3 6 3
ANSWER: Domain: {9, 6} Range: {–3, 6, 3} The relation is not a function because each x-value does not correspond to exactly one of y-value.
11.
PROBLEM:
SOLUTION: The coordinates marked are (–4, 1), (1, 2), and (2, 3). ANSWER: Domain: {–4, 1, 2} Range: {1, 2, 3} The relation is a function because each x-value corresponds to exactly one of yvalue.
12.
PROBLEM:
SOLUTION: The coordinates marked are (–3, –2), (1, –2), and (3, –2). ANSWER: Domain: {–3, 1, 3} Range: {–2} The relation is a function because each x-value corresponds to exactly one of yvalue.
13.
PROBLEM:
SOLUTION: The coordinates marked are (–2, 3), (2, 2), and (2, 5). ANSWER: Domain: {–2, 2} Range: {3, 2, 5} The relation is not a function because each x-value does not correspond to exactly one of y-value.
14.
PROBLEM:
SOLUTION: The coordinates marked are (4, –3), (4, 1), and (4, 3). ANSWER: Domain: {4} Range: {–3, 1, 3} The relation is not a function because each x-value does not correspond to exactly one of y-value.
15.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that 2 is the maximum y-value and any y-value lesser than this is not represented in the relation. Hence, the range is 2. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is {2}. The relation is a function because it passes the vertical line test.
16.
PROBLEM:
SOLUTION: The graph indicates that −3 is the minimum x-value and any xvalue greater than that is not represented in the relation. Hence, the domain is equal to −3. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will not intersect the graph only once; therefore it is not a function. ANSWER: The domain is {–3} and the range is all real numbers R = (–∞,∞). The relation is not a function because it does not pass the vertical line test.
17.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is all real numbers too R = (–∞,∞). The relation is a function because it passes the vertical line test.
18.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that −3 is the minimum y-value and any y-value greater than that is represented in the relation. Hence, the range consists of all yvalues greater than or equal to −3, or in interval notation[−3, ∞) . Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is [–3,∞). The relation is a function because it passes the vertical line test.
19.
PROBLEM:
SOLUTION: The graph indicates that −2 is the minimum x-value and any xvalue greater than that is represented in the relation. Hence, the domain consists of all values greater than or equal to −2 or the interval notation [ −2, ∞ ) . If we
choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph twice; therefore it is not a function. ANSWER: The domain is [ −2, ∞ ) and the range is all real numbers R = (–
∞,∞).The relation is not a function because it does not pass the vertical line test.
20.
PROBLEM:
SOLUTION: The graph indicates that 1 is the maximum x-value and any x-value lesser than that is represented in the relation. Hence, the domain consists of all xvalues lesser than or equal to 1, or in interval notation ( −∞,1] . If we choose any
y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph twice; therefore it is not a function. ANSWER: The domain is ( −∞,1] and the range is all real numbers R = (–
∞,∞).The relation is not a function because it does not pass the vertical line test. 21.
PROBLEM:
SOLUTION: The graph indicates that −4 is the minimum x-value and any xvalue greater than that is represented in the relation. Hence, the domain consists of all values greater than or equal to −4 or the interval notation [ −4, ∞ ) . If we
choose any positive y-value, then we will be able to find a corresponding point on the graph; therefore, the range is greater than or equal to zero or the interval notation [0, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is [ −4, ∞ ) and the range is all positive integers [0,
∞).The relation is a function because it passes the vertical line test.
22.
PROBLEM:
SOLUTION: The graph indicates that 3 is the maximum x-value and any x-value lesser than that is represented in the relation. Hence, the domain consists of all xvalues lesser than or equal to 3, or in interval notation ( −∞,3] . If we choose any
negative y-value, then we will be able to find a corresponding point on the graph; therefore, the range is lesser than or equal to zero or in the interval notation (–∞, 0]. Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞,3] and the range is (–∞, 0].The relation is a
function because it passes the vertical line test. 23.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any positive y-value, then we will be able to find a corresponding point on the graph; therefore, the range is greater than or equal to zero or the interval notation [0, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞, ∞ ) and the range is all positive integers [0,
∞).The relation is a function because it passes the vertical line test.
24.
PROBLEM:
SOLUTION: If we choose any x-value we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any positive y-value we will be able to find a corresponding point on the graph; therefore, the range is greater than or equal to zero or in the interval notation [ 0, ∞ ) . Any vertical line will intersect the graph once; therefore it is a
function. ANSWER: The domain is ( −∞, ∞ ) and the range is all positive
integers [ 0, ∞ ) .The relation is a function because it passes the vertical line test. 25.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The minimum value of y is 2 and any value greater than that is represented in the relation; therefore, the range is greater than or equal to 2 or the interval notation [2, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞, ∞ ) and the range is [2, ∞).The relation is a function because it passes the vertical line test.
26.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The minimum value of y is –3 and any value greater than that is represented in the relation; therefore, the range is greater than or equal to –3 or the interval notation [–3, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞, ∞ ) and the range is [–3, ∞).The relation is a
function because it passes the vertical line test. Part B: Function Notation Given the following functions find the function values.
27.
PROBLEM: f ( x) = 3x find f (−2) SOLUTION: f ( x) = 3 x
f (−2) = 3 ⋅ ( −2 ) = −6 ANSWER: −6
28.
PROBLEM: f ( x) = −5 x + 1 find f (−1) SOLUTION: f ( x) = −5 x + 1
f ( −1) = −5 ⋅ ( −1) + 1 = 5 + 1 = 6 ANSWER: 6
29.
PROBLEM: f ( x) = 53 x − 4 find f (15) SOLUTION: f ( x) = 53 x − 4
f (15) = 53 ⋅15 − 4 = 9 − 4 = 5 ANSWER: 5
30.
PROBLEM: f ( x) = 52 x − 15 find f (3) SOLUTION: f ( x) = 52 x − 15 f (3) =
2 1 5 ⋅ ( 3) − = = 1 5 5 5
ANSWER: 1
31.
PROBLEM: f ( x) = 52 x − 13 find f (− 13 ) SOLUTION: f ( x) = 52 x − 13
5 ⎛ 1⎞ 1 7 ⋅⎜ − ⎟ − = − 2 ⎝ 3⎠ 3 6 7 ANSWER: − 6 f ( x) =
32.
PROBLEM: f ( x) = −6 find f (7) SOLUTION: f ( x) = −6
f ( 7 ) = −6 ANSWER: −6
33.
PROBLEM: g ( x) = 5 find g (−4) SOLUTION: g ( x) = 5 g (−4) = 5 ANSWER: 5
34.
PROBLEM: g ( x) = −5 x find g (−3) SOLUTION: g ( x) = −5 x
g (−3) = −5 ( −3) = 15 ANSWER: 15
35.
PROBLEM: g ( x) = − 18 x + 85 find g ( 85 ) SOLUTION: 1 5 g ( x) = − x + 8 8 1 ⎛ 5 ⎞ 5 35 ⎛5⎞ g ⎜ ⎟ = − ⋅⎜ ⎟ + = 8 ⎝ 8 ⎠ 8 64 ⎝8⎠ 35 ANSWER: 64
36.
PROBLEM: g ( x) = 53 x − 5 find g ( 3) SOLUTION: 5 g ( x) = x − 5 3 5 g ( 3) = ⋅ 3 − 5 = 0 3 ANSWER: 0
37.
PROBLEM: f ( x) = 5 x − 9 find x when f ( x) = 1 SOLUTION: f ( x) = 5 x − 9 When f(x) = 1, f ( x) = 5 x − 9
1 = 5x − 9 10 5 = x 5 5 x=2 ANSWER: x = 2
38.
PROBLEM: f ( x) = −7 x + 2 find x when f ( x) = 0 SOLUTION: f ( x) = −7 x + 2 When f ( x) = 0, f ( x) = −7 x + 2 0 = −7 x + 2 2 = 7x
2 7 = x 7 7 2 x= 7 ANSWER: x =
2 7
39.
PROBLEM: f ( x) = − 75 x − 2 find x when f ( x) = −9 SOLUTION: f ( x) = − 75 x − 2 When f ( x) = −9 , 7 −9 = − x − 2 5 ⎛ 5 ⎞ ⎛ 7 ⎞⎛ 5 ⎞ −7 ⎜ − ⎟ = ⎜ − x ⎟ ⎜ − ⎟ ⎝ 7 ⎠ ⎝ 5 ⎠⎝ 7 ⎠ x=5 ANSWER: x = 5
40.
PROBLEM: f ( x) = − x − 4 find x when f ( x) = SOLUTION: f ( x) = − x − 4 When f ( x) = 12 1 −x − 4 = 2 1 −x − 4 + 4 = + 4 2 9 −x = 2 9 x=− 2 ANSWER: x = −
9 2
1 2
41.
PROBLEM: g ( x) = x find x when g ( x) = 12 SOLUTION: g ( x) = x When g ( x) = 12 , 12 = x x = 12 ANSWER: x = 12
42.
PROBLEM: g ( x) = − x + 1 find x when g ( x) =
2 3
SOLUTION: g ( x) = − x + 1 When g ( x) = 23 2 = −x +1 3 1 x= 3 ANSWER: x =
43.
1 3
PROBLEM: g ( x) = −5 x + 13 find x when g ( x) = − 12 SOLUTION: g ( x) = −5 x + 13 When g ( x) = − 12 , 1 1 − = −5 x + 2 3 5 5x = 6 5 5 x= 5 6⋅5 1 x= 6 1 ANSWER: x = 6
44.
PROBLEM: g ( x) = − 85 x + 3 find x when g ( x) = 3 SOLUTION: g ( x) = − 85 x + 3 When g ( x) = 3 5 3= − x+3 8 5 x=0 8 x=0 ANSWER: x = 0
Given f ( x) = 23 x − 1 and g ( x) = −3 x + 2 calculate the following (#46 ‐ #54):
45.
PROBLEM: f (6) SOLUTION: 2 f ( x) = x − 1 3 2 f ( 6) = ( 6) −1 = 3 3 ANSWER: 3
46.
PROBLEM: f (−12) SOLUTION: 2 f ( x) = x − 1 3 2 f ( −12 ) = ( −12 ) − 1 = −9 3 ANSWER: −9
47.
PROBLEM: f (0) SOLUTION: 2 f ( x) = x − 1 3 2 f ( 0 ) = ( 0 ) − 1 = −1 3
48.
49.
ANSWER: −1 PROBLEM: f (1) SOLUTION: 2 f ( x) = x − 1 3 2 1 f (1) = (1) − 1 = − 3 3 1 ANSWER: − 3 PROBLEM: g ( 23 ) SOLUTION: g ( x ) = −3 x + 2
⎛2⎞ ⎛2⎞ g ⎜ ⎟ = −3 ⎜ ⎟ + 2 = 0 ⎝3⎠ ⎝3⎠ ANSWER: 0 50.
PROBLEM: g (0) SOLUTION: g ( x ) = −3 x + 2
g ( 0 ) = −3 ( 0 ) + 2 = 2 ANSWER: 2
51.
PROBLEM: g ( −1) SOLUTION: g ( x) = −3 x + 2
g ( −1) = −3 ( −1) + 2 = 5 ANSWER: 5 52.
PROBLEM: g ( − 12 ) SOLUTION: g ( x) = −3x + 2
7 ⎛ 1⎞ ⎛ 1⎞ g ⎜ − ⎟ = −3 ⎜ − ⎟ + 2 = 2 ⎝ 2⎠ ⎝ 2⎠ 53.
ANSWER:
7 2
PROBLEM: Find x when f ( x) = 0 . SOLUTION: 2 f ( x) = x − 1 3 When f ( x) = 0 2 0 = x −1 3 3 3 2 ⋅1 = ⋅ x 2 2 3 3 x= 2 ANSWER: x =
3 2
54.
55.
PROBLEM: Find x when f ( x) = −3 . SOLUTION: 2 f ( x) = x − 1 3 When f ( x) = −3 2 −3 = x − 1 3 3 3 2 ⋅ ( −2 ) = ⋅ x 2 2 3 x = −3 ANSWER: x = −3 PROBLEM: Find x when g ( x) = −1 . SOLUTION: g ( x) = −3x + 2 When g ( x) = −1 −1 = −3 x + 2
− 3 = −3 x −3 −3 = x −3 −3 x =1 ANSWER: x = 1
56.
PROBLEM: Find x when g ( x) = 0 . SOLUTION: g ( x ) = −3 x + 2 When g ( x) = 0 0 = −3 x + 2 −2 = −3 x −2 −3 x = −3 −3 2 x= 3
ANSWER: x =
2 3
Given the graph find the function values.
57.
PROBLEM: Given the graph of f ( x) find f (−4) , f (−1) , f (0) and f (2) .
SOLUTION: From the graph when x = –4, y = – 3. Hence, f (−4) = −3. From the graph when x = –1, y = – 0. Hence, f (−1) = 0. From the graph when x = 0, y =1. Hence, f (0) = 1. From the graph when x = 2, y = 3. Hence, f (2) = 3. ANSWER: f (−4) = −3. , f (−1) = 0. , f (0) = 1. , f (2) = 3.
58.
PROBLEM: Given the graph of g ( x ) find g (−3) , g (−1) , g (0) and g (1) .
ANSWER: From the graph when x = –3, y = 4. Hence, From the graph when x = –1, y = 0. Hence, From the graph when x = 0, y = –2. Hence, From the graph when x = 1, y = –4. Hence,
g (−3) = 4. g (−1) = 0. g (0) = −2. g (1) = −4.
59.
PROBLEM: Given the graph of f ( x) find f (−4) , f (−1) , f (0) and f (2) .
SOLUTION: From the graph when x = –4, y = –4 . Hence, f (−4) = −4. From the graph when x = –1, y = –4 . Hence, f (−1) = −4. From the graph when x = 0, y = –4 . Hence, f (0) = −4. From the graph when x = 2, y = –4 . Hence, f (2) = −4. ANSWER: f (−4) = −4. , f (−1) = −4. , f (0) = −4. , f (2) = −4.
60.
PROBLEM: Given the graph of g ( x ) find g (−4) , g (−1) , g (0) and g (2) .
ANSWER: From the graph when x = –4, y = 0. Hence, g (−4) = 0. From the graph when x = –1, y = 0. Hence, g (−1) = 0. From the graph when x = 0, y = 0. Hence, g (0) = 0. From the graph when x = 2, y = 0. Hence, g (2) = 0.
61.
PROBLEM: Given the graph of f ( x ) find f (−1) , f (0) , f (1) ,and f (3) .
SOLUTION: From the graph when x = –1, y = 1 . Hence, f (−1) = 1. From the graph when x = 0, y = –2 . Hence, f (0) = −2. From the graph when x = 1, y = –3 . Hence, f (1) = −3. From the graph when x = 3, y = 1 . Hence, f (3) = 1. ANSWER: f (−1) = 1. , f (0) = −2. , f (1) = −3. , f (3) = 1.
62.
PROBLEM: Given the graph of g ( x ) find g (−2) , g (0) , g (2) and g (6) .
ANSWER: From the graph when x = –2, y = 0. Hence, g (−2) = 0. From the graph when x = 0, y = 12. Hence, g (0) = 12. From the graph when x = 2, y = 16. Hence, g (2) = 16. From the graph when x = 6, y = 0. Hence, g (6) = 0.
63.
PROBLEM: Given the graph of g ( x ) find g (−4) , g (−3) , g (0) and g (4) .
SOLUTION: From the graph when x = –4, y = 0. Hence, g (−4) = 0. From the graph when x = –3, y = 1. Hence, g (−3) = 1. From the graph when x = 0, y = 1. Hence, g (0) = 2. From the graph when x = 4, y = 3. Hence, g (4) = 3. ANSWER: g (−4) = 0. , g (−3) = 1. , g (0) = 2. , g (4) = 3.
64.
PROBLEM: Given the graph of f ( x ) find f (−4) , f (0) , f (1) ,and f (3) .
ANSWER: From the graph when x = –4, y = 1 . Hence, f (−4) = 1. From the graph when x = 0, y = –3 . Hence, f (0) = −3. From the graph when x = 1, y = –2 . Hence, f (1) = −2. From the graph when x = 3, y = 0 . Hence, f (3) = 0.
Given the graph find the x‐values.
65.
PROBLEM: Given the graph of f ( x) find x when f ( x) = 3 , f ( x) = 1 and f ( x) = −3 .
SOLUTION: When f ( x) = 3 , y = 3. From the graph when y = 3, x = –1. When f ( x) = 1 , y = 1. From the graph when y = 1, x = 0. When f ( x) = –3 , y = –3. From the graph when y = –3, x = 2. ANSWER: x = –1, x = 0, x = 2
66.
PROBLEM: Given the graph of g ( x ) find x when g ( x) = −1 , g ( x) = 0 and g ( x) = 1 .
ANSWER: When g ( x) = −1 , y = –1. From the graph when y = –1, x = 1. When g ( x) = 0 , y = 0. From the graph when y = 0, x = 0. When g ( x) = 1 , y = 1. From the graph when y = 1, x = –1.
67.
PROBLEM: Given the graph of f ( x) find x when f ( x) = 3 .
ANSWER: When f ( x) = 3 , y = 3. From the graph when y = 3, x = 0. (Answers may vary)
68.
PROBLEM: Given the graph of g ( x) find x when g ( x) = −4 , g ( x) = 0 and g ( x) = 4 .
ANSWER: When g ( x) = −4 , y = –4. From the graph when y = –4, x = –-6. When g ( x) = 0 , y = 0. From the graph when y = 0, x = –2. When g ( x) = 4 , y = 4. From the graph when y = 4, x = 2.
69.
PROBLEM: Given the graph of f ( x) find x when f ( x) = −16 , f ( x) = −12 and f ( x) = 0 .
SOLUTION: When f ( x) = −16 , y = –16 . From the graph when y = –16, x = –4. When f ( x) = −12 , y = –12. From the graph when y = –12, x = –2 and x = –6. When f ( x) = 0 , y = 0. From the graph when y = 0, x = 0 and x = –8. ANSWER: x = –4, x = –2 and x = –6, x = 0 and x = –8
70.
PROBLEM: Given the graph of g ( x) find x when g ( x) = −3 , g ( x ) = 0 and g ( x) = 1 .
ANSWER: When g ( x) = −3 , y = –3. From the graph when y = –3, x = –2 and x = 2. When g ( x) = 0 , y = 0. From the graph when y = 0, x = –1 and x = 1. When g ( x) = 1 , y = 1. From the graph when y = 1, x = 0.
71.
PROBLEM: Given the graph of f ( x) find x when f ( x) = −4 , f ( x) = 0 and f ( x) = −2 .
SOLUTION: When f ( x) = −4 , y = –4. From the graph when y = –4, x = –4 and x = 4. When f ( x) = 0 , y = 0. From the graph when y = 0, x = 0. When f ( x) = −2 , y = –2. From the graph when y = –2, x = –2 and x = 2. ANSWER: x = –4 and x = 4, x = 0, x = –2 and x = 2
72.
PROBLEM: Given the graph of g ( x) find x when g ( x) = 5 , g ( x) = 3 and g ( x) = 2 .
ANSWER: When g ( x) = 5 , y = 5. From the graph when y = 5, x = –3 and x = 3. When g ( x) = 3 , y = 3. From the graph when y = 3, x = –1 and x = 1. When g ( x) = 2 , y = 2. From the graph when y = 2, x = 0.
73.
PROBLEM: The cost, in dollars, of producing pens with a company logo is given by the function C ( x) = 1.65 x + 120 where x is the number of pens produced. Use the function to calculate the cost of producing 200 pens. SOLUTION: Cost of producing a pen is given by the equation C ( x) = 1.65 x + 120. ANSWER: Cost of producing 200 pens, C (200) = 1.65 ( 200 ) + 120 = $450.
74.
PROBLEM: The revenue, in dollars, from selling sweatshirts is given by the function R( x) = 29.95 x where x is the number of sweatshirts sold. Use the function to determine the revenue if 20 sweatshirts are sold. SOLUTION: Cost of producing a single sweatshirt is given by the function, R( x) = 29.95 x. ANSWER: Cost of producing 20 sweatshirts is R(20) = 29.95 ( 20 ) = $599.
75.
PROBLEM: The value of a new car, in dollars, is given by the function V (t ) = −2500t + 18000 where t represents the age of the car in years. Use the function to determine the value of the car when it is 5 years old. What was the value of the car new? SOLUTION: Value of the car is given by the function, V (t ) = −2500t + 18000.
Value of the car after 5 years, V (5) = −2500 ( 5) + 18000 = $5,500. ANSWER: Value of the new car, V (0) = −2500 ( 0 ) + 18000 = $18, 000.
76.
PROBLEM: The monthly income, in dollars, of a commissioned car salesman is given by the function I (n) = 550n + 1250 where n represents the number of cars sold in the month. Use the function to determine the salesman’s monthly income if he sells 3 cars this month. What is his income if he does not sell any cars in one a month? SOLUTION: Monthly income of the salesman based on the number of cars sold is given by the function, I (n) = 550n + 1250.
Monthly income if he sells 3 cars, I (3) = 550 ( 3) + 1250 = $2,900. ANSWER: Monthly income if he does not sell any car, I (0) = 550 ( 0 ) + 1250 = $1, 250.
77.
PROBLEM: The perimeter of an isosceles triangle with base measuring 10 cm is given by the function P ( x) = 2 x + 10 where x represents the length of each of the equal sides. Find the length of each side if the perimeter measure 40 cm. SOLUTION: P(x) = 40 cm P(x) = 2x + 10 40 = 2x + 10 2x = 30 2 30 x= 2 2 x = 15 cm ANSWER: The length of each side measures 15 cm.
78.
PROBLEM: The perimeter of a square depends on the length of each side s and is modeled by the function P( s ) = 4s . If the perimeter of a square measures 140 meters, then use the function to calculate the length of each side. SOLUTION: P(s) = 4s P(s) = 140 meters 4s = 140 meters 4 140 s= meters 4 4 s = 35 meters ANSWER: The length of each side is 35 meters.
79.
PROBLEM: A certain cellular phone plan charges $18 per month and $0.10 per minute of use which is modeled by the function C ( x) = 0.10 x + 18 where x represents the number of minutes of usage per month. Determine the minutes of usage if the cost for the month was $36. SOLUTION: C ( x) = 0.10 x + 18 C(x) = $36 36 = 0.10x + 18 0.10x = 18 0.10 18 x= 0.10 0.10 x = 180 minutes ANSWER: If the cost for the month is $36, the minutes of usage is 180.
80.
PROBLEM: The monthly revenue generated by selling subscriptions to a tutoring website is given by the function R( x) = 29 x where x represents the number of subscription sales per month. How many subscriptions were sold if the revenues for the month totaled $1,508? SOLUTION: R( x) = 29 x
R( x) = $1,508 29 x = 1508 29 1508 x= 29 29 x = 52 ANSWER: For the month totaled $1,508, 52 subscriptions were sold. Graph the linear function and state the domain and range.
81.
PROBLEM: f ( x) = − 52 x + 10 SOLUTION: f ( x) = − 52 x + 10
y = − 52 x + 10 Set y = 0 y = − 52 x + 10 0 = − 52 x + 10 5 x = 10 2 2 5 2 ⋅ x = ⋅10 5 2 5 x=4 x-intercept = (4, 0) Set x = 0 y = − 52 x + 10 5 ( 0 ) + 10 2 y = 10 y-intercept = (0, 10) y=−
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any yvalue, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 82.
PROBLEM: f ( x) = 53 x − 10 SOLUTION: f ( x) = 53 x − 10
y = 53 x − 10 Set y = 0 y = 53 x − 10 0 = 53 x − 10 3 x = 10 5 5 3 5 ⋅ x = ⋅10 3 5 3 2 x = 16 3 ⎛ 2 ⎞ x-intercept = ⎜16 , 0 ⎟ ⎝ 3 ⎠
Set x = 0 y = 53 x − 10 3 ( 0 ) + 10 5 y = 10 y-intercept = (0, 10) y=
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any yvalue, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 83.
PROBLEM: g ( x) = 6 x + 2 SOLUTION: g ( x) = 6 x + 2 Set g ( x) = 0 g ( x) = 6 x + 2
0 = 6( x) + 2 6 2 x=− 6 6 1 x=− 3 ⎛ 1 ⎞ x-intercept = ⎜ − , 0 ⎟ ⎝ 3 ⎠ Set x = 0 y = 6x + 2 y = 6 ( 0) + 2 y=2 g ( x) - intercept = (0, 2)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any g(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 84.
PROBLEM: g ( x ) = −4 x + 6 SOLUTION: g ( x) = −4 x + 6 Set g ( x) = 0 g ( x ) = −4 x + 6
0 = −4 x + 6 4x = 6 4 6 x= 4 4 1 x =1 2 ⎛ 1 ⎞ x-intercept = ⎜1 , 0 ⎟ ⎝ 2 ⎠ Set x = 0 g ( x) = −4 x + 6 g ( x) = −4 ( 0 ) + 6 g ( x) = 6 g(x) intercept = (0, 6)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any g(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 85.
PROBLEM: h(t ) = 12 t − 3 SOLUTION: h(t ) = 12 t − 3 Set h(t) = 0 h(t ) = 12 t − 3
0 = 12 t − 3 1 2
t =3
t =6 t-intercept = ( 6, 0 ) Set t = 0 h(t ) = 12 t − 3 h(t ) =
1 2
( 0) − 3
h(t ) = −3 h(t)-intercept = (0, –3 )
ANSWER:
If we choose any t-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any h(t)value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 86.
PROBLEM: h(t ) = − 34 t + 3 SOLUTION: h(t ) = − 34 t + 3 Set h(t) = 0 h(t ) = − 34 t + 3
0 = − 34 t + 3 3 t =3 4 4 3 4 ⋅ t = ⋅3 3 4 3 t=4 t-intercept = ( 4, 0 ) Set t = 0 h(t ) = − 34 t + 3
h(t ) = − 34 ( 0 ) + 3
h(t ) = 3 h(t)-intercept = (0, 3)
ANSWER:
If we choose any t-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any h(t)value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 87.
PROBLEM: C ( x) = 100 + 50 x SOLUTION: C ( x) = 100 + 50 x Set C ( x) = 0 C ( x) = 100 + 50 x
0 = 100 + 50 x 50 100 x=− 50 50 x = −2 x-intercept = ( −2, 0 ) Set x = 0 C ( x) = 100 + 50 x C ( x) = 100 + 5 ( 0 ) C ( x) = 100 C ( x) -intercept = (0, 100)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any C(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 88.
PROBLEM: C ( x) = 50 + 100 x SOLUTION: C ( x) = 50 + 100 x Set C ( x) = 0 C ( x) = 50 + 100 x
0 = 50 + 100 x 100 x = −50 100 50 x=− 100 100 1 x=− 2 ⎛ 1 ⎞ x-intercept = ⎜ − , 0 ⎟ ⎝ 2 ⎠ Set x = 0 C ( x) = 50 + 100 x C ( x) = 50 + 100 ( 0 ) C ( x) = 50 C ( x) -intercept = (0, 50)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any C(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ Part C: Discussion Board Topics
89.
PROBLEM: Is a vertical line a function? What are the domain and range of a vertical line? ANSWER: The vertical line is not a function. E.g:
The graph indicates that −3 is the minimum x-value and any x-value greater than that is not represented in the relation. Hence, the domain is equal to −3. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will not intersect the graph only once; therefore it is not a function. The domain is {–3} and the range is all real numbers R = (–∞,∞). The relation is not a function because it does not pass the vertical line test.
90.
PROBLEM: Is a horizontal line a function? What are the domain and range of a horizontal line? ANSWER: The horizontal line is a function. E.g.:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that 2 is the minimum y-value and any y-value greater than that is not represented in the relation. Hence, the range is 2. Any vertical line will intersect the graph only once; therefore it is a function. The domain is all real numbers R = (–∞,∞) and the range is {2}. The relation is a function because it passes the vertical line test. 91.
PROBLEM: Come up with your own correspondence between real world sets. Explain why it represents a function or not? ANSWER: 30 NBA teams from 30 different cities: This represents a function because each NBA team corresponds exactly to one city.
Coffee produced is the highest in Brazil and Mexico: This does not represent a function because coffee does not correspond exactly to one country.
3.8 Linear Inequalities (two variables)
Part A: Solutions to Linear Inequalities (two variables) Is the ordered pair a solution to the given inequality?
1.
PROBLEM: y < 5 x + 1 ; (0, 0) SOLUTION: y < 5x + 1
0 < 5 ( 0) + 1 0 − x − 4 ; (0, −2) 2 SOLUTION: 1 y > − x−4 2 1 −2 > − ( 0 ) − 4 2 −2 > −4
3.
1 ANSWER: (0, −2) is a solution to the inequality y > − x − 4. 2 PROBLEM: 2 y ≤ x + 1 ; (6, 5) 3 SOLUTION: 2 y ≤ x +1 3 2 5 ≤ ( 6) + 1 3 5≤5 ANSWER: (6, 5) is a solution to the inequality y ≤
2 x + 1. 3
4.
PROBLEM: 1 y ≥ − x − 5 ; (−3, −8) 3 SOLUTION: 1 y ≥ − x −5 3 1 −8 ≥ − ( −3) − 5 3 −8 ≥ −4 ( Incorrect )
1 ANSWER: (−3, −8) is not a solution to the inequality y ≥ − x − 5. 3 5.
PROBLEM: 1 1 ⎛ 1 ⎞ y < x − ; ⎜ − , −1 ⎟ 5 3 ⎝ 3 ⎠ SOLUTION: 1 1 y < x− 5 3 1⎛ 1⎞ 1 −1 < ⎜ − ⎟ − 5⎝ 3⎠ 3 6 −1 < − 15 1 1 ⎛ 1 ⎞ ANSWER: ⎜ − , −1⎟ is a solution to the inequality y < x − . 5 3 ⎝ 3 ⎠
6.
PROBLEM: 4 x − 3 y ≤ 2 ; (−2, −1) SOLUTION: 4x − 3 y ≤ 2
4 ( −2 ) − 3 ( −1) ≤ 2 −5 ≤ 2 ANSWER: (−2, −1) is a solution to the inequality 4 x − 3 y ≤ 2.
7.
PROBLEM: − x + 4 y > 7 ; (0, 0) SOLUTION: −x + 4 y > 7
0+0 > 7 0 > 7 ( Incorrect ) ANSWER: (0, 0) is not a solution to the inequality − x + 4 y > 7.
8.
PROBLEM: 7 x − 3 y < 21 ; (5, −3) SOLUTION: 7 x − 3 y < 21
7 ( 5) − 3 ( −3) < 21 44 < 21 (Incorrect ) ANSWER: (5, −3) is not a solution to the inequality 7 x − 3 y < 21.
9.
PROBLEM: y > −5 ; (−3, −1) SOLUTION: y > −5 −1 > −5
ANSWER: (−3, −1) is a solution to the inequality y > −5.
10.
PROBLEM: x ≤ 0 ; (0, 7) SOLUTION: x≤0 0≤0
ANSWER: (0, 7) is a solution to the inequality x ≤ 0.
Part B: Graphing solutions to Linear Inequalities Graph the solution set.
11.
PROBLEM: y < −3x + 3 SOLUTION: y = −3x + 3 Set y = 0 y = −3x + 3
0 = −3x + 3 3x = 3 x =1 x − intercept: (1, 0 ) Set x = 0 y = −3x + 3 y = −3 ( 0 ) + 3 y =3 y − intercept: ( 0, 3) Test Point
y < −3x + 3
(0, 0)
0 < 0+3 0 −5 + 3 1 > −2
We shade the region that contains the test point. ANSWER:
17.
PROBLEM: y > −x + 4 SOLUTION: y = −x + 4 Set y = 0 y = −x + 4 0 = −x + 4 x=4
x − intercept: ( 4, 0 ) Set x = 0
y = −x + 4 y = 0+4 y=4 y − intercept: ( 0, 4 ) Test Point (3 , 3)
y > −x + 4 y > −x + 4
3 > −3 + 4 3 >1
We shade the region that contains the test point. ANSWER:
18.
PROBLEM: y > x−2 SOLUTION: y = x−2 Set y = 0 y = x−2 0 = x−2 x=2
x − intercept : ( 2, 0 ) Set x = 0
y = x−2 0 = x−2 x=2 y − intercept = (0, 2) Test Point
y > x−2
(1, 1)
y > x−2 1 > 1− 2 1 > −1
We shade the region that contains the test point. ANSWER:
19.
PROBLEM: y ≥ −1 SOLUTION: y = −1 Set y = 0 x − intercept : none
Set x = 0 y = −1 y − intercept: (0, − 1) Test Point (0, 0)
y ≥ −1
0 ≥ −1
We shade the region that contains the test point. ANSWER:
20.
PROBLEM: y < −3 SOLUTION: y = −3 Set y = 0 x − intercept : none
Set x = 0 y = −3 y − intercept: (0, − 3) Test Point
y < −3
(–4 , –4)
−4 < −3
We shade the region that contains the test point.
ANSWER:
21.
PROBLEM: x− x+ 2 2 3 5 2 > − ( 2) + 2 2 1 2>− 2
We shade the region that contains the test point. ANSWER:
25.
PROBLEM: −2 x + 3 y > 6 SOLUTION: −2 x + 3 y = 6 Set y = 0 −2 x + 3 y = 6
−2 x + 3 ( 0 ) = 6 −2 6 x= = −3 −2 −2 x − intercept : ( −3, 0 ) Set x = 0 −2 x + 3 y = 6 −2 ( 0 ) + 3 y = 6 3y = 6 3 6 y= 3 3 y=2 y − intercept: ( 0, 2 ) Test Point (–2 , 2)
−2 x + 3 y > 6
−2 x + 3 y > 6 −2 ( −2 ) + 3 ( 2 ) > 6
10 > 6 We shade the region that contains the test point.
ANSWER:
26.
PROBLEM: 7 x − 2 y > 14 SOLUTION: 7 x − 2 y = 14 Set y = 0 7 x − 2 y = 14 7 x − 0 = 14 7 x = 14 7 14 x= 7 7 x=2 x − intercept : ( 2, 0 )
Set x = 0 7 x − 2 y = 14 0 − 2 y = 14 −2 14 y= −2 −2 y = −7 y − intercept: ( 0, −7 )
Test Point (3 , 2)
7 x − 2 y > 14 7 x − 2 y > 14 7 ( 3) − 2 ( 2 ) > 14 21 − 4 > 14 17 > 14
We shade the region that contains the test point. ANSWER:
27.
PROBLEM: 5 x − y < 10 SOLUTION: 5 x − y = 10 Set y = 0 5 x − y = 10 5 x − 0 = 10 5 x = 10 5 10 x= 5 5 x=2 x − intercept : ( 2, 0 )
Set x = 0
5 x − y = 10 0 − y = 10 y = −10 y − intercept: ( 0, −10 ) Test Point (0 , 0)
5 x − y < 10
5 x − y < 10 0 − 0 < 10 0 < 10
We shade the region that contains the test point. ANSWER:
28.
PROBLEM: x− y 0
47.
PROBLEM: Write an inequality that describes all ordered pairs whose y-coordinate is at least 2. SOLUTION: y ≥ 2 describes all ordered pairs whose y-coordinate is at least 2. ANSWER: y ≥ 2
48.
PROBLEM: Write an inequality that describes all ordered pairs whose x-coordinate is at most 5. SOLUTION: x ≤ 5 describes all ordered pairs whose x-coordinate is at most 5. ANSWER: x ≤ 5
3.9 Review Exercises and Sample Exam Chapter 3 Review Exercises
3.1 Rectangular Coordinate System Graph the given set of ordered pairs.
1.
PROBLEM: {( −3, 4 ) , ( −4, 0 ) , ( 0,3) , ( 2, 4 )} SOLUTION: The coordinates x = − 3 and y = 4 , indicate that we should mark a point 3 units left of and 4 units above the origin. The coordinates x = − 4 and y = 0 , indicate that we should mark a point 4 units on the x-axis left from the origin. The coordinates x = 0 and y = 3 , indicate that we should mark a point 3 units on the y-axis above the origin. The coordinates x = 2 and y = 4 , indicate that we should mark a point 2 units right of and 4 units above the origin.
ANSWER:
2.
PROBLEM: {( −5, 5) , ( −3, − 1) , ( 0, 0 ) , ( 3, 2 )} SOLUTION: The coordinates x = − 5 and y = 5 , indicate that we should mark a point 5 units left of and 5 units above the origin. The coordinates x = − 3 and y = −1 , indicate that we should mark a point 3 units left of and 1 unit below the origin. The coordinates x = 0 and y = 0 , indicate that we should mark a point on the origin. The coordinates x = 3 and y = 2 , indicate that we should mark a point 3 units right of and 2 units above the origin. ANSWER:
3.
PROBLEM: Graph the points (−3, 5), (−3, −3) and (3, −3) on a rectangular coordinate plane. Connect the points and calculate the area of the shape. SOLUTION: The coordinates x = − 3 and y = 5 , indicate that we should mark a point 3 units left of and 5 units above the origin. The coordinates x = − 3 and y = −3 , indicate that we should mark a point 3 units left of and 3 units below the origin. The coordinates x = 3 and y = −3 , indicate that we should mark a point 3 units right of and 3 units below the origin. ANSWER:
Base of the triangle is formed by the coordinates, (−3, −3) and (3, −3). Base = b − a = 3 − ( − 3 ) = 6
Height of the triangle is formed by the coordinates, (−3, −3) and (−3, 5). Height = b − a = 5 − ( − 3 ) = 8
1 × Base × Height 2 1 = ×6×8 2 = 24 sq.units
Area =
4.
PROBLEM: Graph the points (−4, 1), (0, 1), (0,−2) and (−4, −2) on a rectangular coordinate plane. Connect the points and calculate the area of the shape. SOLUTION: The coordinates x = − 4 and y = 1 , indicate that we should mark a point 4 units left of and 1 unit above the origin. The coordinates x = 0 and y = 1 , indicate that we should mark a point 1 unit above the origin on the y-axis. The coordinates x = 0 and y = −2 , indicate that we should mark a point two units below on the y-axis. The coordinates x = − 4 and y = −2 , indicate that we should mark a point 4 units left of and 2 units below the origin. ANSWER:
Length is shown by the coordinates (0, 1) and (–4, 1). Length = b − a = − 4 − 0 = 4
Breadth is formed by the coordinates (0, 1) and (0, –2). Breadth = b − a = −2 − 1 = 3
Area = length × breadth =4×3 = 12 sq.units
5.
PROBLEM: Graph the points (1, 0), (4, 0), (1,−5) and (4, −5) on a rectangular coordinate plane. Connect the points and calculate the perimeter of the shape. SOLUTION: The coordinates x = 1 and y = 0 , indicate that we should mark a point 1 unit right on the x-axis. The coordinates x = 4 and y = 0 , indicate that we should mark a point 4 units right on the x-axis. The coordinates x = 1 and y = −5 , indicate that we should mark a point 1 units right of and 5 units below the origin. The coordinates x = 4 and y = −5 , indicate that we should mark a point 4 units right of and 5 units below the origin. ANSWER:
Perimeter = 2(length + breadth) Length is formed by the cordinates (1, 0) and (1, –5). Length = b − a = −5 − 0 = 5
Breadth is formed by the coordinates (1, 0) and (4, 0). Breadth = b − a = 4 − 1 = 3 Perimeter = 2 ( length + breadth ) = 2 ( 5 + 3 ) = 2 ⋅8 = 16 units
6.
PROBLEM: Graph the points (−5, 2), (−5, −3), (1, 2) and (1, −3) on a rectangular coordinate plane. Connect the points and calculate the perimeter of the shape. SOLUTION: The coordinates x = − 5 and y = 2 , indicate that we should mark a point 5 units left of and 2 units above the origin. The coordinates x = − 5 and y = −3 , indicate that we should mark a point 5 units left of and 3 units below the origin. The coordinates x = 1 and y = 2 , indicate that we should mark a point 1 unit right of and 2 units above the origin. The coordinates x = 1 and y = −3 , indicate that we should mark a point 1 units right of and 3 unit below the origin. ANSWER:
Perimeter = 2(length + breadth) Length is formed by the cordinates (1, 2) and (–5, 2). Length = b − a = −5 − 1 = 6
Breadth is formed by the coordinates (1, 2) and (1, –3). Breadth = −3 − 2 = 5 Perimeter = 2 ( length + breadth ) = 2 ( 6 + 5 ) = 2 ⋅ 11 = 22 units
Calculate the distance between the given two points.
7.
PROBLEM: (−1, −2) and (5, 6) SOLUTION: Distance = =
( x2 − x1 )
2
+ ( y 2 − y1 )
2
( 5 − ( − 1) ) + ( 6 − ( − 2 ) ) 2
2
= 36 + 64 = 100 =10 units
ANSWER: 10 units
8.
PROBLEM: (2, −5) and (−2, −2) SOLUTION: Distance = =
( x2 − x1 ) ( −2 − 2 )
2
+ ( y 2 − y1 )
2
+ ( −2 − ( −5 ) )
2
2
= 16 + 9 = 25 =5 units
ANSWER: 5 units
9.
PROBLEM: (−9, −3) and (−8, 4) SOLUTION: Distance = =
( x2 − x1 )
2
+ ( y 2 − y1 )
2
( −8 − ( −9 ) ) + ( 4 − ( −3 ) )
= 1 + 49 = 50 = 25 ⋅ 2 =5 2 units
ANSWER: 5 2 units
2
2
10.
PROBLEM: (−1, 3) and (1, −3) SOLUTION: Distance = =
( x2 − x1 )
2
+ ( y 2 − y1 )
(1 − ( −1) ) + ( −3 − 3 ) 2
2
2
= 4 + 36 = 40 = 10 ⋅ 4 =2 10 units
ANSWER: 2 10 units Calculate the midpoint between the given points.
11.
PROBLEM: (−1, 3) and (5, −7) SOLUTION: ⎛ x + x2 y1 + y 2 Midpoint = ⎜ 1 , 2 ⎝ 2 = ( 2, − 2 )
⎞ ⎛ 5 − 1 −7 + 3 ⎞ , ⎟=⎜ ⎟ 2 ⎠ ⎠ ⎝ 2
ANSWER: ( 2, −2 )
12.
PROBLEM: (6, −3) and (−8, −11) SOLUTION: ⎛ x + x2 y1 + y 2 Midpoint = ⎜ 1 , 2 ⎝ 2 = ( − 1, − 7 )
ANSWER: ( −1, −7 )
⎞ ⎛ 6 − 8 − 3 − 11 ⎞ , ⎟=⎜ ⎟ 2 ⎠ ⎠ ⎝ 2
13.
PROBLEM: (7, −2) and (−6, −1) SOLUTION: ⎛ x + x2 y1 + y 2 Midpoint = ⎜ 1 , 2 ⎝ 2 ⎛1 3⎞ = ⎜ ,− ⎟ ⎝2 2⎠
⎞ ⎛ 7 − 6 −1 − 2 ⎞ , ⎟=⎜ ⎟ 2 ⎠ ⎠ ⎝ 2
⎛1 3⎞ ANSWER: ⎜ , − ⎟ ⎝2 2⎠ 14.
PROBLEM: (−6, 0) and (0, 0) SOLUTION: ⎛ x + x2 y1 + y 2 ⎞ ⎛ − 6 − 0 0 − 0 ⎞ Midpoint = ⎜ 1 , , ⎟=⎜ ⎟ 2 ⎠ ⎝ 2 2 ⎠ ⎝ 2 = ( − 3, 0 )
ANSWER: ( −3, 0 )
15.
PROBLEM: Show that the points (−1, −1), (1, −3), and (2, 0) form an isosceles triangle. SOLUTION: Points: (−1, −1) and (1, −3) a=
=
( x2 − x1 )
2
+ ( y 2 − y1 )
2
(1 − ( −1) ) + ( −3 − ( −1) ) 2
= 4+4 = 2⋅4 = 2 2units
Points: (1, −3) and (2, 0) b=
=
( x2 − x1 ) ( 2 − 1)
2
2
+ ( y 2 − y1 )
+ ( 0 − ( −3 ) )
2
= 1+ 9 = 10 units
Points: (2, 0) and (−1, −1)
2
2
c=
( x2 − x1 )
2
+ ( y 2 − y1 )
=
( −1 − 2 )
2
+ ( −1 − 0 )
2
2
= 9 +1 = 10 units
ANSWER: a = c. Hence, the points form an isosceles triangle.
16.
PROBLEM: Show that the points (2, −1), (6, 1), and (5, 3) form a right triangle. SOLUTION: Points: (2, −1) and (6, 1) a=
=
( x2 − x1 ) (6 − 2)
2
+ ( y 2 − y1 )
2
+ (1 − ( − 1) )
2
2
= 16 + 4 = 20 =
4 ⋅5
= 2 5units
Points: (6, 1) and (5, 3) b=
=
( x2 − x1 ) (5 − 6 )
2
2
+ ( y 2 − y1 )
+ ( 3 − 1)
2
2
= 1+ 4 = 5 units
Points: (5, 3) and (2, −1) c=
=
( x2 − x1 ) (2 − 5)
2
2
+ ( y 2 − y1 )
+ ( −1 − 3 )
2
2
= 9 + 16 = 25 = 5 units
Now to check, a2 + b2 = c2
(2 5 ) + ( 5 ) 2
2
= 52
20 + 5 = 25 25 = 25
ANSWER: Hence, the points form the vertices of a right triangle.
3.2 Graph by Plotting Points Determine whether or not the given point is a solution.
17.
PROBLEM: −5 x + 2 y = 7 ; (1, −1) SOLUTION: −5 x + 2 y = 7
−5 (1) + 2 ( −1) = 7 −5 − 2 = 7 −7 ≠ 7 ANSWER: (1, −1) is not a solution to −5 x + 2 y = 7 .
18.
PROBLEM: 6 x − 5 y = 4 ; ( −1, −2 ) SOLUTION: 6x − 5 y = 4
6 ( −1) − 5 ( −2 ) = 4 4=4 ANSWER: ( −1, −2 ) is a solution to 6 x − 5 y = 4 .
19.
PROBLEM: 3 y = x + 1 ; ( − 23 , 12 ) 4 SOLUTION: 3 y = x +1 4 1 3⎛ 2⎞ = ⎜ − ⎟ +1 2 4⎝ 3⎠ 1 1 = 2 2 ANSWER: ( − 23 , 12 ) is a solution to y =
3 x +1 . 4
20.
PROBLEM: 3 y = − x − 2 ; (10, −8 ) 5 SOLUTION: 3 y = − x−2 5 3 −8 = − (10 ) − 2 5 −8 = −8
3 ANSWER: (10, −8 ) is a solution to y = − x − 2 . 5 Find at least 5 ordered pair solutions and graph.
21.
PROBLEM: y = −x + 2 ANSWER:
x-value
y-value y = − x + 2 = − ( −2 ) + 2 = 4
Solution ( −2, 4)
x = −2 x = −1
y = − x + 2 = − ( −1) + 2 = 3
( −1,3)
y = − x + 2 = −0 + 2 = 2
( 0, 2)
x =1
y = − x + 2 = −1 + 2 = 1
(1,1)
x=2
y = − x + 2 = −2 + 2 = 0
( 2,0)
x=0
22.
PROBLEM: y = 2x − 3 ANSWER:
x-value
y-value y = 2 x − 3 = 2 ( −2 ) − 3 = −7
Solution ( −2, −7 )
x = −2 x = −1
y = 2 x − 3 = 2 ( −1) − 3 = −5
( −1, −5)
y = 2 x − 3 = 0 − 3 = −3
( 0, −3)
x =1
y = 2 x − 3 = 2 (1) − 3 = −1
(1, −1)
x=2
y = 2x − 3 = 2 ( 2) − 3 = 1
( 2,1)
x=0
23.
PROBLEM: 1 y = x−2 2 ANSWER:
x-value
x = −4
y-value 1 1 y = x − 2 = ( −4 ) − 2 = −4 2 2
Solution ( −4, −4)
x = −2
y=
1 1 x − 2 = ( −2 ) − 2 = −3 2 2
x=0
y=
1 1 x − 2 = ( 0 ) − 2 = −2 2 2
( 0, −2)
y=
1 1 x − 2 = ( 2 ) − 2 = −1 2 2
( 2, −1)
y=
1 1 x − 2 = ( 4) − 2 = 0 2 2
( 4,0)
x=2
x=4
( −2, −3)
24.
PROBLEM: 2 y=− x 3 ANSWER:
x-value
x = −3
y-value 2 2 y = − x = − ( −3) = 2 3 3
Solution ( −3, 2)
x=0
2 2 y = − x = − ( 0) = 0 3 3
x=3
2 2 y = − x = − ( 3 ) = −2 3 3
( 3, −2)
x=6
2 2 y = − x = − ( 6 ) = −4 3 3
( 6, −4)
x=9
2 2 y = − x = − ( 9 ) = −6 3 3
( 9, −6)
( 0,0)
25.
PROBLEM: y=3 ANSWER:
x-value
y-value
Solution
x = −2
y=3
( −2,3)
x = −1
y=3
( −1,3)
x=0
y=3
( 0,3)
x =1
y=3
(1,3)
x=2
y=3
( 2,3)
26.
PROBLEM: x = −3 ANSWER:
x-value
y-value
Solution
x = −3
y = −2
( −3, −2)
x = −3
y = −1
( −3, −1)
x = −3
y=0
( −3,0)
x = −3
y =1
( −3,1)
x = −3
y=2
( −3, 2)
27.
PROBLEM: x − 5 y = 15 SOLUTION: x − 5 y = 15 x − x − 5 y = 15 − x 5 y = x − 15 5 1 15 y = x− 5 5 5 1 y = x−3 5 ANSWER:
x-value
x=0
x=5
x = 10
x = 15
x = 20
y-value
Solution
y=
1 1 x − 3 = ( 0 ) − 3 = −3 5 5
( 0, −3)
y=
1 1 x − 3 = ( 5) − 3 = −2 5 5
( 5, −2)
y=
1 1 x − 3 = (10 ) − 3 = −1 5 5
(10, −1)
y=
1 1 x − 3 = (15) − 3 = 0 5 5
(15,0)
y=
1 1 x − 3 = ( 20 ) − 3 = 1 5 5
( 20,1)
28.
PROBLEM: 2 x − 3 y = 12 SOLUTION: 2 x − 3 y = 12 2 x − 2 x − 3 y = 12 − 2 x 3 y = 2 x − 12 3 2 12 y = x− 3 3 3 2 y = x−4 3 ANSWER:
x-value x = −3
x=0
x=3
x=6
x=9
y-value
Solution
y=
2 2 x − 4 = ( −3) − 4 = −6 3 3
( −3, −6)
y=
2 2 x − 4 = ( 0 ) − 4 = −4 3 3
( 0, −4)
y=
2 2 x − 4 = ( 3) − 4 = −2 3 3
( 3, −2)
y=
2 2 x − 4 = ( 6) − 4 = 0 3 3
( 6,0)
y=
2 2 x − 4 = (9) − 4 = 2 3 3
( 9, 2)
3.3 Graph using Intercepts Given the graph find the x‐ and y‐ intercepts
29.
PROBLEM:
ANSWER: The graph intersects the x-axis at –4. Hence the x-intercept is –4. The graph intersects the y-axis at –2. Hence the y-intercept is –2.
30.
PROBLEM:
SOLUTION: The graph intersects the x-axis at 1. Hence the x-intercept is 1. The graph intersects the y-axis at –3. Hence the y-intercept is –3. ANSWER: x-intercept is 1, y-intercept is –3
31.
PROBLEM:
ANSWER: The graph intersects the x-axis at 5. Hence the x-intercept is 5. The graph does not intersect the y-axis. Hence there is no y-intercept.
32.
PROBLEM:
SOLUTION: The graph intersects the x-axis at 0. Hence the x-intercept is 0. The graph intersects the y-axis at 0. Hence the y-intercept is 0. ANSWER: x-intercept is 0, y-intercept is 0. Find the intercepts and graph.
33.
PROBLEM: 3x − 4 y = 12 SOLUTION: Set y = 0 3x − 4 y = 12
3x − 4 ( 0 ) = 12 3x = 12 3 12 x= 3 3 x=4 x − intercept: ( 4, 0 ) Set x = 0 3 x − 4 y = 12 3 ( 0 ) − 4 y = 12 −4 y = 12 −4 12 y= −4 −4 y = −3 y − intercept: ( 0, −3)
ANSWER: ( 4, 0 ) , ( 0, −3)
34.
PROBLEM: 2 x − y = −4 SOLUTION: Set y = 0 2 x − y = −4 2 x − 0 = −4
2 x = −4 2 4 x=− 2 2 x = −2 x − intercept: ( −2, 0 ) Set x = 0 2 x − y = −4 0 − y = −4 y=4 y − intercept: ( 0, 4 )
ANSWER: ( −2, 0 ) , ( 0, 4 )
35.
PROBLEM: 1 1 x − y =1 2 3 SOLUTION: Set y = 0 1 1 x − y =1 2 3 1 x −0 =1 2 1 x =1 2 x=2 x − intercept: ( 2, 0 )
Set x = 0 1 1 x − y =1 2 3 1 1 ( 0) − y = 1 2 3 1 − y =1 3 y = −3 y − intercept: ( 0, −3)
ANSWER: ( 2, 0 ) , ( 0, −3)
36.
PROBLEM: 1 2 − x+ y =2 2 3 SOLUTION: Set y = 0 1 2 − x+ y = 2 2 3 1 − x+0 = 2 2 1 − x=2 2 x = −4 x − intercept: ( −4, 0 )
Set x = 0 1 2 − x+ y =2 2 3 2 0+ y = 2 3 2 y=2 3 3 2 3 ⋅ y = ⋅2 2 3 2 y=3 y − intercept: ( 0,3)
ANSWER: ( −4, 0 ) , ( 0,3)
37.
PROBLEM: 5 y = − x+5 3 SOLUTION: Set y = 0 5 y = − x+5 3 5 0 = − x+5 3 5 x=5 3 3 5 3 ⋅ x = ⋅5 5 3 5 x=3 x − intercept: ( 3, 0 )
Set x = 0 5 y = − x+5 3 y = 0+5
y =5 y − intercept: ( 0,5 )
ANSWER: ( 3, 0 ) , ( 0,5 )
38.
PROBLEM: y = −3 x + 4 SOLUTION: Set y = 0 y = −3 x + 4 0 = −3 x + 4 3x = 4 3 4 x= 3 3 1 x =1 3
⎛ 1 ⎞ x − intercept: ⎜ 1 , 0 ⎟ ⎝ 3 ⎠ Set x = 0 y = −3 x + 4 y = 0+4 y=4 y − intercept: ( 0, 4 )
⎛ 1 ⎞ ANSWER: ⎜1 , 0 ⎟ , ( 0, 4 ) ⎝ 3 ⎠
3.4 Graph using the y‐intercept and Slope Given the graph, determine the slope and y‐intercept.
39.
PROBLEM:
SOLUTION: y = mx + b The points shown in the graph are, (0, 1) and (2, –3). y − y −3 − 1 2 Slope, m = 2 1 = = − = −1 x2 − x1 2 − 0 2 Substitute (0, 1) in the above equation. y = −x + b 1= 0+b 1= b
y − intercept: ( 0,1) ANSWER: −1 , y − intercept: ( 0,1)
40.
PROBLEM:
SOLUTION: y = mx + b The points shown in the graph are, (0, –4) and (2, 2).
y2 − y1 2 − ( −4 ) 6 = = =3 x2 − x1 2−0 2 Substitute (2, 2) in the above equation. y = 3x + b 2 = 6+b −4 = b Slope, m =
y − intercept: ( 0, −4 ) ANSWER: 3 , ( 0, −4 ) Determine the slope given two points.
41.
PROBLEM: (−3, 8) and (5, −6) SOLUTION: y −y −6 − 8 14 7 Slope, m = 2 1 = =− =− x2 − x1 5 − ( −3) 8 4 ANSWER: −
7 4
42.
PROBLEM: (0, −5) and (−6, 3) SOLUTION: y − y 3 − ( −5 ) 8 4 = =− Slope, m = 2 1 = x2 − x1 −6 − 0 −6 3 ANSWER: −
43.
4 3
PROBLEM: (1/2, −2/3) and (1/4, −1/3) SOLUTION:
1 ⎛ 2⎞ 1 − −⎜− ⎟ y2 − y1 4 3 ⎝ 3⎠ Slope, m = = = 3 =− 1 1 1 x2 − x1 3 − − 4 2 4 4 ANSWER: − 3 44.
PROBLEM: (5, −3/4) and (2, −3/4) SOLUTION:
3 ⎛ 3⎞ − −⎜− ⎟ y −y 4 ⎝ 4⎠ 0 Slope, m = 2 1 = = =0 x2 − x1 2−5 −3 ANSWER: 0
Express in slope‐intercept form and identify the slope and y‐intercept.
45.
PROBLEM: 12 x − 4 y = 8 SOLUTION: 12 x − 4 y = 8 12 x − 12 x − 4 y = 8 − 12 x 4 y = 12 x − 8 4 12 8 y = x− 4 4 4 y = 3x − 2 m = 3; b = –2
Slope, m = 3; y − intercept : ( 0, −2 ) ANSWER: Slope, m = 3; y − intercept : ( 0, −2 )
46.
PROBLEM: 3x − 6 y = 24 SOLUTION: 3x − 6 y = 24 3x − 3x − 6 y = 24 − 3x 6 y = 3x − 24 6 3 24 y = x− 6 6 6 1 y = x−4 2 1 m = ; b = –4 2 1 Slope, m = ; y − intercept : ( 0, −4 ) 2 ANSWER: y =
1 1 x − 4 , Slope, m = ; y − intercept : ( 0, −4 ) 2 2
47.
PROBLEM: 1 3 − x + y =1 3 4 SOLUTION: 1 3 − x + y =1 3 4 1 1 3 1 − x + x + y = 1+ x 3 3 4 3 3 1 y = 1+ x 4 3 4 3 4 4 1 ⋅ y = ⋅1 + ⋅ x 3 4 3 3 3 4 4 y = x+ 9 3 4 4 m = ;b = 9 3 4 ⎛ 4⎞ Slope, m = ; y − intercept : ⎜ 0, ⎟ 9 ⎝ 3⎠ 4 ⎛ 4⎞ ANSWER: Slope, m = ; y − intercept : ⎜ 0, ⎟ 9 ⎝ 3⎠
48.
PROBLEM: −5 x + 3 y = 0 SOLUTION: −5 x + 3 y = 0 −5 x + 5 x + 3 y = 0 + 5 x 3 y = 5x 3 5 y= x 3 3 5 y= x 3 5 m = ;b = 0 3 5 Slope, m = ; y − intercept : ( 0, 0 ) 3 ANSWER: y =
5 5 x , Slope, m = ; y − intercept : ( 0, 0 ) 3 3
Graph using the slope and y‐intercept.
49.
PROBLEM: y = −x + 3 SOLUTION: y = −x + 3 m = −1; b = 3 −1 rise Slope, m = ; y − intercept : ( 0,3) = 1 run Starting from the point (0, 3) we will use the slope to mark another point down 1 unit and to the right 1 unit. Join the two points with a straight line. ANSWER:
50.
PROBLEM: y = 4x −1 SOLUTION: y = 4x −1 m = 4; b = −1 4 rise Slope, m = = ; y − intercept : ( 0, −1) 1 run Starting from the point (0, –1) we will use the slope to mark another point up 4 units and to the right 1 unit. Join the two points with a straight line.
ANSWER:
51.
PROBLEM: y = −2 x SOLUTION: y = −2 x m = −2; b = 0 2 Slope, m = − ; y − intercept : ( 0, 0 ) 1 Starting from the point (0, 0) we will use the slope to mark another point down 2 units and to the right 1 unit. Join the two points with a straight line. ANSWER:
52.
PROBLEM: 5 y = − x+3 2 SOLUTION: 5 y = − x+3 2 5 m = − ;b = 3 2 5 rise Slope, m = − = ; y − intercept : ( 0,3) 2 run Starting from the point (0, 3) we will use the slope to mark another point down 5 units and to the right 2 units. Join the two points with a straight line. ANSWER:
53.
PROBLEM: 2x − 3 y = 9 SOLUTION: 2x − 3 y = 9 2x − 2x − 3y = 9 − 2x 3y = 2x − 9 3 2 9 y = x− 3 3 3 2 y = x −3 3 2 m = ; b = −3 3 2 Slope, m = ; y − intercept : ( 0, −3) 3
Starting from the point (0, –3) we will use the slope to mark another point up 2 units and to the right 3 units. Join the two points with a straight line. ANSWER:
54.
PROBLEM: 3 2x + y = 3 2 SOLUTION: 3 2x + y = 3 2 3 2x − 2x + y = 3 − 2x 2 3 y = −2 x + 3 2 2 3 2 2 ⋅ y = − ⋅ 2x + ⋅ 3 3 2 3 3 4 y =− x+2 3 4 m = − ;b = 2 3 4 rise Slope, m = − = ; y − intercept : ( 0, 2 ) 3 run Starting from the point (0, 2) we will use the slope to mark another point down 4 units and to the right 3 units. Join the two points with a straight line.
ANSWER:
55.
PROBLEM: y=0 SOLUTION: y=0 Slope, m = 0; y − intercept : ( 0, 0 )
Starting from the point (0, 0) we will draw an horizontal line with no slope. The xaxis will be the line. ANSWER:
56.
PROBLEM: x − 4y = 0 SOLUTION: x − 4y = 0 x − x − 4 y = −x 4y = x 4 1 y= x 4 4 1 y= x 4 1 m = ;b = 0 4 1 rise Slope, m = = ; y − intercept : ( 0, 0 ) 4 run Starting from the point (0, 0) we will use the slope to mark another point up 1 unit and to the right 4 units. Join the two points with a straight line. ANSWER:
3.5 Finding Linear Equations Given the graph, determine the equation of the line.
57.
PROBLEM:
SOLUTION: The points shown in the graph are (0, 1) and (2, –3). y = mx + b y − y −3 − 1 −4 = = −2 Slope, m = 2 1 = 2 x2 − x1 2 − 0 y = −2 x + b
Substitute (0, 1) in the equation above, y = −2 x + b 1= 0+b 1= b ANSWER: Completing the equation, y = −2 x + 1 .
58.
PROBLEM:
SOLUTION: The points shown in the graph are (2, 2) and (0, –4). y = mx + b y − y −4 − 2 −6 = =3 Slope, m = 2 1 = x2 − x1 0 − 2 −2 y = 3x + b
Substitute (2, 2) in the equation above, y = 3x + b 2 = 3( 2) + b 2 = 6+b −4 = b ANSWER: Completing the equation, y = 3 x − 4 .
59.
PROBLEM:
SOLUTION: The points shown in the graph are (0, –5 ) and (1, –5). y = mx + b y − y −5 − ( −5 ) 0 Slope, m = 2 1 = = =0 x2 − x1 1− 0 1 y = 0+b y=b
Substitute (0, –5 ) in the equation above, y=b −5 = b ANSWER: Completing the equation, y = −5 .
60.
PROBLEM:
SOLUTION: The points shown in the graph are (4, 0) and (4, 1). y = mx + b y − y 1− 0 = undefined Slope, m = 2 1 = x2 − x1 4 − 4 ANSWER: The slope is undefined. Hence, the equation of the line is a vertical line given by x = 4. Find the equation of a line given the slope and a point on the line.
61.
PROBLEM: m = 1/2; (−4, 8) SOLUTION: y = mx + b 1 y = x+b 2 Substitute (–4, 8 ) in the equation above, 1 y = x+b 2 1 8 = ( −4 ) + b 2 10 = b Completing the equation, y = mx + b 1 ANSWER: y = x + 10 2
62.
PROBLEM: m = −1/5; (−5, −9) SOLUTION: y = mx + b 1 y = − x+b 5 Substitute (−5, −9) in the above equation, 1 y = − x+b 5 1 −9 = − ( −5 ) + b 5 −9 − 1 = b −10 = b Completing the equation, y = mx + b 1 ANSWER: y = − x − 10 5
63.
PROBLEM: m = 2/3; (1, −2) SOLUTION: y = mx + b 2 y = x+b 3 Substitute (1, −2) in the equation above, 2 y = x+b 3 2 −2 = + b 3 8 − =b 3 Completing the equation, y = mx + b 2 8 ANSWER: y = x − 3 3
64.
PROBLEM: m = −3/4; (2, −3) SOLUTION: y = mx + b 3 y = − x+b 4 Substitute (2, −3) in the above equation, 3 y = − x+b 4 3 −3 = − ( 2 ) + b 4 3 − =b 2 Completing the equation, y = mx + b 3 3 ANSWER: y = − x − 4 2
Find the equation of the line given two points on the line.
65.
PROBLEM: (−5, −5) and (10, 7) SOLUTION: y = mx + b 7 − ( −5 ) 12 4 y −y = = Slope, m = 2 1 = x2 − x1 10 − ( −5 ) 15 5
4 x+b 5 Substitute (10, 7) in the equation above, 4 y = x+b 5 4 7 = (10 ) + b 5 −1 = b Completing the equation y = mx + b , 4 ANSWER: y = x − 1 5 y=
66.
PROBLEM: (−6, 12) and (3, −3) SOLUTION: y = mx + b y −y 15 5 −3 − 12 Slope, m = 2 1 = =− =− 9 3 x2 − x1 3 − ( −6 )
5 y = − x+b 3 Substitute (3, −3) in the equation above, 5 y = − x+b 3 5 −3 = − ( 3 ) + b 3 2=b Completing the equation y = mx + b , 5 ANSWER: y = − x + 2 3 67.
PROBLEM: (2, −1) and (−2, 2) SOLUTION: y = mx + b y − y 2 − ( −1) 3 =− Slope, m = 2 1 = −2 − 2 x2 − x1 4 3 y = − x+b 4 Substitute(2, −1) in the equation above, 3 y = − x+b 4 3 −1 = − ( 2 ) + b 4 1 =b 2 Completing the equation y = mx + b , 3 1 ANSWER: y = − x + 4 2
68.
PROBLEM: (5/2, −2) and (−5, 5/2) SOLUTION: y = mx + b 5 9 − −2 y2 − y1 2 ( ) 3 = =− 2 =− Slope, m = 5 15 x2 − x1 5 −5 − 2 2 3 y = − x+b 5 Substitute (−5, 5/2) in the equation above, 3 y = − x+b 5 5 3 = − ( −5 ) + b 2 5 1 − =b 2 Completing the equation y = mx + b , 3 1 ANSWER: y = − x − 5 2
69.
PROBLEM: (7, −6) and (3, −6) SOLUTION: y = mx + b y − y −6 − ( −6 ) =0 Slope, m = 2 1 = x2 − x1 3−7 y =b Substitute (3, −6) in the equation above, y=b −6 = b Completing the equation y = mx + b ,
ANSWER: y = −6
70.
PROBLEM: (10, 1) and (10, −3) SOLUTION: y = mx + b −3 − 1 y −y 4 = − = undefined Slope, m = 2 1 = x2 − x1 10 − 10 0 ANSWER: The slope is undefined. Hence the line is a vertical line x =10.
3.6 Parallel and Perpendicular Lines Determine if the lines are parallel, perpendicular or neither.
71.
PROBLEM: ⎧−3x + 7 y = 14 ⎨ ⎩ 6 x − 14 y = 42 SOLUTION: −3x + 7 y = 14 −3x + 3 x + 7 y = 14 + 3x 7 y = 3x + 14
7 3 14 y = x+ 7 7 7 3 y = x+2 7 3 m= 7
6 x − 14 y = 42 6 x − 6 x − 14 y = 42 − 6 x 14 y = 6 x − 42 14 6 42 y = x− 14 14 14 3 y = x −3 7 3 m= 7
ANSWER: Slopes are equal; hence the lines are parallel to each other.
72.
PROBLEM: ⎧2 x + 3 y = 18 ⎨ ⎩ 2 x − 3 y = 36 SOLUTION: 2 x + 3 y = 18 2 x − 2 x + 3 y = 18 − 2 x
3 y = −2 x + 18 3 2 18 y = − x+ 3 3 3 2 y = − x+6 3 2 m=− 3
2 x − 3 y = 36 2 x − 2 x − 3 y = 36 − 2 x 3 y = 2 x − 36 3 2 36 y = x− 3 3 3 2 y = x − 12 3 2 m= 3
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
73.
PROBLEM: ⎧ x + 4y = 2 ⎨ ⎩8 x − 2 y = −1 SOLUTION: x + 4y = 2 x − x + 4y = 2 − x
8 x − 2 y = −1 8 x − 8 x − 2 y = −1 − 8 x 2 y = 8x + 1 2 8 1 y = x+ 2 2 2 1 y = 4x + 2 m=4
4 y = −x + 2 4 1 2 y =− x+ 4 4 4 1 1 y =− x+ 4 2 1 m=− 4
ANSWER: Slopes are negative reciprocals; hence the lines are perpendicular to each other.
74.
PROBLEM: ⎧y = 2 ⎨ ⎩x = 2 SOLUTION: y=2
x=2
m=0
m = undefined
ANSWER: Slopes are negative reciprocals; hence the lines are perpendicular to each other. Find the equation of the line in slope‐intercept form:
75.
PROBLEM: Parallel to 5 x − y = 15 passing through (−10, −1). SOLUTION: 5 x − y = 15
5 x − 5 x − y = 15 − 5 x y = 5 x − 15 Here the given line has slope m = 5 and thus m& = 5 . We substitute this slope and the given point into point-slope form.
Point (−10, −1)
Slope m& = 5
y − y1 = m ( x − x1 )
y − ( −1) = 5 ( x − ( −10 ) ) y + 1 = 5 x + 50 y = 5 x + 49 ANSWER: y = 5 x + 49
76.
PROBLEM: Parallel to x − 3 y = 1 passing through (2, −2). SOLUTION: x − 3y = 1 x − x − 3y = 1− x
3y = x −1 3 1 1 y = x− 3 3 3 1 1 y = x− 3 3 1 1 and thus m& = . We substitute this slope and 3 3 the given point into point-slope form.
Here the given line has slope m =
Point
Slope 1 (2, −2) m& = 3 y − y1 = m ( x − x1 ) 1 ( x − 2) 3 1 2 y+2= x− 3 3 1 8 y = x− 3 3 y − ( −2 ) =
ANSWER: y =
1 8 x− 3 3
77.
PROBLEM: Perpendicular to 8 x − 6 y = 4 passing through (8, −1). SOLUTION: 8x − 6 y = 4
8x − 8x − 6 y = 4 − 8x 6 y = 8x − 4 6 8 4 y = x− 6 6 6 4 2 y = x− 3 3 4 1 3 and thus m⊥ = − = − . We substitute this 4 3 4 3 slope and the given point into point-slope form.
Here the given line has slope m =
Point
Slope 3 (8, −1) m⊥ = − 4 y − y1 = m ( x − x1 ) y − ( −1) = −
3 ( x − 8) 4
3 y +1 = − x + 6 4 3 y = − x+5 4
3 ANSWER: y = − x + 5 4 78.
PROBLEM: Perpendicular to 7 x + y = 14 passing through (5, 1). SOLUTION: 7 x + y = 14
7 x − 7 x + y = 14 − 7 x y = −7 x + 14 Here the given line has slope m = −7 a and thus m⊥ = and the given point into point-slope form.
1 . We substitute this slope 7
Point
Slope 1 m⊥ = ( 5, 1) 7 y − y1 = m ( x − x1 ) 1 ( x − 5) 7 1 2 y = x+ 7 7 y −1 =
ANSWER: y =
79.
1 2 x+ 7 7
PROBLEM: Parallel to y = 1 passing through (4, −1). SOLUTION: Here the given line has slope m = 0 and thus m& = 0 . We substitute this slope and
the given point into point-slope form. Point (4, −1)
Slope m& = 0
y − y1 = m ( x − x1 )
y − ( −1) = 0 ( x − 4 ) y = −1 ANSWER: y = −1
80.
PROBLEM: Perpendicular to y = 1 passing through (4, −1). SOLUTION: Here the given line has slope m = 0 a and thus m⊥ = undefined . We substitute
this slope and the given point into point-slope form. Point (4, −1)
Slope m⊥ = undefined
y − y1 = m ( x − x1 )
y − ( −1) = undefined ( x − 4 ) x−4 =0 x=4
ANSWER: x = 4 3.7 Introduction to Functions Determine the domain and range and state whether it is a function or not.
81.
PROBLEM: {(−10, −1), (−5, 2), (5, 2)} SOLUTION:
−10 −5
−1 2
5 ANSWER: Domain:{–10, –5, 5} Range: {–1, 2} The relation is a function because each x-value corresponds to exactly one of yvalue.
82.
PROBLEM: {(−12, 4), (−1, −3), (−1, −2)} SOLUTION:
−12 −1
4 –3 –2
ANSWER: Domain:{–12, –1} Range: {4, –3, –2} The relation is not a function because each x-value does not correspond to exactly one of y-value.
83.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is all real numbers R = (–∞,∞). The relation is a function because it passes the vertical line test.
84.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that 2 is the maximum y-value and any x-value greater than that is not represented in the relation. Hence, the domain is equal to 2. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is R = (–∞,∞) and the range is {2}. The relation is a function because it passes the vertical line test.
85.
PROBLEM:
SOLUTION: The graph indicates that −3 is the minimum x-value and any x-value greater than that is represented in the relation. Hence, the domain is greater than or equal to −3 or the interval notation [−3, ∞) . If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will not intersect the graph only once; therefore it is not a function. ANSWER: The domain is [−3, ∞) and the range is all real numbers R = (–∞,∞). The relation is not a function because it does not pass the vertical line test.
86.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that −4 is the minimum y-value and any y-value greater than that is represented in the relation. Hence, the range is greater than or equal to −4 or the interval notation [−4, ∞) .Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is [−3, ∞). The relation is a function because it passes the vertical line test. Given the following:
87.
PROBLEM: f ( x) = 9 x − 4 find f (−1) . SOLUTION: f ( x) = 9 x − 4
f (−1) = 9 ( −1) − 4 = −9 − 4 = −13 ANSWER: −13
88.
PROBLEM: f ( x) = −5 x + 1 find f (−3) . SOLUTION: f ( x) = −5 x + 1
f (−3) = −5 ( −3) + 1 = 15 + 1 = 16 ANSWER: 16
89.
PROBLEM: 1 1 g ( x) = x − find g ( − 13 ) . 2 3 SOLUTION: 1 1 g ( x) = x − 2 3 1⎛ 1⎞ 1 1 g ( − 13 ) = ⎜ − ⎟ − = − 2⎝ 3⎠ 3 2 1 ANSWER: − 2
90.
PROBLEM: 3 1 ⎛2⎞ g ( x) = − x + find g ⎜ ⎟ . 4 3 ⎝3⎠ SOLUTION: 3 1 g ( x) = − x + 4 3 3⎛2⎞ 1 1 ⎛2⎞ g⎜ ⎟ = − ⎜ ⎟+ = − 4⎝3⎠ 3 6 ⎝3⎠ ANSWER: −
91.
1 6
PROBLEM: f ( x) = 9 x − 4 find x when f ( x) = 0 . SOLUTION: f ( x) = 9 x − 4 When f ( x) = 0 0 = 9x − 4 9x = 4 9 4 x= 9 9 4 x= 9 ANSWER: x =
4 9
92.
PROBLEM: f ( x) = −5 x + 1 find x when f ( x) = 2 . SOLUTION: f ( x) = −5 x + 1 When f ( x) = 2 f ( x) = −5 x + 1 2 = −5 x + 1 5x = 1 − 2 5 x = −1 5 1 x=− 5 5 1 x=− 5 ANSWER: x = −
93.
1 5
PROBLEM: 1 1 g ( x) = x − find x when g ( x) = 1 . 2 3 SOLUTION: 1 1 g ( x) = x − 2 3 When g ( x) = 1 1 1 1= x − 2 3 1 4 x= 2 3 8 x= 3 ANSWER: x =
8 3
94.
PROBLEM: 3 1 g ( x) = − x + find x when g ( x) = −1 . 4 3 SOLUTION: 3 1 g ( x) = − x + 4 3 When g ( x) = −1 3 1 −1 = − x + 4 3 3 4 x= 4 3 4 3 4 4 ⋅ x= ⋅ 3 4 3 3 16 x= 9 16 ANSWER: x = 9
Given the graph of a function f ( x) determine the following:
95.
PROBLEM: f (3) SOLUTION: From the graph when x = 3, y = – 2. Hence, f (3) is – 2. ANSWER: – 2
96.
PROBLEM: x when f ( x) = 4 ANSWER: If f ( x) = 4 , y = 4. From the graph when y = 4 , x = 0.
3.8 Linear Inequalities (two variables) Is the ordered pair a solution to the given inequality?
97.
PROBLEM: 6 x − 2 y ≤ 1 ; (−3, −7) SOLUTION: 6x − 2 y ≤ 1
6 ( −3) − 2 ( −7 ) ≤ 1 −18 + 14 ≤ 1 −4 ≤ 1 ANSWER: (−3, −7) is a solution to the inequality 6 x − 2 y ≤ 1 .
98.
PROBLEM: −3x + y > 2 ; (0, 2) SOLUTION: −3x + y > 2 0+2 > 2 2 > 2 Incorrect ANSWER: (0, 2) is not a solution to the inequality −3x + y > 2 .
99.
PROBLEM: 6 x − 10 y < −1 ; (5, −3) SOLUTION: 6 x − 10 y < −1
6 ( 5 ) − 10 ( −3) < −1 30 + 30 < −1 60 < −1
Incorrect
ANSWER: (5, −3) is not a solution to the inequality 6 x − 10 y < −1 .
100.
PROBLEM: 1 x − y > 0 ; (1, 4) 3 SOLUTION: 1 x− y >0 3 1 1 − ( 4) > 0 3 1 − > 0 Incorrect 3
1 ANSWER: (1, 4) is not a solution to the inequality x − y > 0 . 3 101.
PROBLEM: y > 0 ; (−3, −1) SOLUTION: y>0 −1 > 0 Incorrect ANSWER: (−3, −1) is not a solution to the inequality 6 x − 10 y < −1 .
102.
PROBLEM: x ≤ −5 ; (−6, 4) SOLUTION: x ≤ −5 −6 ≤ −5 ANSWER: (−6, 4) is as solution to the inequality x ≤ −5 .
Graph the solution set.
103.
PROBLEM: y ≥ −2 x + 1 SOLUTION: y = −2 x + 1 Set y = 0 y = −2 x + 1 0 = −2 x + 1
2x = 1 1 x= 2 ⎛1 ⎞ x − intercept : ⎜ , 0 ⎟ ⎝2 ⎠ Set x = 0 y = −2 x + 1 y = 0 +1 y =1 y − intercept = (0, 1)
Test Point
y ≥ −2 x + 1
(1, 1)
y ≥ −2 x + 1 1 ≥ −2 + 1 1 ≥ −1
We shade the region that contains the test point.
ANSWER:
104.
PROBLEM: y < 3x − 4 SOLUTION: y = 3x − 4 Set y = 0 y = 3x − 4 0 = 3x − 4
3x = 4 4 x= 3 ⎛ 1 ⎞ x − intercept : ⎜1 , 0 ⎟ ⎝ 3 ⎠ Set x = 0 y = 3x − 4 y = 0−4 y = −4 y − intercept : (0, − 4)
Test Point
y < 3x − 4
(2, –1)
y < 3x − 4 −1 < 6 − 4 −1 < 2
We shade the region that contains the test point. ANSWER:
105.
PROBLEM: −x + y ≤ 3 SOLUTION: −x + y = 3 Set y = 0 −x + y = 3
−x + 0 = 3 x = −3 x − intercept : ( −3, 0 ) Set x = 0 −x + y = 3 0+ y = 3 y =3 y − intercept = (0, 3)
Test Point
−x + y ≤ 3
(1, 1)
−x + y ≤ 3 −1 + 1 ≤ 3 0≤3
We shade the region that contains the test point.
ANSWER:
106.
PROBLEM: 5 1 x+ y ≤2 2 2 SOLUTION: 5 1 x+ y =2 2 2 Set y = 0 5 1 x+ y =2 2 2 5 x+0 = 2 2 4 x= 5
⎛4 ⎞ x − intercept : ⎜ , 0 ⎟ ⎝5 ⎠ Set x = 0 5 1 x+ y =2 2 2 1 0+ y = 2 2 y=4 y − intercept: (0, 4)
Test Point
5 1 x+ y ≤2 2 2
(–2, –2)
5 1 x+ y ≤2 2 2 5 1 ( −2 ) + ( −2 ) ≤ 2 2 2 −6 ≤ 2
We shade the region that contains the test point. ANSWER:
107.
PROBLEM: 3x − 5 y > 0 SOLUTION: 3x − 5 y = 0 Set y = 0 3x − 5 y = 0
3x − 0 = 0 x=0 x − intercept : ( 0, 0 ) Set x = 0 3x − 5 y = 0 0 − 5y = 0 y=0 y − intercept = (0, 0)
Test Point
3x − 5 y > 0
(1, –1 )
3x − 5 y > 0 3+5 > 0 8>0
We shade the region that contains the test point. ANSWER:
108.
PROBLEM: y>0 SOLUTION: y=0 x − intercept : none
y − intercept : (0, 0) Test Point
y>0
(1, –1)
y>0 1> 0
We shade the region that contains the test point.
ANSWER:
Chapter 3 Sample Exam
1.
PROBLEM: Graph the points (−4, −2), (−4, 1) and (0, −2) on a rectangular coordinate plane. Connect the points and calculate the area of the shape. SOLUTION: The coordinates x = − 4 and y = 2 , indicate that we should mark a point 4 units left of and 2 units above the origin. The coordinates x = − 4 and y = 1 , indicate that we should mark a point 4 units left of and 1 unit below the origin. The coordinates x = 0 and y = −2 , indicate that we should mark a point 2 units below the origin on the y-axis. ANSWER:
The shape formed is a , Triangle
1 × base × height 2 Base is formed by the coordinates (0, − 2) and (4, − 2). Area =
Base = 4 − 0 = 4 Height is formed by the coordinates ( − 4,1) and ( − 4, − 2) Height = 1 − ( −2 ) = 3 Area =
2.
1 × 4 × 3 = 6 sq.units 2
PROBLEM: Is (−2, 4) a solution to 3x − 4 y = −10 ? Justify your answer. SOLUTION: 3x − 4 y = −10
3 ( −2 ) − 4 ( 4 ) = −10 −6 − 16 = −10 −22 ≠ −10 ANSWER: (−2, 4) is not a solution to 3x − 4 y = −10 . Given the set of x‐values {−2, −1, 0, 1, 2} find the corresponding y‐values and graph the following:
3.
PROBLEM: y = x −1 ANSWER:
x-value
y-value y = x − 1 = −2 − 1 = −3
Solution ( −2, −3)
x = −2 x = −1
y = x − 1 = −1 − 1 = −2
( −1, −2)
y = x − 1 = 0 − 1 = −1
( 0, −1)
x =1
y = x −1 = 1 −1 = 0
(1,0)
x=2
y = x −1 = 2 −1 = 1
( 2,1)
x=0
4.
PROBLEM: y = −x +1 ANSWER:
x-value
y-value y = − x + 1 = − ( −2 ) + 1 = 3
Solution ( −2,3)
x = −2 x = −1
y = − x + 1 = − ( −1) + 1 = 2
( −1, 2)
y = −x +1 = 0 +1 = 1
( 0,1)
x =1
y = − x + 1 = −1 + 1 = 0
(1,0)
x=2
y = − x + 1 = −2 + 1 = −1
( 2, −1)
x=0
5.
PROBLEM: On the same set of axes graph y = 4 and x = −3 . Give the point where they intersect. SOLUTION: x-value
y-value y=4
Solution ( −2, 4)
x = −2 x = −1
y=4
( −1, 4)
y=4
( 0, 4)
x =1
y=4
(1, 4)
x=2
y=4
( 2, 4)
x=0
x-value
y-value y = −2
Solution ( −2, −3)
x = −3 x = −3
y = −1
( −1, −3)
y=0
( 0, −3)
x = −3
y =1
(1, −3)
x = −3
y=2
( 2, −3)
x = −3
ANSWER: The lines intersect at ( −3, 4 ) . Find the x‐ and y‐intercepts and use those points to graph the following:
6.
PROBLEM: 2x − y = 8 SOLUTION: Set y = 0 2x − y = 8
2x − 0 = 8 2x = 8 x=4 x − intercept : ( 4, 0 ) Set x = 0 2x − y = 8 2 ( 0) − y = 8 y = −8 y − intercept : ( 0, −8)
ANSWER: ( 4, 0 ) , ( 0, −8 )
7.
PROBLEM: 12 x + 5 y = 15 SOLUTION: Set y = 0 12 x + 5 y = 15 12 x = 15 15 5 1 x = = =1 12 4 4 ⎛ 1 ⎞ x − intercept : ⎜1 , 0 ⎟ ⎝ 4 ⎠ Set x = 0 12 x + 5 y = 15 0 + 5 y = 15 5 y = 15 y =3
y − intercept : ( 0,3)
⎛ 1 ⎞ ANSWER: ⎜1 , 0 ⎟ , ( 0,3) ⎝ 4 ⎠
8.
PROBLEM: Calculate the slope of the line passing through (−4, −5) and (−3, 1). SOLUTION: 1 − ( −5 ) y −y 6 = =6 Slope,m = 2 1 = x2 − x1 −3 − ( −4 ) 1 ANSWER: 6
Determine the slope and y‐intercept. Use them to graph the following:
9.
PROBLEM: 3 y = − x+6 2 SOLUTION: 3 y = − x+6 2 3 m = − ;b = 6 2 3 rise ; y − intercept : ( 0, 6 ) m=− = 2 run Starting from the point (0, 6) we will use the slope to mark another point down 3 units and to the right 2 units. Join the two points with a straight line. ANSWER: m = −
3 rise = ; y − intercept : ( 0, 6 ) 2 run
10.
PROBLEM: 5x − 2 y = 6 SOLUTION: 5x − 2 y = 6
5x − 5x − 2 y = 6 − 5x 2 y = 5x − 6 2 5 6 y = x− 2 2 2 5 y = x −3 2 5 m = ; b = −3 2 5 rise ; y − intercept : ( 0, −3) m= = 2 run Starting from the point (0, –3) we will use the slope to mark another point up 5 units and to the right 2 units. Join the two points with a straight line.
ANSWER: m =
11.
5 rise = ; y − intercept : ( 0, −3) 2 run
PROBLEM: Given m = −3 , determine m⊥ . SOLUTION: m = −3 1 1 m⊥ = = − m 3 ANSWER: −
12.
1 3
PROBLEM:
Are the given lines parallel, perpendicular, or neither?
SOLUTION: −2 x + 3 y = −12 −2 x + 2 x + 3 y = −12 + 2 x
3 y = 2 x − 12 2 y = x−4 3 2 m= 3
⎧−2 x + 3 y = −12 ⎨ ⎩ 4 x − 6 y = 30
4 x − 6 y = 30 4 x − 4 x − 6 y = 30 − 4 x 6 y = 4 x − 30 2 y = x −5 3 2 m= 3
ANSWER: Slopes are equal; hence the lines are parallel to each other.
13.
PROBLEM: Determine the slope of the given lines: a. y = −2 b. x = 13 c. Are these lines parallel, perpendicular, or neither? ANSWER: a. y = −2 m=0
b. x = 13 m = undefined c. The slopes are negative reciprocals. Hence, the lines are perpendicular to each other. 14.
PROBLEM: Given the graph, determine slope-intercept form of the line:
SOLUTION: y = mx + b The points shown in the graph are, (–4, 0) and (0, 6). y −y 6−0 6 3 = = Slope,m = 2 1 = x2 − x1 0 − ( −4 ) 4 2
3 x+b 2 Substitute (0, 6) in the above equation, 3 6 = (0) + b 2 6=b Completing the equation y = mx + b , y=
ANSWER: y =
3 x+6 2
15.
PROBLEM: Determine the equation of the line with slope m = − 34 passing through (8, 1). SOLUTION: y = mx + b m = − 34 ; Point : ( 8,1)
3 y = − x+b 4 Substitute ( 8,1) in the above equation, 1= −
3 (8) + b 4
7=b Completing the equation, y = mx + b 3 ANSWER: y = − x + 7 4 16.
PROBLEM: Find the equation to the line passing through (−2, 3) and (4, 1). SOLUTION: y = mx + b Points are (−2, 3) and (4, 1) y −y −2 1− 3 1 = =− Slope,m = 2 1 = x2 − x1 4 − ( −2 ) 6 3
1 y = − x+b 3 Substitute (4, 1) in the above equation, 1 y = − x+b 3 1 1 = − ( 4) + b 3 7 =b 3 Completing the equation y = mx + b , 1 7 ANSWER: y = − x + 3 3
17.
PROBLEM: Find the equation of the line parallel to 5 x − y = 6 passing through (−1, −2). SOLUTION: 5x − y = 6
5x − 5x − y = 6 − 5x y = 5x − 6 Here the given line has slope m = 5 and thus m& = 5 . We substitute this slope and the given point into point-slope form. Point (−1, −2)
Slope m& = 5
y − y1 = m ( x − x1 )
y − ( −2 ) = 5 ( x − ( −1) ) y + 2 = 5x + 5 y = 5x + 3 ANSWER: y = 5 x + 3
18.
PROBLEM: Find the equation of the line perpendicular to − x + 2 y = 4 passing through (1/2, 5). SOLUTION: −x + 2 y = 4 −x + x + 2 y = 4 + x
2y = x + 4 1 y = x+2 2 1 1 and thus m⊥ = − = −2 . We substitute this 1 2 2 slope and the given point into point-slope form.
Here the given line has slope m =
Point ⎛1 ⎞ ⎜ ,5 ⎟ ⎝2 ⎠
Slope m⊥ = −2
y − y1 = m ( x − x1 ) 1⎞ ⎛ y − 5 = −2 ⎜ x − ⎟ 2⎠ ⎝ y − 5 = −2 x + 1 y = −2 x + 6 ANSWER: y = −2 x + 6
4 Given a linear function f ( x) = − x + 2 determine the following: 5 19.
20.
PROBLEM: f (10) SOLUTION: 4 f ( x) = − x + 2 5 4 f (10 ) = − (10 ) + 2 = −6 5 ANSWER: −6 PROBLEM: x where f(x) = 0 SOLUTION: 4 f ( x) = − x + 2 5 Where f(x) = 0 4 0=− x+2 5 4 x=2 5 5 x= 2 ANSWER: x =
5 2
21.
PROBLEM: Graph the solution set: 3x − 4 y > 4 SOLUTION: 3x − 4 y = 4 Set y = 0 3x − 0 = 4
3x = 4 4 x= 3 ⎛ 1 ⎞ x − intercept : ⎜1 , 0 ⎟ ⎝ 3 ⎠ Set x = 0 3x − 4 y = 4 0 − 4y = 4 y = −1 y − intercept : ( 0, −1) Test Point (1, –1)
3x − 4 y > 4 3+ 4 > 4 7>4
We shade the region that contains the test point. ANSWER:
22.
PROBLEM: Graph the solution set: y − 2 x ≥ 0 SOLUTION: y − 2x = 0 Set y = 0 y − 2x = 0
0 − 2x = 0 x=0 x − intercept : ( 0, 0 ) Set x = 0 y − 2x = 0 y−0 = 0 y=0 y − intercept : ( 0, 0 ) Test Point
y − 2x ≥ 0
(1, –1)
1+ 2 ≥ 0 3≥0
We shade the region that contains the test point. ANSWER:
23.
PROBLEM: A rental car company charges $32.00 to rent a car plus $0.52 per mile driven. Write an equation that gives the cost of renting the car in terms of the number of miles driven. Use the formula to determine the cost of renting the car and driving it 46 miles. SOLUTION: Let the number of miles driven by the car be represented by x. Total cost = $32 + $0.52x f(x) = $32 + $0.52x f(46) = 32 + 0.52(46) = $55.92 ANSWER: Cost of renting the car and driving it 46 miles is $55.92.
24.
PROBLEM: A car was purchased new for $12,000 and was sold 5 years later for $7,000. Write a linear equation that gives the value of the car in terms of its age in years. SOLUTION:
Given: (0,12000) and (5,7000) 7000 − 12000 5000 Depreciation= =− 5−0 5 = −1000 Total value = Original value – Depreciating amount × Number of years ANSWER: Total value = $12,000 – $1,000x
25.
PROBLEM: The area of a rectangle is 72 square meters. If the width measures 4 meters, then determine the length of the rectangle. SOLUTION: Length of the rectangle is represented by x. Area = Length × Width 72 sq. meters = x × 4 meters x = 18 meters ANSWER: Length of the rectangle is 18 meters.
Give the coordinates of points A, B, C, D and E. 1.
PROBLEM:
SOLUTION: The coordinates of A are The coordinates of B are The coordinates of C are The coordinates of D are The coordinates of E are
x = 3 and y = 5 . x = − 2 and y = 3 . x = − 5 and y = 0 . x = 1 and y = −3 . x = − 3 and y = −4 .
ANSWER: The coordinates of A are (3, 5). The coordinates of B are (–2, 3). The coordinates of C are (–5, 0). The coordinates of D are (1, –3). The coordinates of E are (–3, –4).
2.
PROBLEM:
ANSWER: The coordinates of A are x = 2 and y = 4 , (2, 4). The coordinates of B are x = 0 and y = 0 , ( 0, 0 ) .
The coordinates of C are x = − 5 and y = −2 , ( −5, −2 ) . The coordinates of D are x = 0 and y = −4 , ( 0, −4 ) . The coordinates of E are x = 5 and y = −5 , ( 0, −5 ) . 3.
PROBLEM:
SOLUTION: The coordinates of A are x = 0 and y = 6 . The coordinates of B are x = − 4 and y = 3 . The coordinates of C are x = − 8 and y = 0 .
The coordinates of D are x = − 6 and y = −6 . The coordinates of E are x = 8 and y = −9 . ANSWER: The coordinates of A are (0, 6). The coordinates of B are (–4, 3). The coordinates of C are (–8, 0). The coordinates of D are (–6, –6). The coordinates of E are (8, –9).
4.
PROBLEM:
ANSWER: The coordinates of A are x = 0 and y = 9 , ( 0,9 ) .
The coordinates of B are x = 6 and y = 9 , ( 6,9 ) . The coordinates of C are x = − 2 and y = 3 , ( −2,3) . The coordinates of D are x = 4 and y = −3 , ( 4, −3) . The coordinates of E are x = − 8 and y = −6 , ( −8, −6 ) . 5.
PROBLEM:
SOLUTION: The coordinates of A are The coordinates of B are The coordinates of C are The coordinates of D are The coordinates of E are
x = − 10 and y = 25 . x = 30 and y = 20 . x = 0 and y = 10 . x = 15 and y = 0 . x = 25 and y = −10 .
ANSWER: The coordinates of A are (–10, 25). The coordinates of B are (30, 20). The coordinates of C are (0, 10). The coordinates of D are (15, 0). The coordinates of E are (25, –10).
6.
PROBLEM:
ANSWER: The coordinates of A are x = 0 and y = 15 , ( 0,15 ) .
The coordinates of B are x = 10 and y = 15 , (10,15 ) . The coordinates of C are x = − 10 and y = 5 , ( −10,5 ) . The coordinates of D are x = 40 and y = 0 , ( 40, 0 ) . The coordinates of E are x = − 10 and y = −15 , ( −10, −15 ) . Graph the given set of ordered pairs.
7.
PROBLEM: {( −4, 5) , ( −1, 1) , ( −3, −2 ) , ( 5, −1)} SOLUTION: The coordinates x = − 4 and y = 5 , indicate that we should mark a point 4 units left of and 5 units above the origin.
The coordinates x = − 1 and y = 1 , indicate that we should mark a point 1 unit left of and 1 unit above the origin. The coordinates x = − 3 and y = −2 , indicate that we should mark a point mark a point 3 units left of and 2 units below the origin. The coordinates x = 5 and y = −1 , indicate that we should mark a point 5 units right of and 1 unit below the origin. ANSWER:
8.
PROBLEM: {( −15, −10 ) , ( −5, 10 ) , (15,10 ) , ( 5, −10 )} SOLUTION: The coordinates x = − 15 and y = −10 , indicate that we should mark a point 15 units left of and 10 units below the origin. The coordinates x = − 5 and y = 10 , indicate that we should mark a point 5 units left of and 10 units above the origin. The coordinates x = 15 and y = 10 , indicate that we should mark a point mark a point 15 units right of and 10 units above the origin. The coordinates x = 5 and y = −10 , indicate that we should mark a point 5 units right of and 10 units below the origin. ANSWER:
9.
PROBLEM: {( −2, 5) , (10, 0 ) , ( 2, −5) , ( 6, −10 )} SOLUTION: The coordinates x = − 2 and y = 5 , indicate that we should mark a point 2 units left of and 5 units above the origin. The coordinates x = 10 and y = 0 , indicate that we should mark a point 10 units right of the origin. The coordinates x = 2 and y = −5 , indicate that we should mark a point 2 units right on the x-axis and 5 below the origin. The coordinates x = 6 and y = −10 , indicate that we should mark a point 6 units right of and 10 units below the origin. ANSWER:
10.
PROBLEM: {( −8, 3) , ( −4, 6 ) , ( 0, − 6 ) , ( 6, 9 )} SOLUTION: The coordinates x = − 8 and y = 3 , indicate that we should mark a point 8 units left of and 3 units above the origin. The coordinates x = − 4 and y = 6 , indicate that we should mark a point 4 units left of and 6 units above the origin. The coordinates x = 0 and y = −6 , indicate that we should mark a point 6 units below the origin on the y-axis. The coordinates x = 6 and y = 9 , indicate that we should mark a point 6 units right of and 9 units above the origin.
ANSWER:
11.
PROBLEM: {( −10, 5 ) , ( 20, −10 ) , ( 30,15 ) , ( 50, 0 )} SOLUTION: The coordinates x = − 10 and y = 5 , indicate that we should mark a point 10 units left of and 5 units above the origin. The coordinates x = 20 and y = −10 , indicate that we should mark a point 20 units right of and 10 units below the origin. The coordinates x = 30 and y = 15 , indicate that we should mark a point 30 units right of and 15 units above the origin. The coordinates x = 50 and y = 0 , indicate that we should mark a point 50 units right of the origin on the x-axis. ANSWER:
12.
PROBLEM: ⎧⎛ 5 1 ⎞ ⎛ 1 1 ⎞ ⎛ 2 ⎞ ⎛ 5 ⎞⎫ ⎨⎜ − , − ⎟ , ⎜ − , ⎟ , ⎜ , −1⎟ , ⎜ ,1⎟ ⎬ ⎠ ⎝ 3 ⎠⎭ ⎩⎝ 3 2 ⎠ ⎝ 3 2 ⎠ ⎝ 3 SOLUTION: The coordinates x = −
1 5 and y = − , indicate that we should mark 2 3
2 1 a point 1 units left of and unit below the origin. 3 2
The coordinates x = −
left of and
1 1 1 and y = , indicate that we should mark a point unit 2 3 3
1 unit above the origin. 2
The coordinates x =
2 2 and y = −1 , indicate that we should mark a point unit to 3 3
the right and 1 unit below the origin The coordinates x =
2 5 and y = 1 , indicate that we should mark a point 1 units 3 3
right of and 1 unit above the origin. ANSWER:
13.
PROBLEM: ⎧⎛ 3 4 ⎞ ⎛ 2 4 ⎞ ⎛ ⎫ 2⎞ ⎨⎜ − , − ⎟ , ⎜ , ⎟ , ⎜1, − ⎟ , ( 0,1) ⎬ 3⎠ ⎩⎝ 5 3 ⎠ ⎝ 5 3 ⎠ ⎝ ⎭ SOLUTION:
The coordinates x = −
4 3 3 and y = − , indicate that we should mark a point unit 3 5 5
1 units below the origin. 3 4 2 2 The coordinates x = and y = , indicate that we should mark a point unit 3 5 5 1 right of and 1 units above the origin. 3 2 The coordinates x = 1 and y = − , indicate that we should mark a point 1 unit 3 2 right of and unit below the origin. 3 The coordinates x = 0 and y = 1 , indicate that we should mark a point 1 unit above the origin on the y-axis. left of and 1
ANSWER:
14.
PROBLEM: {( −3.5, 0 ) , ( −1.5, 2 ) , ( 0,1.5) , ( 2.5, −1.5)} SOLUTION: The coordinates x = − 3.5 and y = 0 , indicate that we should mark a point 3.5 units left of the origin on the x-axis. The coordinates x = − 1.5 and y = 2 , indicate that we should mark a point 1.5 units left of and 2 unit above the origin. The coordinates x = 0 and y = 1.5 , indicate that we should mark a point 1.5 units above the origin on the y-axis. The coordinates x = 2.5 and y = −1.5 , indicate that we should mark a point 2.5 units right of and 1.5 units below the origin. ANSWER:
15.
PROBLEM: {( −0.8, 0.2 ) , ( −0.2, −0.4 ) , ( 0, −1) , ( 0.6, −0.4 )} SOLUTION: The coordinates x = − 0.8 and y = 0.2 , indicate that we should mark a point 0.8 unit left of and 0.2 unit above the origin. The coordinates x = − 0.2 and y = −0.4 , indicate that we should mark a point 0.2 unit left of and 0.4 unit below the origin.
The coordinates x = 0 and y = −1 , indicate that we should mark a point 1 unit below the origin on the y-axis. The coordinates x = 0.6 and y = −0.4 , indicate that we should mark a point 0.6 unit right and 0.4 unit below the origin. ANSWER:
16.
PROBLEM: {( −1.2, −1.2 ) , ( −0.3, −0.3) , ( 0, 0 ) , ( 0.6, 0.6 ) , (1.2,1.2 )} SOLUTION: The coordinates x = − 1.2 and y = −1.2 , indicate that we should mark a point 1.2 units left of and 1.2 units below the origin. The coordinates x = − 0.3 and y = −0.3 , indicate that we should mark a point 0.3 unit left of and 0.3 unit below the origin. The coordinates x = 0 and y = 0 , indicate that the point is marked on the origin. The coordinates x = 0.6 and y = 0.6 , indicate that we should mark a point 0.6 unit right and 0.6 unit above the origin. The coordinates x = 1.2 and y = 1.2 , indicate that we should mark a point 1.2 units right and 1.2 units above the origin. ANSWER:
State the quadrant in which the given point lies.
17.
PROBLEM: (−3, 2) SOLUTION: The point is plotted in quadrant two (QII) because the x-coordinate is negative and the y-coordinate is positive. ANSWER: Quadrant two (QII)
18.
PROBLEM: (5, 7) SOLUTION: The point is plotted in quadrant one (QI) because the x-coordinate is positive and the y-coordinate is positive. ANSWER: quadrant one (QI)
19.
PROBLEM: (−12, −15) SOLUTION: The point is plotted in quadrant three (QIII) because the x-coordinate is negative and the y-coordinate is negative. ANSWER: Quadrant three (QIII)
20.
PROBLEM: (7, −8) SOLUTION: The point is plotted in quadrant four (QIV) because the xcoordinate is positive and the y-coordinate is negative. ANSWER: quadrant four (QIV)
21.
PROBLEM: (−3.8, 4.6) SOLUTION: The point is plotted in quadrant two (QII) because the x-coordinate is negative and the y-coordinate is positive. ANSWER: Quadrant two (QII)
22.
PROBLEM: (17.3, 1.9) SOLUTION: The point is plotted in quadrant one (QI) because the x-coordinate is positive and the y-coordinate is positive. ANSWER: quadrant one (QI)
23.
PROBLEM: ( − 18 , − 85 ) SOLUTION: The point is plotted in quadrant three (QIII) because the x-coordinate is negative and the y-coordinate is negative. ANSWER: Quadrant three (QIII)
24.
PROBLEM: ( 34 , − 14 ) SOLUTION: The point is plotted in quadrant four (QIV) because the xcoordinate is positive and the y-coordinate is negative. ANSWER: quadrant four (QIV)
25.
PROBLEM:
x > 0 and y < 0
SOLUTION: The point is plotted in quadrant four (QIV) because the x-coordinate is positive and the y-coordinate is negative. ANSWER: Quadrant four (QIV)
26.
PROBLEM:
x < 0 and y < 0
SOLUTION: The point is plotted in quadrant three (QIII) because the xcoordinate is negative and the y-coordinate is negative. ANSWER: quadrant three (QIII)
27.
PROBLEM:
x < 0 and y > 0
SOLUTION: The point is plotted in quadrant two (QII) because the x-coordinate is negative and the y-coordinate is positive. ANSWER: Quadrant two (QII)
28.
PROBLEM:
x > 0 and y > 0
SOLUTION: The point is plotted in quadrant one (QI) because the x-coordinate is positive and the y-coordinate is positive. ANSWER: quadrant one (QI) The average price of a gallon of regular unleaded gasoline in US cities is given in the following line graph. Use the graph to answer the following questions. (Data: Bureau of Labor Statistics)
Scale: x-axis:5 years = 1unit y-axis: $1 = 5 units 29.
PROBLEM: What was the average price of a gallon of unleaded gasoline in 2004? SOLUTION: Year 2004 corresponds to 7 on the x-coordinate. The corresponding y-value is 9. This value corresponds to $1.8. ANSWER: In the year 2004, the average price of a gallon of unleaded gasoline is $1.8.
30.
PROBLEM: What was the average price of a gallon of unleaded gasoline in 1976? SOLUTION: Year 1976 corresponds to 0 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.6. ANSWER: In the year 1976, the average price of a gallon of unleaded gasoline is $0.6.
31.
PROBLEM: In what years was the average price of a gallon of unleaded gasoline $1.20? SOLUTION: $1.20 corresponds to y = 6. This occurs in three data points, when x = 1, x = 2, and x = 5. These values correspond to 1980, 1984, and 1996 respectively. ANSWER: The average price of a gallon of unleaded gasoline was $1.20 in 1980, 1984, and 1996.
32.
PROBLEM: What is the price increase of a gallon on gasoline from 1980 to 2008? SOLUTION: Year 1980 corresponds to 1 on the x-coordinate. The corresponding y-value is 6. This value corresponds to $1.2. Year 2008 corresponds to 8 on the x-coordinate. The corresponding y-value is 16. This value corresponds to $3.2. From 1980 to 2008, the price per gallon of gasoline increases from $1.2 to $3.2. ANSWER: Hence the increase in the price is $3.2 – $1.2 = $2.0.
33.
PROBLEM: What was the percentage increase in the price of a gallon of unleaded gasoline from 1976 to 1980? SOLUTION: Year 1976 corresponds to 0 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.6. Year 1980 corresponds to 1 on the x-coordinate. The corresponding y-value is 6. This value corresponds to $1.2. From 1976 to 1980, the price per gallon of gasoline increases from $0.6 to $1.2. Hence the increase in the price is $1.2 – $0.6 = $0.6.
Increase in price × 100 Initial price $0.6 = × 100 $0.6 = 100%
% Increase in price =
ANSWER: The percentage increase in the price of a gallon of unleaded gasoline from 1976 to 1980 is 100%.
34.
PROBLEM: What was the percentage increase in the price of a gallon of unleaded gasoline from 2000 to 2008? SOLUTION: Year 2000 corresponds to 6 on the x-coordinate. The corresponding y-value is 7.5. This value corresponds to $1.5. Year 2008 corresponds to 8 on the x-coordinate. The corresponding y-value is 16. This value corresponds to $3.2. From 2000 to 2008, the price per gallon of gasoline increases from $1.5 to $3.2. Hence the increase in the price is $3.2 – $1.5 = $1.7. Increase in price % Increase in price = × 100 Initial price $1.7 × 100 = $1.5 = 113% ANSWER: The percentage increase in the price of a gallon of unleaded gasoline from 2000 to 2008 is 113%.
The average price of all purpose white flour in US cities from 1980 to 2008 is given in the following line graph. Use the graph to answer the following questions. (Data: Bureau of Labor Statistics)
Scale: x-axis: 1 unit = 4 years y-axis: 1 unit = $0.1 35.
PROBLEM: What was the average price per pound of all purpose white flour in 2000? SOLUTION: Year 2000 corresponds to 6 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.3. ANSWER: In the year 2000, the average price per pound of all purpose white flour is $0.3.
36.
PROBLEM: What was the average price per pound of all purpose white flour in 2008? SOLUTION: Year 2008 corresponds to 7 on the x-coordinate. The corresponding y-value is 5. This value corresponds to $0.5. ANSWER: In the year 2008, the average price per pound of all purpose white flour is $0.5.
37.
PROBLEM: In what year did the price of flour average $0.25 per pound? SOLUTION: $0.25 corresponds to y = 2.5. This occurs in one data point, when x = 3. This value corresponds to the year 1992. ANSWER: The price of the flour averages $0.25 per pound in the year 1992.
38.
PROBLEM: In what years did the price of flour average $0.20 per pound? SOLUTION: $0.20 corresponds to 2 on the y-coordinate. This data point occurs in three places, when x = 0, x = 1, and x = 2. These values correspond to 1980, 1984, and 1988 respectively. ANSWER: The price of flour averages $0.20 per pound in the years 1980, 1984, and 1988.
39.
PROBLEM: What was the percentage increase in flour from the year 2000 to 2008? SOLUTION: Year 2000 corresponds to 5 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.3. Year 2008 corresponds to 7 on the x-coordinate. The corresponding y-value is 5. This value corresponds to $0.5. From 2000 to 2008, the price per pound of flour increases from $0.3 to $0.5. Hence the increase in the price is $0.5 – $0.3 = $0.2. Increase in price % Increase in price = × 100 Initial price $0.2 = × 100 $0.3 = 67% ANSWER: The percentage increase in the price per pound of flour from 2000 to 2008 is 67%.
40.
PROBLEM: What was the percentage increase in flour from the year 1992 to 2000? SOLUTION: Year 1992 corresponds to 4 on the x-coordinate. The corresponding y-value is 2.5. This value corresponds to $0.25. Year 2000 corresponds to 5 on the x-coordinate. The corresponding y-value is 3. This value corresponds to $0.3. From 1992 to 2000, the price per pound of flour increases from $0.25 to $0.3. Hence the increase in the price is $0.3 – $0.25 = $0.05. Increase in price % Increase in price = × 100 Initial price $0.05 = × 100 $0.25 = 20% ANSWER: The percentage increase in the price per pound of flour from 1992 to 2000 is 20%.
Given the following data create a line graph.
41.
PROBLEM: The percentage of total high school graduates who enroll in college. (Source: Digest of Education Statistics) Year Percentage 1969 36% 1979 40% 1989 47% 1999 42% ANSWER:
42.
PROBLEM: The average daily temperature given in degrees Fahrenheit in May. Exam 8:00 am 12:00 pm 4:00 pm 8:00 pm 12:00am 4:00 am
Temperature 60 72 75 67 60 55
ANSWER:
Calculate the area of the shape formed by connecting the following set of vertices.
43.
PROBLEM: {(0, 0), (0, 3), (5, 0), (5, 3)} SOLUTION:
When the vertices are connected a rectangle is formed. Area of rectangle = length × breadth Length = 5 – 0 = 5 units. Breadth = 3 – 0 = 3 units. ANSWER: Area = 5 units × 3 units = 15 sq. units
44.
PROBLEM: {(−1, −1), (−1, 1), (1, −1), (1, 1)} SOLUTION:
When the vertices are connected a square is formed. Area of square = (side)2 side = 1 – (–1) = 2 units. ANSWER: Area = (2 units)2 = 4 sq.units
45.
PROBLEM: {(−2, −1), (−2, 3), (5, 3), (5, −1)} SOLUTION:
When the vertices are connected a rectangle is formed. Area of rectangle = length × breadth Length = 5 – (– 2) = 7 units. Breadth = 3 – (–1) = 4 units. ANSWER: Area = 7 units × 4 units = 28 sq. units
46.
PROBLEM: {(−5, −4), (−5, 5), (3, 5), (3, −4)} SOLUTION:
When the vertices are connected a rectangle is formed. Area of rectangle = length × breadth Length = 3 – (– 5) = 8 units. Breadth = 5 – (–4) = 9 units. ANSWER: Area = 8 units × 9 units = 72 sq.units
47.
PROBLEM: {(0, 0), (4, 0), (2, 2)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 4 – 0 = 4 units. Height = 2 – 0 = 2 units. ANSWER: 1 Area = × 4 units × 2 units = 4 sq.units 2
48.
PROBLEM: {(−2, −2), (2, −2), (0, 2)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 2 – (– 2) = 4 units. Height = 2 – (– 2) = 4 units. 1 ANSWER: Area = × 4 units × 4 units = 8 sq.units 2
49.
PROBLEM: {(0, 0), (0, 6), (3, 4)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 6 – 0 = 6 units. Height = 3 – 0 = 3 units. ANSWER: 1 Area = × 6 units × 3 units = 9 sq.units 2
50.
PROBLEM: {(−2, 0), (5, 0), (3, −3)} SOLUTION:
When the vertices are connected a triangle is formed. 1 Area of triangle = × base × height 2 Base = 5 – (– 2) = 7 units. Height = 0 – (– 3) = 3 units. 1 1 ANSWER: Area = × 7 units × 3 units = 10 sq.units 2 2
Part B: Distance Formula Calculate the distance between the given two points.
51.
PROBLEM: (−5, 3) and (−1, 6) SOLUTION: x2 = −1, x1 = −5, y2 = 6, y1 = 3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣ −1 − ( −5 ) ⎤⎦ + ( 6 − 3 )
=
[ −1 + 5] + ( 6 − 3 )
=
[ 4] + ( 3)
2
2
2
2
2
2
= 16 + 9 =
25
= 5 units
ANSWER: 5 units
52.
PROBLEM: (6, −2) and (−2, 4) SOLUTION: x2 = −2, x1 = 6, y2 = 4, y1 = −2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
( −2 − 6 )
2
+ ⎡⎣ 4 − ( −2 ) ⎤⎦
=
( −2 − 6 )
2
+ [ 4 + 2]
=
( −8 )
=
64 + 36
2
2
+ [6]
2
= 100 = 10 units
ANSWER: 10 units
2
53.
PROBLEM: (0, 0) and (5, 12) SOLUTION: x2 = 5, x1 = 0, y2 = 12, y1 = 0 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 5 − 0 ) + (12 − 0 )
=
( 5 ) + (12 )
=
25 + 144
2
2
2
2
2
= 169 = 13 units
ANSWER: 13 units
54.
PROBLEM: (−6, −8) and (0, 0) SOLUTION: x2 = 0, x1 = −6 , y2 = 0, y1 = −8 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣ 0 − ( −6 ) ⎤⎦ + ⎣⎡ 0 − ( −8 ) ⎦⎤
=
( −6 )
2
2
+ ( −8 )
2
= 36 + 64 = 100 = 10 units
ANSWER: 10 units
2
55.
PROBLEM: (−7, 8) and (5, −1) SOLUTION: x2 = 5, x1 = −7 , y2 = −1, y1 = 8 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣5 − ( −7 ) ⎤⎦ + ( −1 − 8 )
=
[ 5 + 7 ] + ( −1 − 8 )
=
[12]
2
2
2
+ ( −9 )
2
2
2
= 144 + 81 =
225
= 15 units
ANSWER: 15 units
56.
PROBLEM: (−1, −2) and (9, 22) SOLUTION: x2 = 9, x1 = −1 , y2 = 22, y1 = −2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣9 − ( −1) ⎤⎦ + ⎡⎣ 22 − ( −2 ) ⎤⎦
=
[9 + 1]
=
[10]
2
2
2
+ ( 22 + 2 )
+ [ 24 ]
2
= 100 + 576 =
676
= 26 units
ANSWER: 26 units
2
2
57.
PROBLEM: (−1, 2) and (−7/2, −4) SOLUTION: x2 = −7/2 , x1 = −1 , y2 = −4, y1 = 2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
=
2 ⎡ 7 ⎤ ⎢⎣ − 2 − ( −1) ⎥⎦ + ( −4 − 2 )
=
2 ⎡ 7 ⎤ ⎢⎣ − 2 + 1⎥⎦ + ( −4 − 2 )
=
2 ⎡ 5 ⎤ ⎢⎣ − 2 ⎥⎦ + ( −6 )
=
25 + 36 4
2
2
169 4 13 = units 2 =
ANSWER: 13 units 2
58.
PROBLEM: ⎛ 1 1⎞ ⎜ − , ⎟ and ⎝ 2 3⎠
⎛ 5 11 ⎞ ⎜ ,− ⎟ ⎝2 3 ⎠
SOLUTION: x2 = d=
5 1 11 1 , x1 = − , y2 = − , y1 = 2 2 3 3
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
=
⎡ 5 ⎛ 1 ⎞⎤ ⎛ 11 1 ⎞ ⎢ 2 − ⎜ − 2 ⎟⎥ + ⎜ − 3 − 3 ⎟ ⎝ ⎠⎦ ⎝ ⎠ ⎣
=
⎡5 1⎤ ⎛ 11 1 ⎞ ⎢⎣ 2 + 2 ⎥⎦ + ⎜⎝ − 3 − 3 ⎟⎠
=
⎡6⎤ ⎛ 12 ⎞ ⎢⎣ 2 ⎥⎦ + ⎜⎝ − 3 ⎟⎠
=
[3]
2
2
2
+ ( −4 )
2
= 9 + 16 =
25
= 5 units
ANSWER: 5 units
2
2
2
59.
PROBLEM: ⎛ 1 2⎞ ⎛ 1⎞ ⎜ − , ⎟ and ⎜1,− ⎟ ⎝ 3 3⎠ ⎝ 3⎠ SOLUTION: 1 1 2 x2 =1, x1 = − , y2 = − , y1 = 3 3 3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
=
⎡ ⎛ 1 ⎞⎤ ⎛ 1 2⎞ ⎢1 − ⎜ − 3 ⎟ ⎥ + ⎜ − 3 − 3 ⎟ ⎠⎦ ⎝ ⎠ ⎣ ⎝
=
⎡ 1⎤ ⎛ 1 2⎞ ⎢1 + 3 ⎥ + ⎜ − 3 − 3 ⎟ ⎣ ⎦ ⎝ ⎠
=
⎡ 3 + 1⎤ ⎛ 1 2⎞ ⎢⎣ 3 ⎥⎦ + ⎜⎝ − 3 − 3 ⎟⎠
=
⎡4⎤ ⎛ −3 ⎞ ⎢⎣ 3 ⎦⎥ + ⎝⎜ 3 ⎠⎟
=
16 9 + 9 9
=
25 9
2
2
2
=
5 units 3
ANSWER: 5 units 3
2
2
2
2
60.
PROBLEM: ⎛1 3⎞ ⎜ , − ⎟ and ⎝2 4⎠
⎛3 1⎞ ⎜ , ⎟ ⎝2 4⎠
SOLUTION: x2 = d=
3 1 1 3 , x1 = , y2 = , y1 = − 2 2 4 4
( x2 − x1 ) + ( y2 − y1 ) 2
2
2
⎡ 1 ⎛ 3 ⎞⎤ ⎛3 1⎞ = ⎜ − ⎟ + ⎢ − ⎜ − ⎟⎥ ⎝2 2⎠ ⎣ 4 ⎝ 4 ⎠⎦ 2
⎛3 1⎞ ⎡1 3⎤ = ⎜ − ⎟ + ⎢ + ⎥ ⎣4 4⎦ ⎝2 2⎠ 2
⎛2⎞ ⎛4⎞ = ⎜ ⎟ +⎜ ⎟ ⎝2⎠ ⎝4⎠ =
(1)
=
2 units
2
+ (1)
2
2
2
ANSWER: 2 units
61.
PROBLEM: (1, 2) and (4, 3) SOLUTION: x2 = 4, x1 = 1, y2 = 3, y1 = 2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 4 − 1)
=
( 3)
2
2
+ (3 − 2)
+ (1)
= 9+1 = 10 units
ANSWER: 10 units
2
2
2
2
62.
PROBLEM: (2,−4) and (−3, −2) SOLUTION: x2 = −3, x1 = 2, y2 = −2, y1 = −4 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( −3 − 2 )
=
( −3 − 2 )
=
( −5 )
=
25 + 4
=
29 units
2
+ ⎡⎣ −2 − ( −4 ) ⎦⎤ + [ −2 + 4 ]
2
2
+ ( 2)
2
2
2
ANSWER: 29 units
63.
PROBLEM: (−1, 5) and (1, −3) SOLUTION: x2 = 1, x1 = −1, y2 = −3, y1 = 5 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣1 − ( −1) ⎤⎦ + ( −3 − 5 )
=
[1 + 1]
=
[ 2]
=
68
=
4 × 17
2
2
2
+ ( −3 − 5 )
+ ( −8 )
= 2 17 units
ANSWER: 2 17 units
2
2
2
2
64.
PROBLEM: (1, −7) and (5, −1) SOLUTION: x2 = 5, x1 = 1, y2 = −1, y1 = −7 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 5 − 1)
=
( 5 − 1)
=
( 4)
2
+ ⎡⎣ −1 − ( −7 ) ⎤⎦
2
2
2
2
+ [ −1 + 7 ]
2
+ (6)
2
= 16 + 36 = 52 =
4 × 13
= 2 13 units
ANSWER: 2 13 units
65.
PROBLEM: (−7, −3) and (−1, 6) SOLUTION: x2 = −1, x1 = −7, y2 = 6, y1 = −3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣ −1 − ( −7 ) ⎤⎦ + ⎡⎣ 6 − ( −3 ) ⎦⎤
=
[ −1 + 7 ]
=
[6]
2
2
2
+ [ 6 + 3]
2
+ [9 ]
= 36 + 81 = 117 = 13 × 9 = 3 13 units
ANSWER: 3 13 units
2
2
66.
PROBLEM: (0, 1) and (1, 0) SOLUTION: x2 = 1, x1 = 0, y2 = 0, y1 = 1 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
(1 − 0 )
=
(1)
2
2
+ ( 0 − 1)
+ (1)
2
2
2
= 1+1 =
2 units
ANSWER: 2 units
67.
PROBLEM: (−0.2, −0.2) and (1.8, 1.8) SOLUTION: x2 = 1.8, x1 = −0.2, y2 = 1.8, y1 = −0.2 d=
( x2 − x1 ) + ( y2 − y1 ) 2
2
=
⎡⎣1.8 − ( −0.2 ) ⎤⎦ + ⎡⎣1.8 − ( −0.2 ) ⎤⎦
=
[1.8 + 0.2]
= =
2
2
[ 2.0]
2
4+4
= 8 = 2.8 units
ANSWER: 2.8 units
+ [1.8 + 0.2 ]
2
+ [ 2.0 ]
2
2
68.
PROBLEM: (1.2, −3.3) and (2.2, −1.7) SOLUTION: x2 = 2.2, x1 = 1.2, y2 = −1.7, y1 = −3.3 d=
( x2 − x1 ) + ( y2 − y1 ) 2
=
( 2.2 − 1.2 )
=
( 2.2 − 1.2 )
=
(1.0 )
2
2
2
+ ⎡⎣ −1.7 − ( −3.3 ) ⎤⎦
2
+ [ −1.7 + 3.3]
2
+ [1.6 ]
2
2
= 1.0 + 2.56 = 3.56 = 1.9 units
ANSWER: 1.9 units Show that the three points form a right triangle.
69.
PROBLEM: (−3, −2), (0, −2), and (0,4) SOLUTION: Points: (−3, −2) and (0, −2)
a = ⎡⎣0 − ( −3) ⎤⎦ + ⎡⎣ −2 − ( −2 ) ⎤⎦ 2
=
[3] + [0] 2
2
= 9 =3 Points: (0, −2) and (0, 4)
b=
= =
( 0 − 0)
[4
+ ⎡⎣ 4 − ( −2 ) ⎤⎦
2
2
+ 2]
( 6)
2
2
= 36 = 6 Points: (0, 4) and (−3, −2)
2
( −3 − 0 ) + ( −2 − 4 ) 2
c=
( −3)
= =
+ ( −6 )
2
2
2
9 + 36
= 45
Now we check to see if a 2 + b 2 = c 2 . ANSWER: a 2 + b2 = c2
( 3) + ( 6 ) 2
2
=
(
9 + 36 = 45 45 = 45 45 = 45
45
)
2
70.
PROBLEM: (7,12), (7, −13), and (−5, −4) SOLUTION: Points: (7, 12) and (7, −13)
( 7 − 7 ) + ( −13 − 12 ) 2
a=
( 0)
=
2
+ ( −25)
2
2
= 625 = 25 Points: (7, −13) and (−5, −4) b=
( −5 − 7 )
2
+ ⎡⎣ −4 − ( −13) ⎤⎦
=
( −5 − 7 )
2
+ [ −4 + 13]
=
( −12 )
2
2
2
+ [ 9]
2
= 144 + 81 = 225 = 15 Points: (−5, −4) and (7, 12) c = ⎡⎣7 − ( −5) ⎤⎦ + ⎡⎣12 − ( −4 ) ⎤⎦ 2
[ 7 + 5]
2
=
+ (12 + 4 )
(12 ) + (16 ) 2
=
2
2
2
= 144 + 256 = 400 = 20 Now we check to see if b 2 + c 2 = a 2 . b2 + c2 = a 2
(15 ) + ( 20 ) 2
2
= ( 25 )
225 + 400 = 625 625 = 625
ANSWER:
2
71.
PROBLEM: (−1.4, 0.2), (1, 2) and (1, −3) SOLUTION: Points: (−1.4, −0.2) and (1, 2)
a = ⎡⎣1 − ( −1.4 ) ⎤⎦ + ( 2 − 0.2 ) 2
=
2 2 [1 + 1.4] + ( 2 − 0.2 )
=
[ 2.4]
2
+ (1.8 )
2
2
= 5.76 + 3.24 = 9 =3 Points: (1, 2) and (1, −3)
(1 − 1) + ( −3 − 2 ) 2
b=
2
[ −5]
2
=
= 25 = 5 Points: (1, −3) and (−1.4, 0.2)
( −1.4 − 1)
c=
2
+ ⎡⎣0.2 − ( −3) ⎤⎦
2
( −1.4 − 1) + [0.2 + 3] 2
=
2
( −2.4 ) + [3.2] 2
=
2
= 5.76 + 10.24 = 16 =4
Now we check to see if a 2 + c 2 = b 2 . ANSWER: a 2 + c2 = b2
( 3) + ( 4 ) 2
2
= ( 5)
9 + 16 = 25 25 = 25
2
72.
PROBLEM: (2,−1), (−1,2), and (6,3) SOLUTION: Points: (2,−1) and ( −1,2)
a=
( −1 − 2 )
2
+ ⎡⎣ 2 − ( −1) ⎤⎦
=
( −1 − 2 ) + [ 2 + 1]
=
( −3) + [3]
2
2
2
2
= 9+9 = 18 Points: (−1,2) and (6,3)
2
b = ⎡⎣6 − ( −1) ⎤⎦ + ( 3 − 2 ) 2
=
[6 + 1]
=
[7]
2
+ (3 − 2)
+ (1)
2
2
2
2
= 49 + 1 = 50 Points: (6,3) and (2,−1) c=
( 2 − 6 ) + ( −1 − 3) 2
=
( −4 )
=
16 + 16
2
+ ( −4 )
2
2
= 32 Now we check to see if a 2 + c 2 = b 2 . a 2 + c2 = b2
( 18 ) + ( 2
32
) =( 2
18 + 32 = 50 50 = 50
ANSWER:
50
)
2
73.
PROBLEM: (−5,2), (−1, −2), and (−2,5) SOLUTION: Points: (−5,2) and (−1, −2)
a = ⎡⎣ −1 − ( −5) ⎤⎦ + ( −2 − 2 ) 2
=
2 2 [ −1 + 5] + ( −2 − 2 )
=
2 2 [ 4] + ( −4 )
2
= 16 + 16 = 32 Points: (−1, −2) and (−2,5) b = ⎡⎣ −2 − ( −1) ⎤⎦ + ⎡⎣5 − ( −2 ) ⎤⎦ 2
=
2
[ −2 + 1] + [5 + 2] 2
2
=
[ −1] + [7]
=
1 + 49
2
2
= 50 Points: (−2,5) and (−5,2) c = ⎡⎣ −5 − ( −2 ) ⎤⎦ + ( 2 − 5) 2
= =
2
2 2 [ −5 + 2] + ( 2 − 5) 2 2 [ −3] + ( −3)
= 9+9 = 18 Now we check to see if a 2 + c 2 = b 2 . ANSWER: a 2 + c2 = b2
(
32
) + ( 18 ) = ( 2
2
32 + 18 = 50 50 = 50
50
)
2
74.
PROBLEM: (1,−2), (2,3), and (−3,4) SOLUTION: Points: (1,−2) and ( 2,3)
a=
( 2 − 1)
2
+ ⎡⎣3 − ( −2 ) ⎤⎦
=
( 2 − 1) + [3 + 2]
=
(1) + [5]
2
2
2
2
= 1 + 25 = 26 Points: (2,3) and (−3,4)
2
( −3 − 2 ) + ( 4 − 3) 2
b=
( −5)
= =
2
+ (1)
2
2
25 + 1
= 26
Points: (−3,4) and (1,−2) c = ⎡⎣1 − ( −3) ⎤⎦ + ( −2 − 4 ) 2
=
2
2 2 [1 + 3] + ( −2 − 4 )
=
2 2 [ 4] + ( −6 )
= 16 + 36 = 52 Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c2
(
26
) +( 2
26
) =( 2
26 + 26 = 52 52 = 52
ANSWER:
52
)
2
Isosceles triangles have two legs of equal length. Show that the following points form an isosceles triangle. 75.
PROBLEM: (1, 6), (−1, 1) and (3, 1) SOLUTION: Points: (1,6) and (−1, 1) a= =
( −1 − 1) + (1 − 6 ) 2
( −2 ) + ( −5) 2
2
2
= 4 + 25 = 29
Points: (−1, 1) and (3,1)
b = ⎡⎣3 − ( −1) ⎤⎦ + (1 − 1) 2
= =
2
2 2 [3 + 1] + (1 − 1) 2 2 [ 4] + ( 0 )
= 16 =4 Points: (3,1) and (1,6) c= = =
(1 − 3) + ( 6 − 1) 2
( −2 ) + ( 5) 2
2
2
4 + 25
= 29
ANSWER: The edges a and c are equal. Hence the following points form an isosceles triangle.
76.
PROBLEM: (−6, −2), (−3, −5) and (−9, −5) SOLUTION: Points: (−6, −2) and (−3, −5)
a = ⎡⎣ −3 − ( −6 ) ⎤⎦ + ⎡⎣ −5 − ( −2 ) ⎤⎦ 2
=
[ −3 + 6] + [ −5 + 2]
=
[3] + [ −3]
2
2
2
2
2
= 9+9 = 18 Points: (−3, −5 ) and (-9,-5) b = ⎡⎣ −9 − ( −3) ⎤⎦ + ⎡⎣ −5 − ( −5) ⎤⎦ 2
=
[ −9 + 3] + [ −5 + 5]
=
[ −6] + [0]
2
= 36 =6
2
2
2
2
Points: (−9, −5 ) and (−6, −2) c = ⎡⎣ −6 − ( −9 ) ⎤⎦ + ⎡⎣ −2 − ( −5) ⎤⎦ 2
=
[ −6 + 9] + [ −2 + 5]
=
[ −3] + [3]
2
2
2
2
2
= 9+9 = 18 ANSWER: The edges a and c are equal. Hence the following points form an isosceles triangle.
77.
PROBLEM: (−3, 0), (0, 3) and (3, 0) SOLUTION: Points: (−3, 0) and (0, 3)
a = ⎡⎣0 − ( −3) ⎤⎦ + ( 3 − 0 ) 2
=
2
2 2 [3] + ( 3)
= 9+9 = 18 Points: (0, 3) and (3, 0) b= =
( 3 − 0 ) + ( 0 − 3) 2
( 3) + ( −3) 2
=
9+9
=
18
2
2
Points: (3, 0) and (−3, 0) c= =
( −3 − 3) + ( 0 − 0 ) 2
( −6 ) + ( 0 ) 2
2
2
= 36 =6 ANSWER: The edges a and b are equal. Hence the following points form an isosceles triangle.
78.
PROBLEM: (0, −1), (0, 1) and (1, 0) SOLUTION: Points: (0, −1) and (0, 1)
a=
( 0 − 0)
2
+ ⎡⎣1 − ( −1) ⎦⎤
=
( 0 − 0 ) + [1 + 1]
=
[ 2]
2
2
2
2
= 4 =2 Points: (0, 1) and (1, 0) b= =
(1 − 0 ) + ( 0 − 1) 2
(1) + (1) 2
2
2
= 1+1 = 2
Points: (1, 0) and (0, −1) c= =
( 0 − 1) + ( −1 − 0 ) 2
( −1) + ( −1) 2
2
2
= 1+1 = 2
ANSWER: The edges b and c are equal. Hence the following points form an isosceles triangle.
Calculate the area and the perimeter of the triangles formed by the following set of vertices.
79.
PROBLEM: {(−4, −5), (−4, 3), (2, 3)} SOLUTION: Points: (−4, −5) and (−4, 3)
a = ⎡⎣ −4 − ( −4 ) ⎤⎦ + ⎡⎣3 − ( −5) ⎤⎦ 2
=
[ −4 + 4] + [3 + 5]
=
[0] + [8]
2
2
2
2
2
= 64 =8 Points: (−4, 3) and (2, 3) b = ⎡⎣ 2 − ( −4 ) ⎤⎦ + ( 3 − 3) 2
= =
2
2 2 [ 2 + 4] + ( 0 )
[ 6]
2
= 36 = 6 Points: (2, 3) and (−4, −5) c= = =
( −4 − 2 ) + ( −5 − 3) 2
( −6 ) + ( −8) 2
2
2
36 + 64
= 100 = 10 ANSWER:
Perimeter = a + b + c = 8 + 6 + 10 = 24 units Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c2
(8) + ( 6 ) 2
2
= (10 )
2
64 + 36 = 100 100 = 100
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge c = 10 is the hypotenuse. 1 Area = × base × height 2 1 = × a×b 2 1 = ×8× 6 2 = 24 sq.units 80.
PROBLEM: {(−1, 1), (3, 1), (3, −2)} SOLUTION: Points: (−1, 1) and (3, 1)
a = ⎡⎣3 − ( −1) ⎤⎦ + (1 − 1) 2
=
2 2 [3 + 1] + (1 − 1)
=
2 2 [ 4] + ( 0 )
2
= 16 = 4 Points: (3, 1) and (3, −2) b= =
( 3 − 3) + ( −2 − 1) 2
( 0 ) + ( −3) 2
2
2
= 9 = 3 Points: (3, −2) and (−1, 1)
( −1 − 3)
c=
2
+ ⎡⎣1 − ( −2 ) ⎤⎦
2
( −1 − 3) + [1 + 2] 2
=
2
( −4 ) + [3] 2
=
2
= 16 + 9 = 25 = 5 Perimeter = a + b + c = 3 + 4 + 5 = 12 units Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c 2
( 3) + ( 4 ) 2
2
= ( 5)
2
9 + 16 = 25 25 = 25
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge c = 5 is the hypotenuse. 1 Area = × base × height 2 1 = × a×b 2 1 = × 4×3 2 = 6 sq.units ANSWER: 6 sq.units , 12 units
81.
PROBLEM: {(−3, 1), (−3, 5), (1, 5)} SOLUTION: Points: (−3, 1) and (−3, 5)
a = ⎡⎣ −3 − ( −3) ⎤⎦ + ( 5 − 1) 2
=
2 2 [ −3 + 3] + ( 5 − 1)
=
2 2 [ 0] + ( 4 )
2
= 16 =4 Points: (−3, 5) and (1, 5) b = ⎡⎣1 − ( −3) ⎤⎦ + ( 5 − 5) 2
2
2 2 [1 + 3] + ( 5 − 5)
=
2 2 [ 4] + ( 0 )
=
= 16 = 4 Points: (1, 5) and (−3, 1)
( −3 − 1) + (1 − 5) 2
c= =
( −4 ) + ( −4 )
=
16 + 16
2
2
2
= 32 = 2 ×16 =4 2 ANSWER: Perimeter = a + b + c = 4 + 4 + 4 2 = 8+ 4 2 units
Now we check to see if a 2 + b 2 = c 2 . a 2 + b2 = c2
( 4) + ( 4) 2
2
(
= 4 2
16 + 16 = 32 32 = 32
)
2
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge c = 4 2 is the hypotenuse. 1 Area = × base × height 2 1 = × a×b 2 1 = × 4× 4 2 = 8 sq.units 82.
PROBLEM: {(−3, −1), (−3, 7), (1, −1)} SOLUTION: Points: (−3, −1) and (−3, 7)
a = ⎡⎣ −3 − ( −3) ⎤⎦ + ⎡⎣7 − ( −1) ⎤⎦ 2
=
[ −3 + 3] + [7 + 1]
=
[ 0 ] + [ 8]
2
2
2
2
= 64 =8 Points: (−3, 7) and (1, −1) b = ⎡⎣1 − ( −3) ⎦⎤ + ( −1 − 7 ) 2
=
2
2 2 [1 + 3] + ( −1 − 7 )
=
2 2 [ 4] + ( −8)
=
16 + 64
= 80 = 16 × 5 =4 5 Points: (1, −1) and (−3, −1)
2
( −3 − 1)
c=
2
+ ⎡⎣ −1 − ( −1) ⎤⎦
2
( −3 − 1) + [ −1 + 1] 2
=
2
( −4 ) + [0] 2
=
2
= 16 =4 Perimeter = a + b + c = 8 + 4 5 + 4 = 12 + 4 5 units Now we check to see if a 2 + c 2 = b 2 . a 2 + c2 = b2
(8) + ( 4 ) 2
2
(
= 4 5
)
2
64 + 16 = 80 80 = 80
The vertices form a right triangle. In a right triangle the edge opposite to the right angle is the hypotenuse and the largest edge is the hypotenuse. The other two edges are the base and height of the triangle. In the problem above the edge b = 4 5 is the hypotenuse. 1 Area = × base × height 2 1 = ×a×c 2 1 = ×8× 4 2 = 16 sq.units ANSWER: 16 sq.units , 12 + 4 5 units
Part C: Midpoint Formula Find the midpoint between the given two points.
83.
PROBLEM: (−1,6) and (−7, −2) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−1, 6) (−7, −2) x1 + x2 −1 − 7 −8 = = = −4 2 2 2 y1 + y2 6 − 2 4 = = =2 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( −4, 2 ) ANSWER: ( −4, 2 )
84.
PROBLEM: (8, 0) and (4,−3) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(8, 0) (4, −3) x1 + x2 8 + 4 12 = = =6 2 2 2 y1 + y2 0 − 3 −3 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 3⎞ ⎛ ⎜ 6, − ⎟ 2⎠ ⎝
3⎞ ⎛ ANSWER: ⎜ 6, − ⎟ 2⎠ ⎝
85.
PROBLEM: (−10,0) and (10,0) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−10, 0) (10, 0) x1 + x2 −10 + 10 0 = = =0 2 2 2 y1 + y2 0 + 0 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 0, 0 ) ANSWER: ( 0, 0 )
86.
PROBLEM: (−3, −6) and (−3,6) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−3, −6) (−3, 6) x1 + x2 −3 − 3 −6 = = = −3 2 2 2 y1 + y2 −6 + 6 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( −3, 0 ) ANSWER: ( −3, 0 )
87.
PROBLEM: (−10,5) and (14,−5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−10, 5) (14, −5) x1 + x2 −10 + 14 4 = = =2 2 2 2 y1 + y2 5 − 5 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 2, 0 ) ANSWER: ( 2, 0 )
88.
PROBLEM: (0, 1) and (2, 2) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(0,1) (2, 2) x1 + x2 0 + 2 2 = = =1 2 2 2 y1 + y2 1 + 2 3 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛ 3⎞ ⎜ 1, ⎟ ⎝ 2⎠ ⎛ 3⎞ ANSWER: ⎜1, ⎟ ⎝ 2⎠
89.
PROBLEM: (5,−3) and (4,−5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( 5, −3) ( 4, −5)
x1 + x2 5 + 4 9 = = 2 2 2 y1 + y2 −3 − 5 −8 = = = −4 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
⎛9 ⎞ ⎜ ,− 4⎟ ⎝2 ⎠ ANSWER: ⎛9 ⎞ ⎜ ,− 4⎟ ⎝2 ⎠
90.
PROBLEM: (0, 0) and (1, 1) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(0, 0) (1,1) x1 + x2 0 + 1 1 = = 2 2 2 y1 + y2 0 + 1 1 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛1 1⎞ ⎜ , ⎟ ⎝2 2⎠ ⎛1 1⎞ ANSWER: ⎜ , ⎟ ⎝2 2⎠
91.
PROBLEM: (−1,−1) and (4,4) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( −1, −1) ( 4, 4 )
x1 + x2 −1 + 4 3 = = 2 2 2 y1 + y2 −1 + 4 3 = = 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
⎛3 3⎞ ⎜ , ⎟ ⎝2 2⎠ ANSWER: ⎛3 3⎞ ⎜ , ⎟ ⎝2 2⎠
92.
PROBLEM: (3, −5) and (3, 5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(3, −5) (3,5) x1 + x2 3 + 3 6 = = =3 2 2 2 y1 + y2 −5 + 5 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 3, 0 )
ANSWER: ( 3, 0 )
93.
PROBLEM: ⎛ 1 1⎞ ⎜ − , − ⎟ and ⎝ 2 3⎠
⎛3 7⎞ ⎜ , ⎟ ⎝2 3⎠
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
⎛ 1 1⎞ ⎛ 3 7⎞ ⎜− ,− ⎟ ⎜ , ⎟ ⎝ 2 3⎠ ⎝ 2 3⎠ 1 3 2 − + x1 + x2 2 1 = 2 2= 2= = 2 2 2 4 2 1 7 6 − + y1 + y2 6 = 3 3 = 3 = =1 2 2 2 6 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛1 ⎞ ⎜ , 1⎟ ⎝2 ⎠ ANSWER: ⎛1 ⎞ ⎜ , 1⎟ ⎝2 ⎠
94.
PROBLEM: ⎛3 2⎞ ⎜ , − ⎟ and ⎝4 3⎠
⎛1 1⎞ ⎜ ,− ⎟ ⎝8 2⎠
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
⎛3 2⎞ ⎛1 1⎞ ⎜ ,− ⎟ ⎜ ,− ⎟ ⎝4 3⎠ ⎝8 2⎠ 3 1 3⋅ 2 +1 6 +1 7 + x1 + x2 4 8 7 = = 8 = 8 =8 = 2 2 2 2 2 16 2 1 −2 ⋅ 2 − 1 ⋅ 3 − 4 − 3 7 − − − y1 + y2 7 6 = 3 2= = 6 = 6 =− 2 2 2 2 2 12 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 7⎞ ⎛7 ⎜ ,− ⎟ ⎝ 16 12 ⎠
7⎞ ⎛7 ANSWER: ⎜ , − ⎟ ⎝ 16 12 ⎠ 95.
PROBLEM: ⎛5 1⎞ ⎛ 1 3⎞ ⎜ , ⎟ and ⎜ − , − ⎟ ⎝ 6 2⎠ ⎝3 4⎠ SOLUTION: ( x1 , y1 ) ( x2 , y2 )
⎛5 1⎞ ⎛ 1 3⎞ ⎜ , ⎟ ⎜− ,− ⎟ ⎝3 4⎠ ⎝ 6 2⎠ 5 1 5 ⋅ 6 − 1 ⋅ 3 30 − 3 27 − x1 + x2 3 6 27 3 18 = = = 18 = 18 = = 2 2 2 2 2 36 4 5 1 3 1− 3⋅ 2 1− 6 − − y1 + y2 4 2 5 = = 4 = 4 = 4 =− 2 2 2 2 2 8 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 5⎞ ⎛3 ⎜ ,− ⎟ 8⎠ ⎝4 ANSWER: 5⎞ ⎛3 ⎜ ,− ⎟ 8⎠ ⎝4
96.
PROBLEM: 1⎞ ⎛ 1 5⎞ ⎛ 7 ⎜ − , − ⎟ and ⎜ , − ⎟ ⎝ 5 2⎠ ⎝ 10 4 ⎠ SOLUTION: ( x1 , y1 ) ( x2 , y2 )
1⎞ ⎛ 1 5⎞ ⎛ 7 ⎜− ,− ⎟ ⎜ ,− ⎟ ⎝ 5 2 ⎠ ⎝ 10 4 ⎠ 1 7 5 − 1 ⋅ 2 + 7 −2 + 7 − + x1 + x2 5 1 10 = 5 10 = = 10 = 10 = = 2 2 2 2 2 20 4 5 1 −5 ⋅ 2 − 1 −10 − 1 11 − − − y1 + y2 11 4 4 = 2 4= = = 4 =− 2 2 2 2 2 8
⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ⎛ 1 11 ⎞ ⎜ ,− ⎟ 8⎠ ⎝4 ⎛ 1 11 ⎞ ANSWER: ⎜ , − ⎟ 8⎠ ⎝4 97.
PROBLEM: Given the right triangle formed by the vertices (0, 0), (0, 6), and (6, 8), show that the midpoints of the sides form a right triangle. SOLUTION: Points: (0,0) and (0,6) ( x1 , y1 ) ( x2 , y2 )
(0, 0) (0, 6) x1 + x2 0 + 0 0 = = =0 2 2 2 y1 + y2 0 + 6 6 = = =3 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 0, 3) Points: (0,6) and (6,8) ( x1 , y1 ) ( x2 , y2 )
( 0, 6 ) ( 6,8)
x1 + x2 0 + 6 6 = = =3 2 2 2 y1 + y2 6 + 8 14 = = =7 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 3, 7 )
Points: (6,8) and (0,0) ( x1 , y1 ) ( x2 , y2 )
( 6,8) ( 0, 0 )
x1 + x2 6 + 0 6 = = =3 2 2 2 y1 + y2 8 + 0 8 = = =4 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 3, 4 )
Considering the midpoints as the vertices, (0,3), (3,7), and (3,4). Points: (0, 3) and (3, 7) a= =
( 3 − 0 ) + ( 7 − 3) 2
( 3) + ( 4 ) 2
2
2
= 9 + 16 = 25 =5 Points: (3, 7) and (3, 4) b= =
( 3 − 3) + ( 4 − 7 ) 2
( 0 ) + ( −3) 2
2
2
= 9 = 3 Points: (3, 4) and (0, 3) c= = =
( 0 − 3) + ( 3 − 4 ) 2
( −3) + ( −1) 2
2
2
9 +1
= 10
Now we check to see if b 2 + c 2 = a 2 . b2 + c2 = a 2
( 10 )
2
+ ( 3) = ( 5 ) 2
2
10 + 9 = 25 19 ≠ 25 ANSWER: The mid points of the vertices do not form a right triangle.
98.
PROBLEM: Given the isosceles triangle formed by the vertices (−10, −12), (0, 12), and (10, −12), show that the midpoints of the sides also forms an isosceles triangle. SOLUTION: Points: (−10, −12) and (0,12) ( x1 , y1 ) ( x2 , y2 )
(−10, −12) (0,12) x1 + x2 −10 + 0 −10 = = = −5 2 2 2 y1 + y2 −12 + 12 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( −5, 0 ) Points: (0, 12) and (10, −12) ( x1 , y1 ) ( x2 , y2 )
( 0, 12 ) (10,
− 12 )
x1 + x2 0 + 10 10 = = =5 2 2 2 y1 + y2 12 − 12 0 = = =0 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( 5, 0 )
Points: (10, −12) and (−10, −12) ( x1 , y1 ) ( x2 , y2 )
(10,
− 12 )
( −10,
− 12 )
x1 + x2 10 − 10 0 = = =0 2 2 2 y1 + y2 −12 − 12 −24 = = = −12 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2 ( 0, − 12 )
Considering the midpoints as the vertices, (−5,0), (5,0), and (0, −12). Points: (−5,0) and (5,0)
a = ⎡⎣5 − ( −5 ) ⎤⎦ + ( 0 − 0 ) 2
=
2 2 [5 + 5] + ( 0 − 0 )
=
[10]
2
2
= 100 = 10 Points: (5,0) and (0, −12)
b= = =
( 0 − 5) + ( −12 − 0 ) 2
( −5) + ( −12 ) 2
2
2
25 + 144
= 169 = 13 Points: (0, −12) and (−5,0) c= = =
( −5 − 0 )
2
+ ⎡⎣0 − ( −12 ) ⎤⎦
2
( −5) + [ −12] 2
2
25 + 144
= 169 = 13
ANSWER: The sides b and c are equal. Hence the midpoints of the sides also form an isosceles triangle.
99.
PROBLEM: Calculate the area of the triangle formed by the vertices (−4, −3), (−1, 1), and (2, −3). (Hint: the vertices form an isosceles triangle.) SOLUTION: Points: (−4, −3) and (−1, 1)
a = ⎣⎡ −1 − ( −4 ) ⎦⎤ + ⎡⎣1 − ( −3) ⎤⎦ 2
=
[ −1 + 4] + [1 + 3]
=
[3] + [ 4]
2
2
2
2
2
= 9 + 16 = 25 =5 Points: (−1, 1) and (2, −3)
b = ⎡⎣ 2 − ( −1) ⎤⎦ + ( −3 − 1) 2
=
2
2 2 [ 2 + 1] + ( −3 − 1)
=
2 2 [3] + ( −4 )
=
9 + 16
= 25 =5 Points: (2, −3) and (−4, −3)
c= = =
( −4 − 2 )
2
+ ⎡⎣ −3 − ( −3) ⎤⎦
2
( −4 − 2 ) + [ −3 + 3] 2
2
( −6 ) + [ −0] 2
2
= 36 =6 The edges a and b are equal. The unequal side of the triangle is the base of the triangle. The height of the triangle is the distance between the mid-point of the base vertices, i.e. (2, −3) and (−4, −3), and the common end of the equal sides, i.e. (−1, 1). Mid-point of (2, −3) and (−4, −3):
Points: (2, −3) and (−4, −3) ( x1 , y1 ) ( x2 , y2 )
( 2,
− 3)
( −4,
− 3)
x1 + x2 2 − 4 −2 = = = −1 2 2 2 y1 + y2 −3 − 3 −6 = = = −3 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
( −1, − 3)
Distance between the (−1, −3) and (−1, 1): Points: (−1, −3) and (−1, 1) Height = ⎡⎣ −1 − ( −1) ⎦⎤ + ⎡⎣1 − ( −3) ⎤⎦ 2
=
[ −1 + 1] + [1 + 3]
=
[ 0] + [ 4 ]
2
2
2
2
2
= 16 =4 ANSWER: 1 Area = × base × height 2 1 = × 6× 4 2 = 12 sq.units
100.
PROBLEM: Calculate the area of the triangle formed by the vertices (−2, 1), (4, 1), and (1, −5). SOLUTION: Points: (−2, 1) and (4, 1)
a = ⎡⎣ 4 − ( −2 ) ⎦⎤ + (1 − 1) 2
=
2 2 [ 4 + 2] + (1 − 1)
=
2 2 [ 6] + ( 0 )
= 36 =6
2
Points: (4, 1) and (1, −5) b= =
(1 − 4 ) + ( −5 − 1) 2
( −3) + ( −6 ) 2
=
9 + 36
=
45
2
2
Points: (1, −5) and (−2, 1)
c= = =
( −2 − 1)
2
+ ⎡⎣1 − ( −5) ⎤⎦
2
( −2 − 1) + [1 + 5] 2
2
( −3) + [ −6] 2
2
= 9 + 36 = 45 The edges b and c are equal. The unequal side of the triangle is the base of the triangle. The height of the triangle is the distance between the mid-point of the base vertices, i.e. (−2, 1) and (4, 1), and the common end of the equal sides, i.e. (1, −5). Mid-point of (−2, 1) and (4, 1): ( x1 , y1 ) ( x2 , y2 )
( −2, 1) ( 4, 1)
x1 + x2 −2 + 4 2 = = =1 2 2 2 y1 + y2 1 + 1 2 = = =1 2 2 2 ⎛ x1 + x2 y1 + y2 ⎞ , ⎜ ⎟ 2 ⎠ ⎝ 2
(1, 1)
Distance between the (1, 1) and (1, −5): Height = =
(1 − 1) + ( −5 − 1) 2
( 0 ) + ( −6 )
= 36 =6
2
2
2
ANSWER: 1 Area = × base × height 2 1 = × 6× 6 2 = 18 sq.units
Part D: Discussion Board Topics
101.
PROBLEM: Research and discuss the life and contributions to mathematics of Rene Descartes. ANSWER: Rene Descartes (31 March 1596 – 11 February 1650), was a French philosopher, mathematician, physicist. He has been dubbed the “Father of Modern Philosophy”, and much subsequent Western philosophy is a response to his writings, which are studied closely to this day. Descartes created analytic geometry, and discovered an early form of the law of conservation of momentum (the term momentum refers to the momentum of a force). He outlined his views on the universe in his Principles of Philosophy. One of Descartes most enduring legacies was his development of Cartesian geometry, which uses algebra to describe geometry. He also “invented”, the notation that uses superscripts to show the powers or exponents, for example the 4 used in x4 to indicate squaring of squaring. (Source: http://en.wikipedia.org/wiki/Ren%C3%A9_Descartes)
102.
PROBLEM: Research and discuss the history of the right triangle and the Pythagorean Theorem. ANSWER: A right triangle is triangle with an angle of 90o(π/2 radians). The sides a, b, and c of such a triangle satisfy the Pythagorean theorem. a2 + b2 = c2 where the largest side is conventionally denoted c and is called hypotenuse. The other two sides of lengths a and b are called legs, or sometimes catheti.
103.
PROBLEM: What is a Pythagorean Triple? Provide some examples. ANSWER: A Pythagorean triple consists of three positive integers a, b, and c such that a2 + b2 = c2. The name is derived from the Pythagorean Theorem, satisfying the formula a2 + b2 = c2; thus, Pythagorean triples describe the three integer side lengths of a right triangle. A Pythagorean triple is commonly written as (a, b, c). Some examples are (3, 4, 5), (5, 12, 13), (7, 24, 25) etc.
104.
PROBLEM: Explain why we cannot use a ruler to calculate distance on a graph. ANSWER: The graph is often scaled down to obtain the representation of large objects. The physical distance on a graph measured using a ruler would not be accurate because of this scaling down process. Hence, it is not possible to measure the distance of an object from its graphical representation using a ruler. However, the physical distance measured using a ruler can be multiplied by the scaling factor to obtain the actual distance.
105.
PROBLEM: How can we use a compass and a straight edge to bisect a line segment? ANSWER: The radius of the compass should be greater than half the length of the line segment. From one end of the line segment draw arcs above and below the line segment. From the other end draw arcs intersecting the previous arcs. Connect the intersecting points of the two arcs using a straight edge. The intersecting point of the line drawn using the straight edge and the line segment is the mid-point.
3.2 Graph by Plotting Points Part A: Solutions to Linear Systems Determine whether or not the given point is a solution.
1.
PROBLEM: 5x – 2y = 4; (−1, 1) SOLUTION: 5x – 2y = 4 5(– 1) – 2(1) = 4 –5– 2 = 4 –7 ≠ 4 ANSWER: The given point (−1, 1) is not a solution of 5x – 2y = 4.
2.
PROBLEM: 3x – 4y = 10; (2, −1) SOLUTION: 3(2) – 4(–1) = 10 6 + 4 = 10 10 = 10 ANSWER: The given point (2, −1) is a solution of 3x – 4y = 10.
3.
PROBLEM: –3x + y = –6 ; (4, 6) SOLUTION: –3(4) + 6 = –6 –12 + 6 = –6 –6 = –6 ANSWER: The given point (4, 6) is a solution of –3x + y = –6.
4.
PROBLEM: –8x – y = 24 ; (−2, −3) SOLUTION: –8(–2) – (–3) = 24 16 + 3 = 24 19 ≠ 24 ANSWER: The given point (−2, −3) is not a solution of –8x – y = 24.
5.
PROBLEM: –x + y = –7; (5, −2) SOLUTION: –(5) + (–2) = –7 –7 = –7 ANSWER: The given point (5, −2) is a solution of –x + y = –7.
6.
PROBLEM: 9x – 3y = 6; (0, −2) SOLUTION: 9x – 3y = 6 9(0) – 3(–2) = 6 0+6=6 6=6 ANSWER: The given point (0, −2) is a solution of 9x – 3y = 6.
7.
PROBLEM: 1 1 1 x + y = − ; (1, −2) 2 3 6 SOLUTION: 1 1 1 x+ y =− 2 3 6 1 1 1 (1) + ( −2 ) = − 2 3 6 1 2 1 − =− 2 3 6 3 ⋅1 − 2 ⋅ 2 1 =− 6 6 3− 4 1 =− 6 6 1 1 − =− 6 6 ANSWER: The given point (1, –2) is a solution of
8.
1 1 1 x+ y =− . 2 3 6
PROBLEM: 3 1 x − y = −1 ; (2, 1) 4 2
SOLUTION: 3 1 ( 2 ) − (1) = −1 4 2 6 1 − = −1 4 2 6 − 1⋅ 2 = −1 4 6−2 = −1 4 4 = −1 4 1 ≠ −1 ANSWER: The given point (2, 1) is not a solution of
3 1 x − y = −1 . 4 2
9.
PROBLEM: ⎛1 1⎞ 4x – 3y = 1; ⎜⎝ 2 , 3 ⎟⎠ SOLUTION: 4x – 3y = 1
⎛1⎞ ⎛1⎞ 4 ⎜ ⎟ − 3⎜ ⎟ = 1 ⎝ 2⎠ ⎝3⎠ 2 −1 = 1 1=1 ⎛1 1⎞ ANSWER: The given point ⎜ , ⎟ is a solution of 4x – 3y = 1. ⎝ 2 3⎠
10.
PROBLEM: 9 −10 x + 2 y = − ; 5
( 15 , 101 )
SOLUTION: 9 5 9 ⎛1⎞ ⎛ 1⎞ −10 ⎜ ⎟ + 2 ⎜ ⎟ = − 5 ⎝5⎠ ⎝ 10 ⎠ 10 2 9 − + =− 5 10 5 −10 ⋅ 2 + 2 9 =− 10 5 −20 + 2 9 = − 10 5 18 9 − =− 10 5 9 9 − = − 5 5 −10 x + 2 y = −
9 ⎛1 1 ⎞ ANSWER: The given point ⎜ , ⎟ is a solution of −10 x + 2 y = − . 5 ⎝ 5 10 ⎠
11.
PROBLEM: 1 y = x + 3 ; (6, 3) 3 SOLUTION: 1 y = x+3 3 1 ( 3) = ( 6 ) + 3 3 3= 2+3 3≠5 ANSWER: The given point (6, 3) is a solution of y =
12.
1 x + 3. 3
PROBLEM: y = −4x + 1 ; (−2, 9) SOLUTION: 9 = −4(−2) + 1 9=8+1 9=9 ANSWER: The given point (−2, 9) is a solution of y = −4x + 1.
13.
PROBLEM: 2 y = x − 3 ; (0, −3) 3 SOLUTION: 2 y = x−3 3 2 −3 = ( 0 ) − 3 3 −3 = 0 − 3 −3 = −3 ANSWER: The given point (0, −3) is a solution of y =
2 x − 3. 3
14.
PROBLEM: 5 y = − x + 1 ; (8, −5) 8 SOLUTION: 5 y = − x +1 8 5 −5 = − ( 8 ) + 1 8 − 5 = −5 + 1 − 5 ≠ −4 5 ANSWER: The given point (8, −5) is not a solution of y = − x + 1. 8
15.
PROBLEM: 1 3 y = − x + ; ( − 12 ,1) 2 4 SOLUTION: 1 3 y =− x+ 2 4 1 1 3 (1) = − ⎛⎜ − ⎞⎟ + 2⎝ 2⎠ 4 1 3 1= + 4 4 1+ 3 1= 4 4 1= 4 1=1 1 3 ANSWER: The given point ( − 12 ,1) is a solution of y = − x + . 2 4
16.
PROBLEM: 1 1 y =− x− ; 3 2
( 12 , − 23 )
SOLUTION: 1 1 y =− x− 3 2 1⎛1⎞ 1 ⎛ 2⎞ ⎜− ⎟ = − ⎜ ⎟− 3⎝ 2⎠ 2 ⎝ 3⎠ 2 1 1 − =− − 6 2 3 2 −1 − 1⋅ 3 − = 3 6 2 −1 − 3 − = 3 6 2 −4 − = 3 6 2 2 − =− 3 3 ANSWER: The given point
17.
( 12 , − 23 ) is a solution of
1 1 y =− x− . 3 2
PROBLEM: y = 2; (−3, 2) SOLUTION: y = 2 2=2 ANSWER: The given point (−3, 2) is a solution of y = 2.
18.
PROBLEM:
y = 4; (4, −4) SOLUTION: y = 4 −4 ≠ 4 ANSWER: The given point (4, −4) is not a solution of y = 4.
19.
PROBLEM: x = 3; (3, −3) SOLUTION: x = 3 3=3 ANSWER: The given point (3, −3) is a solution of x = 3.
20.
PROBLEM: x = 0; (1,0) SOLUTION: x = 0 1≠0 ANSWER: The given point (1, 0) is not a solution of x = 0.
Find the ordered pair solutions given the set of x‐values.
21.
PROBLEM: y = −2x + 4; {−2, 0, 2} SOLUTION:
x-value
y-value
x = −2
y = −2 ( − 2 ) + 4 = 4+4 =8
x=0
y = −2 ( 0 ) + 4 = 0+4
Solution (−2, 8)
(0, 4)
=4
x=2
ANSWER: (−2, 8), (0, 4), (2, 0)
y = −2 ( 2 ) + 4 = −4 + 4 =0
(2, 0)
22.
PROBLEM: 1 y = x − 3 ; {−4, 0, 4} 2 SOLUTION:
x-value
23.
y-value 1 x −3 2 1 = ( −4 ) − 3 2 = −2−3 = −5
x = −4
y=
x=0
y=
x=4
y=
1 x −3 2 1 = (0) − 3 2 = 0−3 = −3
1 x −3 2 1 = ( 4) − 3 2 = 2−3 = −1
Answer (−4, −5)
(0, −3)
(4, −1)
PROBLEM: 3 1 y = − x + ; {−2, 0, 2} 4 2 SOLUTION:
x-value
x = −2
y-value 3 1 3 1 y = − x + = − ( −2 ) + 4 2 4 2 3 1 4 = + = =2 2 2 2
Solution (−2,2)
3 1 3 1 1 y = − x + = − (0) + = 0 + 4 2 4 2 2 1 = 2
x=0
3 1 3 1 3 1 y = − x + = − ( 2) + = − + 4 2 4 2 2 2 2 = − = −1 2
x=2
⎛ 1⎞ ⎜ 0, ⎟ ⎝ 2⎠
(2, −1)
ANSWER: ⎛ 1⎞ (−2,2), ⎜ 0, ⎟ , (2, −1) ⎝ 2⎠
24.
PROBLEM: y = −3 x + 1 ; {−1/2, 0, 1/2}
SOLUTION:
x-value
y-value
Answer
1 x=− 2
3 3 + 1⋅ 2 3 + 2 ⎛ 1⎞ y = −3 x + 1 = − 3 ⎜ − ⎟ + 1 = + 1 = = 2 2 2 ⎝ 2⎠ 5 = 2
⎛ 1 5⎞ ⎜− , ⎟ ⎝ 2 2⎠
x=0
y = −3 x + 1 = − 3 ( 0 ) + 1 = 0 + 1 = 1
(0, 1)
1 x= 2
3 −3 + 1 ⋅ 2 −3 + 2 ⎛1⎞ y = −3 x + 1 = − 3 ⎜ ⎟ + 1 = − + 1 = = 2 2 2 ⎝2⎠ 1 =− 2
⎛1 1⎞ ⎜ ,− ⎟ ⎝2 2⎠
25.
PROBLEM: y = − 4 ; {−3, 0, 3} SOLUTION:
x-value
y-value
Solution
x = −3
y = −4
(−3, −4)
x=0
y = −4
(0, −4 )
x=3
y = −4
(3, −4)
ANSWER: (−3, −4), (0, −4 ), (3, −4)
26.
PROBLEM: 1 3 y = x + ; {−1/4, 0, 1/4} 2 4 ANSWER:
x-value
x=−
1 4
x=0
1 x= 4
y-value 1⎛ 1⎞ 3 1 3 −1 + 3 ⋅ 2 −1 + 6 y = ⎜− ⎟+ =− + = = 2⎝ 4⎠ 4 8 4 8 8 5 = 8 y=
1 3 3 3 (0) + = 0 + = 2 4 4 4
1⎛1⎞ 3 1 3 1+ 3⋅ 2 1+ 6 = y = ⎜ ⎟+ = + = 2⎝4⎠ 4 8 4 8 8 7 = 8
Answer ⎛ 1 5⎞ ⎜− , ⎟ ⎝ 4 8⎠
⎛ 3⎞ ⎜ 0, ⎟ ⎝ 4⎠ ⎛1 7⎞ ⎜ , ⎟ ⎝4 8⎠
27.
PROBLEM: 2 x − 3 y = 1 ; {0, 1, 2} SOLUTION: 2x – 3y = 1 2x – 2x – 3y = 1 – 2x – 3y = 1– 2x 3y = 2x – 1 3y 2x – 1 = 3 3 2 1 y = x− 3 3 x-value
x=0
x =1
x=2
y-value
y=
2 1 2 1 1 x − = (0) − = − 3 3 3 3 3
⎛ 1⎞ ⎜1, ⎟ ⎝ 3⎠
2 1 2 1 4 1 3 x − = ( 2) − = − = 3 3 3 3 3 3 3 =1
(2, 1)
y=
PROBLEM: 3 x − 5 y = − 15 ; {−5, 0, 5}
SOLUTION: 3x – 5y = – 15 3x – 3x – 5y = –15 – 3x – 5y = –15 – 3x −5 y –15 3 = − x −5 −5 −5 3 y = 3+ x 5 ANSWER:
1⎞ ⎛ ⎜ 0, − ⎟ 3⎠ ⎝
2 1 2 1 2 1 x − = (1) − = − 3 3 3 3 3 3 1 = 3
y=
ANSWER: 1⎞ ⎛ 1⎞ ⎛ ⎜ 0, − ⎟ , ⎜1, ⎟ , (2, 1) 3⎠ ⎝ 3⎠ ⎝
28.
Solution
x-value
29.
y-value
Answer
x = −5
3 3 y = 3 + x = 3 + ( −5 ) = 3 − 3 = 0 5 5
(−5, 0)
x=0
3 3 y = 3 + x = 3 + ( 0) = 3 − 0 = 3 5 5
(0, 3)
x=5
3 3 y = 3 + x = 3 + (5) = 3 + 3 = 6 5 5
(5,6)
PROBLEM: – x + y = 3 ; {−5, −1, 0} SOLUTION: –x + y = 3 –x + x + y = 3 + x y=3+x
x-value
y-value
Solution
x = −5
y = 3 + x = 3 + (–5) = 3 – 5 = –2
(–5, –2)
x = −1
y = 3 + x = 3 + (–1) =3–1=2
(– 1, 2)
x=0
y = 3 + x = 3 + (0) =3+0=3
(0, 3)
ANSWER: (–5, –2), (– 1, 2), (0, 3)
30.
PROBLEM: 1 1 x − y = −4 ; {−4, −2, 0} 2 3 SOLUTION: 1 1 x − y = −4 2 3 1 1 1 1 x − x − y = −4 − x 2 2 3 2 1 1 − y = −4 − x 3 2 1 1 ( −3) ⋅ ⎛⎜ − y ⎞⎟ = ( −3) ⋅ ⎛⎜ −4 − x ⎞⎟ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 y = 12 + x 2
ANSWER:
x-value
x = −4
x = −2
y-value 3 3 12 12 ⋅ 2 − 12 24 − 12 y = 12 + x = 12 + ( −4 ) = 12 − = = 2 2 2 2 2 12 = =6 2 y = 12 + =
x=0
3 3 6 12 ⋅ 2 − 6 24 − 6 = x = 12 + ( −2 ) = 12 − = 2 2 2 2 2
Answer (−4, 6)
(−2, 9)
18 =9 2
y = 12 +
3 3 x = 12 + ( 0 ) = 12 + 0 = 12 2 2
(0,12)
31.
PROBLEM: 3 1 x + y = 2 ; {−15, −10, −5} 5 10 SOLUTION: 3 1 x+ y = 2 5 10 3 3 1 3 x− x+ y = 2− x 5 5 10 5 1 3 y = 2− x 10 5 1 3 ⎞ ⎛ 10 ⋅ y = 10 ⋅ ⎜ 2 − x ⎟ 10 5 ⎠ ⎝ y = 20 − 6 x
x-value
y-value
Solution
x = −15
y = 20 – 6x = 20 – 6(–15) = 20 + 90 = 110
(–15, 110)
x = −10
y = 20 – 6x = 20 – 6(–10) = 20 + 60 = 80
(– 10, 80)
x = −5
y = 20 – 6x = 20 – 6(–5) = 20 + 30 = 50
(–5, 50)
ANSWER: (–15, 110), (– 10, 80), (–5, 50)
32.
PROBLEM: x − y = 0 ; {10, 20, 30} SOLUTION: x – y = 0 –y = –x y=x ANSWER:
x-value
y-value
x = 10
y = x = 10
(10, 10)
x = 20
y = x = 20
(20, 20)
x = 30
y = x = 30
(30,30)
Find the ordered pair solutions given the set of y‐values.
33.
Solution
PROBLEM: 1 y = x − 1 ; {−5, 0, 5} 2 SOLUTION: 1 y = x −1 2 1 y +1 = x −1+1 2 1 x = y +1 2 1 2 ⋅ x = 2 ⋅ ( y + 1) 2 x = 2y + 2
y-value
x-value
Solution
y = −5
x = 2y + 2 = 2(–5) + 2 = –10 + 2 = –8
(–8, –5)
y=0
x = 2y + 2 = 2(0) + 2 =0+2 =2
(2, 0)
y =5
x = 2y + 2 = 2(5) + 2 = 10 + 2 = 12
(12, 5)
ANSWER: (–8, –5), (2, 0), (12, 5)
34.
PROBLEM: 3 y = − x + 2 ; {0, 2, 4} 4 SOLUTION: 3 y =− x+2 4 3 y−2 = − x+2−2 4 3 − x = y−2 4 ⎛ 4 ⎞⎛ 3 ⎞ ⎛ 4 ⎞ ⎜ − ⎟ ⎜ − x ⎟ = ⎜ − ⎟ ( y − 2) ⎝ 3 ⎠⎝ 4 ⎠ ⎝ 3 ⎠ 8 4 x= − y 3 3 ANSWER:
y-value
35.
x-value
Solution
y=0
8 4 8 4 8 x = − y = − (0) = 3 3 3 3 3
⎛8 ⎞ ⎜ , 0⎟ ⎝3 ⎠
y=2
8 4 8 4 8 8 x = − y = − ( 2) = − = 0 3 3 3 3 3 3
(0, 2)
y=4
8 4 8 4 8 16 8 x = − y = − ( 4) = − = − 3 3 3 3 3 3 3
⎛ 8 ⎞ ⎜ − , 0⎟ ⎝ 3 ⎠
PROBLEM: 3 x − 2 y = 6 ; {−3, −1, 0} SOLUTION: 3x − 2 y = 6 3x − 2 y + 2 y = 6 + 2 y 3x = 6 + 2 y 3x 6 2 y = + 3 3 3 2 x = 2+ y 3 y-value x-value 2 2 2 2 x = + y = + ( −3) = 2 − 2 = 0 y = −3 3 3
y = −1
x = 2+ =
y=0
ANSWER:
2 2 2 2⋅3 − 2 6 − 2 = y = 2 + ( −1) = 2 − = 3 3 3 3 3
4 3
x = 2+
2 2 y = 2 + ( 0) = 2 + 0 = 2 3 3
Solution (0, –3) ⎛4 ⎞ ⎜ , −1⎟ ⎝3 ⎠
(2, 0)
⎛4 ⎞ (0, –3), ⎜ , −1⎟ , (2, 0) ⎝3 ⎠ 36.
PROBLEM: − x + 3 y = 4 ; {−4, −2, 0} SOLUTION: −x + 3y = 4 −x + 3y − 3y = 4 − 3y −x = 4 − 3y x = 3y − 4 ANSWER:
y-value
x-value
Solution
y = −4
x = 3y – 4 = 3(–4) – 4 = –12 – 4 = –16
(–16, –4)
y = −2
x = 3y – 4 = 3(–2) – 4 = –6 – 4 = –10
(–10, –2)
y=0
x = 3y – 4 = 3(0) – 4 = 0 – 4 = –4
(–4, 0)
37.
PROBLEM: 1 1 x − y = −4 ; {−1, 0, 1} 3 2
SOLUTION: 1 1 x − y = −4 3 2 1 1 1 1 x − y + y = −4 + y 3 2 2 2 1 1 x = −4 + y 3 2 1 1 ⎞ ⎛ 3 ⋅ x = 3 ⋅ ⎜ −4 + y ⎟ 3 2 ⎠ ⎝ 3 x = −12 + y 2 3 x = y − 12 2 y-value x-value
y = −1
y=0
y =1
Solution
x=
3 3 3 −3 − 12 ⋅ 2 −3 − 24 −27 = = y − 12 = ( −1) − 12 = − − 12 = 2 2 2 2 2 2
x=
3 3 y − 12 = ( 0 ) − 12 = 0 − 12 = −12 2 2
x=
3 3 3 3 − 12 ⋅ 2 3 − 24 21 = =− y − 12 = (1) − 12 = − 12 = 2 2 2 2 2 2
ANSWER: ⎛ 27 ⎞ ⎜ − , −1⎟ , (-12, 0), ⎝ 2 ⎠
⎛ 21 ⎜− , ⎝ 2
⎞ 1⎟ ⎠
⎛ 27 ⎞ ⎜ − , −1⎟ ⎝ 2 ⎠
(-12, 0)
⎛ 21 ⎞ ⎜ − , 1⎟ ⎝ 2 ⎠
38.
PROBLEM: 3 1 x + y = 2 ; {−20, −10, −5} 5 10 SOLUTION: 3 1 x+ y = 2 5 10 3 1 1 1 x+ y− y = 2− y 5 10 10 10 3 1 x = 2− y 5 10 5 3 5 ⎛ 1 ⎞ ⋅ x = ⋅⎜ 2 − y ⎟ 3 5 3 ⎝ 10 ⎠ 10 1 x= − y 3 6 ANSWER:
y-value
y = −20 y = −10
y = −5
x-value 10 1 10 1 10 10 20 x = − y = − ( −20 ) = + = 3 6 3 6 3 3 3
x=
10 1 10 1 10 5 15 − y = − ( −10 ) = + = = 5 3 6 3 6 3 3 3
10 1 10 1 10 5 10 ⋅ 2 + 5 20 + 5 − y = − ( −5 ) = + = = 3 6 3 6 3 6 6 6 25 = 6
x=
Solution ⎛ 20 ⎞ ⎜ , −20 ⎟ ⎝ 3 ⎠ (5, −10)
⎛ 25 ⎞ ⎜ , −5 ⎟ ⎝ 6 ⎠
Part B: Graphing Lines Given the set of x‐values {−2, −1, 0, 1, 2} find the corresponding y‐values and graph.
39.
PROBLEM: y=x+1 ANSWER:
x y −2 −1 −1 0 0 1 1 2 2 3
40.
y = x +1 y = −2 + 1 = −1 y = −1 + 1 = 0 y = 0 +1 = 1 y = 1+1 = 2 y = 2 +1 = 3
PROBLEM: y = −x + 1 ANSWER:
x −2
y 3
y = −x + 1
−1
2
y = − ( −1) + 1 = 1 + 1 = 2
0 1
1 0
y = 0 +1 = 1 y = − (1) + 1 = −1 + 1 = 0
2
−1
y = − ( 2 ) + 1 = −2 + 1 = −1
y = − ( −2 ) + 1 = 2 + 1 = 3
41.
PROBLEM: y = 2x − 1 ANSWER:
x y −2 −5
y = 2x − 1
−1 −3
y = 2 ( −1) − 1 = −2 − 1 = −3
0
−1
y = 2 ( 0 ) − 1 = 0 − 1 = −1
1
1
y = 2 (1) − 1 = 2 − 1 = −1
2
3
y = 2 ( 2) −1 = 4 −1 = 3
y = 2 ( −2 ) − 1 = −4 − 1 = −5
42.
PROBLEM: y = −3x + 2 ANSWER:
y = −3x + 2
x −2
y 8
−1
5
y = −3 ( −1) + 2 = 3 + 2 = 5
0
2
y = −3 ( 0 ) + 2 = 0 + 2 = 2
1
−1
y = −3 (1) + 2 = −3 + 2 = −1
2
−4
y = −3 ( 2 ) + 2 = −6 + 2 = −4
43.
y = −3 ( −2 ) + 2 = 6 + 2 = 8
PROBLEM: y = 5x – 10 ANSWER:
x y −2 −20
y = 5x – 10
y = 5 ( −2 ) − 10 = −10 − 10 = −20
−1 −10
y = 5 ( −1) − 10 = −5 − 10 = −15
0
−10
y = 5 ( 0 ) − 10 = 0 − 10 = −10
1
−5
y = 5 (1) − 10 = 5 − 10 = −5
2
0
y = 5 ( 2 ) − 10 = 10 − 10 = 0
44.
PROBLEM: 5 x + y = 15
SOLUTION: 5x + y = 15 5x – 5x + y = 15 – 5x y = 15 – 5x ANSWER:
x y −2 25
y = 15 – 5x
y = 15 − 5 ( −2 ) = 15 + 10 = 25
−1 20
y = 15 − 5 ( −1) = 15 + 5 = 20
0
15
y = 15 − 5 ( 0 ) = 15 − 0 = 15
1
10
y = 15 − 5 (1) = 15 − 5 = 10
2
5
y = 15 − 5 ( 2 ) = 15 − 10 = 5
45.
PROBLEM: 3x – y = 9 SOLUTION: 3x – 3x – y = 9 – 3x y = 3x – 9 ANSWER:
x y −2 −15
y = 3x – 9
y = 3 ( −2 ) – 9 = − 6 − 9 = −15
−1 −12
y = 3 ( −1) – 9 = − 3 − 9 = −12
0
−9
y = 3 ( 0 ) – 9 = 0 − 9 = −9
1
−6
y = 3 (1) – 9 = 3 − 9 = −6
2
−3
y = 3 ( 2 ) – 9 = 6 − 9 = −3
46.
PROBLEM: 6x – 3y = 9 SOLUTION: 6x – 6x – 3y = 9 – 6x – 3y = 9 – 6x 3y = 6x – 9 3y 6x 9 = − 3 3 3 y = 2x – 3
ANSWER:
x y −2 −7
y = 2x – 3
y = 2 ( −2 ) – 3 = −4 − 3 = −7
−1 −5
y = 2 ( −1) – 3 = −2 − 3 = −5
0
−3
y = 2 ( 0 ) – 3 = 0 − 3 = −3
1
−1
y = 2 (1) – 3 = 2 − 3 = −1
2
1
y = 2 ( 2) – 3 = 4 − 3 = 1
47.
PROBLEM: y=–5 ANSWER:
x −2 −1 0 1 2
y −5 −5 −5 −5 −5
48.
PROBLEM: y=3 ANSWER:
x −2 −1 0 1 2
y 3 3 3 3 3
Find at least five ordered pair solutions and graph.
49.
PROBLEM: y = 2x – 1 ANSWER:
y = 2x – 1
x −2
y −5
−1
−3
y = 2 ( −1) − 1 = −2 − 1 = −3
0
−1
y = 2 ( 0 ) − 1 = 0 − 1 = −1
1
1
y = 2 (1) − 1 = 2 − 1 = 1
2
3
y = 2 ( 2) −1 = 4 −1 = 3
y = 2 ( −2 ) − 1 = −4 − 1 = −5
50.
PROBLEM: y = –5x + 3 ANSWER:
y = –5x + 3
x 0
y 3
y = –5 ( 0 ) + 3 = 0 + 3 = 3
0.6
0
y = –5 ( 0.6 ) + 3 = − 3 + 3 = 0
1
−2
y = –5 (1) + 3 = − 5 + 3 = − 2
1.4
−4
y = –5 (1.4 ) + 3 = − 7 + 3 = − 4
1.6
−5
y = –5 (1.6 ) + 3 = − 8 + 3 = − 5
51.
PROBLEM: y = –4x + 2 ANSWER:
y = –4x + 2
x −0.75
y 5
−0.5
4
y = –4 ( −0.5) + 2 = 2 + 2 = 4
0
2
y = –4 ( 0 ) + 2 = 0 + 2 = 2
1
−2
y = –4 (1) + 2 = −4 + 2 = −2
2
−6
y = –4 ( 2 ) + 2 = −8 + 2 = −6
52.
y = –4 ( −0.75) + 2 = 3 + 2 = 5
PROBLEM: y = 10 x − 20
ANSWER:
x y −1 −30
y = 10 x − 20 y = 10 ( − 1) − 20 = − 10 − 20 = − 30
0
−20
y = 10 ( 0 ) − 20 = 0 − 20 = − 20
2
0
y = 10 ( 2 ) − 20 = 20 − 20 = 0
3
10
y = 10 ( 3 ) − 20 = 30 − 20 = 10
4
20
y = 10 ( 4 ) − 20 = 40 − 20 = 20
53.
PROBLEM: 1 y =− x+2 2 ANSWER:
x
y
−4
4
−2
3
0
2
2
1
4
0
1 y =− x+2 2 1 y = − ( −4 ) + 2 = 2 + 2 = 4 2 1 y = − ( −2 ) + 2 = 1 + 2 = 3 2 1 y = − ( 0) + 2 = 0 + 2 = 2 2 1 y = − ( 2 ) + 2 = −1 + 2 = 1 2 1 y = − ( 4 ) + 2 = −2 + 2 = 0 2
54.
PROBLEM: 1 y = x −1 3 ANSWER:
x
y
−6
−3
−3
−2
0
−1
3
0
6
1
55.
1 x −1 3 1 y = ( − 6 ) − 1 = − 2 − 1 = −3 3 1 y = ( −3 ) − 1 = −1 − 1 = − 2 3 1 y = ( 0 ) − 1 = 0 − 1 = −1 3 1 y = ( 3) − 1 = 1 − 1 = 0 3 1 y = (6) −1 = 2 −1 = 1 3 y=
PROBLEM: 2 y = x−6 3 ANSWER:
x
y
−9
−12
−6
−10
2 x−6 3 2 y = ( −9 ) − 6 = −6 − 6 = −12 3 2 y = ( −6 ) − 6 = −4 − 6 = −10 3
y=
−3
−8
0
−6
3
−4
56.
2 ( −3) − 6 = −2 − 6 = −8 3 2 y = ( 0 ) − 6 = 0 − 6 = −6 3 2 y = ( 3) − 6 = 2 − 6 = −4 3 y=
PROBLEM: 2 y =− x+2 3 ANSWER:
x
y
−3
4
0
2
3
0
6
−2
9
−4
2 y =− x+2 3 2 y = − ( −3 ) + 2 = 2 + 2 = 4 3 2 y = − (0) + 2 = 0 + 2 = 2 3 2 y = − ( 3 ) + 2 = −2 + 2 = 0 3 2 y = − ( 6 ) + 2 = −4 + 2 = − 2 3 2 y = − ( 9 ) + 2 = −6 + 2 = − 4 3
57.
PROBLEM: y=x ANSWER:
x −5 −4 −2 0 1
y −5 −4 −2 0 1
58.
PROBLEM: y = −x ANSWER:
x −5 −4 −2 0 1
y 5 4 2 0 −1
59.
PROBLEM:
− 2 x + 5 y = − 15
SOLUTION:
− 2 x + 5 y = − 15
− 2 x + 2 x + 5 y = − 15 + 2 x 5 y = 2 x − 15 5 2 15 y = x− 5 5 5 2 y = x−3 5
ANSWER:
x
y
−5
−5
0
−3
5
−1
10
1
15
3
y= y= y= y= y= y=
2 x −3 5 2 ( − 5 ) − 3 = − 2 − 3 = −5 5 2 ( 0 ) − 3 = 0 − 3 = −3 5 2 ( 5 ) − 3 = 2 − 3 = −1 5 2 (10 ) − 3 = 4 − 3 = 1 5 2 (15 ) − 3 = 6 − 3 = 3 5
60.
PROBLEM: x + 5y = 5
SOLUTION: x + 5y = 5
x − x + 5y = 5 − x 5y = 5 − x 5 5 1 y= − x 5 5 5 1 y = 1− x 5
ANSWER:
x
y
−5
2
0
1
10
−1
15
−2
20
−3
1 y = 1− x 5 1 y = 1 − ( −5 ) = 1 + 1 = 2 5 1 y = 1 − (0) = 1 − 0 = 1 5 1 y = 1 − (10 ) = 1 − 2 = −1 5 1 y = 1 − (15 ) = 1 − 3 = −2 5 1 y = 1 − ( 20 ) = 1 − 4 = −3 5
61.
PROBLEM: 6x − y = 2
SOLUTION:
6x − y = 2 6x − 6x − y = 2 − 6x − y = 2 − 6x y = 6x − 2
x −1
ANSWER: y = 6x − 2 y −8 y = 6 ( −1) − 2 = −6 − 2 = −8
0
−2
y = 6 ( 0 ) − 2 = 0 − 2 = −2
1
4
y = 6 (1) − 2 = 6 − 2 = 4
2
10
y = 6 ( 2 ) − 2 = 12 − 2 = 10
3
16
y = 6 ( 3) − 2 = 18 − 2 = 16
62.
PROBLEM: 4 x + y = 12
SOLUTION:
4 x + y = 12 4 x − 4 x + y = 12 − 4 x y = 12 − 4 x
ANSWER:
y = 12 − 4 x
x −1
y 16
y = 12 − 4 ( −1) = 12 + 4 = 16
0
12
y = 12 − 4 ( 0 ) = 12 − 0 = 12
1
8
y = 12 − 4 (1) = 12 − 4 = 8
2
4
y = 12 − 4 ( 2 ) = 12 − 8 = 4
3
0
y = 12 − 4 ( 3) = 12 − 12 = 0
63.
PROBLEM: −x + 5y = 0
SOLUTION:
−x + 5y = 0 −x + x + 5y = 0 + x 5y = x 5 1 y= x 5 5 1 y= x 5
ANSWER:
x
y
−10
−2
1 x 5 1 y = ( −10 ) = −2 5
y=
−5
−1
0
0
5
1
10
2
64.
1 ( −5) = −1 5 1 y = (0) = 0 5 1 y = ( 5) = 1 5 1 y = (10 ) = 2 5
y=
PROBLEM: x + 2y = 0 SOLUTION: x + 2y = 0 x − x + 2 y = −x 2 y = −x 2 1 y=− x 2 2 1 y=− x 2 ANSWER:
x
y
−4
2
−2
1
0
0
1 y=− x 2 1 y = − ( −4 ) = 2 2 1 y = − ( −2 ) = 1 2 1 y = − ( 0) = 1 2
2
−1
4
−2
65.
1 ( 2 ) = −1 2 1 y = − ( 4 ) = −2 2
y=−
PROBLEM: 1 x− y =3 10 SOLUTION: 1 x− y =3 10 1 1 1 x − x − y = 3− x 10 10 10 1 −y = 3− x 10 1 y = x −3 10
x −20 −10 0 10 20
ANSWER: y 1 y = x−3 10 −5 1 y = ( −20 ) − 3 = −2 − 3 = −5 10 −4 1 y = ( −10 ) − 3 = −1 − 3 = −4 10 −3 1 y = ( 0 ) − 3 = 0 − 3 = −3 10 −2 1 y = (10 ) − 3 = 1 − 3 = −2 10 −1 1 y = ( 20 ) − 3 = 2 − 3 = −1 10
66.
PROBLEM: 3 x + 5 y = 30 2 SOLUTION: 3 x + 5 y = 30 2 3 3 3 x − x + 5 y = 30 − x 2 2 2 3 5 y = 30 − x 2 5 30 3 y= − x 5 5 2⋅5 3 y = 6− x 10 ANSWER:
x
y
−20
12
−10
9
0
6
10
3
20
0
3 x 10 3 y = 6 − ( −20 ) = 6 + 6 = 12 10 3 y = 6 − ( −10 ) = 6 + 3 = 9 10 3 y = 6 − (0) = 6 − 0 = 6 10 3 y = 6 − (10 ) = 6 − 3 = 3 10 3 y = 6 − ( 20 ) = 6 − 6 = 0 10
y = 6−
Part C: Horizontal and Vertical Lines Find at least five ordered pair solutions and graph. 67.
PROBLEM: y =4 ANSWER:
x −5 −4 0 1 2
y 4 4 4 4 4
68.
PROBLEM: y = − 10
ANSWER:
x −5 −4 0 1 2
69.
y −10 −10 −10 −10 −10
PROBLEM: x=4 ANSWER:
x 4 4 4 4 4
y 1 2 3 4 5
70.
PROBLEM: x = −1 ANSWER:
x −1 −1 −1 −1 −1
y 1 2 3 4 5
71.
PROBLEM: y=0
x 1 2 3 4 5
ANSWER: y 0 0 0 0 0
72.
PROBLEM: x=0 ANSWER:
x 0 0 0 0 0
y 1 2 3 4 5
73.
PROBLEM: 3 y= 4 ANSWER:
x −1 −2 0 1 2
y 3 4 3 4 3 4 3 4 3 4
74.
PROBLEM: 5 x=− 4 ANSWER:
x 5 − 4 5 − 4 5 − 4
y −1 −2 0
5 4 5 − 4
1
75.
PROBLEM: Graph the lines y = − 4 and x = 2 on the same set of axes. Where do they intersect?
−
2
ANSWER:
x −1 −2 0 1 2 x=2 2 2 2 2 2
y = −4 −4 −4 −4 −4 −4 y −4 −3 0 3 4
The lines intersect at (2, −4). 76.
PROBLEM: Graph the lines y = 5 and x = −5 on the same set of axes. Where do they intersect? SOLUTION:
x −5 −4 0 4 5
y =5 5 5 5 5 5
x = −5 −5 −5 −5 −5 −5
y =5 −5 −4 0 4 5
ANSWER: The lines intersect at (−5,5).
77.
PROBLEM: What is the equation that describes the x-axis? SOLUTION: If the line is considered to be the x-axis or the line lies completely on the x-axis, the value of y for any value of x is zero. Hence the equation is y = 0. ANSWER: y=0
78.
PROBLEM: What is the equation that describes the y-axis? SOLUTION: If the line is considered to be the y-axis or the line lies completely on the y-axis, the value of x for any value of y is zero. ANSWER: Hence the equation is x = 0.
Part D: Mixed Practice Graph by plotting points
79.
PROBLEM: 3 y = − x+6 5 SOLUTION:
x
y
−5
9
0
6
5
3
10
0
15
−3
3 y = − x+6 5 3 y = − ( −5) + 6 = 3 + 6 = 9 5 3 y = − ( 0) + 6 = 0 + 6 = 6 5 3 y = − ( 5) + 6 = −3 + 6 = 3 5 3 y = − (10 ) + 6 = −6 + 6 = 0 5 3 y = − (15 ) + 6 = −9 + 6 = −3 5
ANSWER:
80.
PROBLEM: 3 y = x −3 5
x
y
−5
−6
0
−3
5
0
10
3
15
6
SOLUTION: 3 y = x −3 5 3 y = ( −5) − 3 = −3 − 3 = −6 5 3 y = ( 0 ) − 3 = 0 − 3 = −3 5 3 y = ( 5) − 3 = 3 − 3 = 0 5 3 y = (10 ) − 3 = 6 − 3 = 3 5 3 y = (15) − 3 = 9 − 3 = 6 5 ANSWER:
81.
PROBLEM: y = −3 SOLUTION:
x −10 −5 0 5 10
y −3 −3 −3 −3 −3 ANSWER:
82.
PROBLEM: x = −5 SOLUTION:
x −5 −5 −5 −5 −5
y −8 −4 0 4 8
ANSWER:
83.
x −4 −2 0 2 4
PROBLEM: 3x − 2 y = 6 SOLUTION: 3x − 2 y = 6 3x − 3x − 2 y = 6 − 3x −2 y = 6 − 3 x 2 y = 3x − 6 2 3 6 y = x− 2 2 2 3 y = x −3 2 y 3 y = x −3 2 −9 3 y = ( −4 ) − 3 = −6 − 3 = −9 2 −6 3 y = ( −2 ) − 3 = −3 − 3 = −6 2 −3 3 y = ( 0 ) − 3 = 0 − 3 = −3 2 0 3 y = ( 2) − 3 = 3 − 3 = 0 2 3 3 y = ( 4) − 3 = 6 − 3 = 3 2
ANSWER:
84.
PROBLEM: −2 x + 3 y = −12
x
y
−6
−8
−3
−6
0
−4
3
−2
6
0
SOLUTION: −2 x + 3 y = −12 −2 x + 2 x + 3 y = −12 + 2 x 3 y = 2 x − 12 3 2 12 y = x− 3 3 3 2 y = x−4 3 2 y = x−4 3 2 y = ( −6 ) − 4 = −4 − 4 = −8 3 2 y = ( −3) − 4 = −2 − 4 = −6 3 2 y = ( 0 ) − 4 = 0 − 4 = −4 3 2 y = ( 3) − 4 = 2 − 4 = −2 3 2 y = ( 6) − 4 = 4 − 4 = 0 3
ANSWER:
Part E: Discussion Board Topics
85.
PROBLEM: Discuss the significance of the relationship between algebra and geometry in describing lines. ANSWER: Many concepts in geometry have their counterpart in algebra. For example, a point in geometry corresponds to an ordered pair (x, y) of numbers in algebra, a line corresponds to a set of ordered pairs satisfying an equation of the form ax + by = c (a, b, c ∈R), the intersection of two lines to the set of ordered pairs that satisfy the corresponding equations.
86.
PROBLEM: Give real world examples relating two unknowns. ANSWER: The cost of a flower and one box of chocolates are $1 and $10, respectively. If John spends $100 on flowers and chocolates for his girlfriend on Valentine’s day, how many flowers and boxes of chocolates did he buy? Let the number of flowers bought be x. Let the number of boxes of chocolates bought be y. Amount spent on flowers = 1⋅x = x Amount spent on chocolates = 10⋅y = 10y Total amount spent = x + 10y x + 10y = 100
The cost of a kilo of fruits and a liter of vanilla ice-cream is $15 and $20, respectively. If Chris, an ice-cream vendor, spends $300 per day on fruits and icecream, to prepare fruit salads with vanilla ice-cram in them, what is the quantity of fruits and ice-cream bought by Chris? Let the number of kilos of fruits bought be x. Let the number of liters of ice-cream bought be y. Amount spent on fruits = 15⋅x = 15x
Amount spent on ice-cream = 20⋅y Total amount spent = 15x + 20y 15x + 20y = 300 15 20 300 x+ y= 5 5 5 3 x + 4 y = 60
3.3 Graph using Intercepts
Part A: Intercepts Given the graph find the x‐ and y‐intercepts.
1.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = −3. Hence the y-intercept is (0, −3). The line intersects the x-axis at x = 4. Hence the x-intercept is (4, 0). ANSWER: x-intercept is (4, 0), y-intercept is (0, −3).
2.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = 1. Hence the y-intercept is (0, 1). The line intersects the x-axis at x = 2. Hence the x-intercept is (2, 0). ANSWER: x-intercept is (2, 0), y-intercept is (0, 1).
3.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = −3. Hence the y-intercept is (0,
−3). The line does not intersect the x-axis. Hence there is no x-intercept. ANSWER: no x-intercept, y-intercept is (0, −3).
4.
PROBLEM:
SOLUTION: The line does not intersect the y-axis. Hence there is no y-intercept. The line intersects the x-axis at x = 6. Hence the x-intercept is (6, 0). ANSWER: x-intercept is (6, 0), no y-intercept.
5.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = 0. Hence the y-intercept is (0,
0). The line intersects the x-axis at x = 0. Hence the x-intercept is (0, 0). ANSWER: x-intercept is (0, 0), y-intercept is (0, 0).
6.
PROBLEM:
SOLUTION: The line intersects the y-axis at y = −1. Hence the y-intercept is (0, −1). The line intersects the x-axis at x = −2. Hence the x-intercept is (−2, 0). ANSWER: x-intercept is (−2, 0), y-intercept is (0, −1). Find the x‐ and y‐intercepts.
7.
PROBLEM: 5 x − 4 y = 20 SOLUTION: Set y = 0 5 x − 4 y = 20
5 x − 4 ( 0 ) = 20 5 x = 20 5 20 x= 5 5 x=4 x-intercept: (4, 0) Set x = 0 5 x − 4 y = 20 5 ( 0 ) − 4 y = 20 −4 y = 20 −4 20 y=− −4 4 y = −5 y-intercept: (0, −5)
ANSWER: (4, 0), (0, −5)
8.
PROBLEM: −2 x + 7 y = −28 SOLUTION: Set y = 0 −2 x + 7 y = −28
−2 x + 7 ( 0 ) = −28 −2 x = −28 −2 −28 x= −2 −2 x = 14 x-intercept: (14, 0) Set x = 0 −2 x + 7 y = −28 −2 ( 0 ) + 7 y = −28 7 y = −28 −28 7 y= 7 7 y = −4 y-intercept: (0, −4)
ANSWER: (14, 0), (0, −4)
9.
PROBLEM: x− y =3 SOLUTION: Set y = 0 x− y =3
x − (0) = 3 x=3 x-intercept: (3, 0) Set x = 0 x− y =3
( 0) − y = 3 y = −3 y-intercept: (0, −3) ANSWER: (3, 0), (0, −3)
10.
PROBLEM: −x + y = 0 SOLUTION: Set y = 0 −x + y = 0
− x + ( 0) = 0 x=0 x-intercept: (0, 0) Set x = 0 − ( 0) + y = 0 y=0 y-intercept: (0, 0)
ANSWER: (0, 0), (0, 0)
11.
PROBLEM: 3x − 4 y = 1 SOLUTION: Set y = 0 3x − 4 y = 1 3x − 4 ( 0 ) = 1 3x = 1 3 1 x= 3 3 1 x= 3
⎛1 ⎞ x-intercept: ⎜ , 0 ⎟ ⎝3 ⎠ Set x = 0 3x − 4 y = 1 3( 0) − 4 y = 1 −4 y = 1 −4 1 y=− −4 4 1 y=− 4
1⎞ ⎛ y-intercept: ⎜ 0, − ⎟ 4⎠ ⎝
1⎞ ⎛1 ⎞ ⎛ ANSWER: ⎜ , 0 ⎟ , ⎜ 0, − ⎟ 4⎠ ⎝3 ⎠ ⎝ 12.
PROBLEM: −2 x + 5 y = 3 SOLUTION: Set y = 0 −2 x + 5 y = 3 −2 x + 5 ( 0 ) = 3 −2 x = 3 3 −2 x=− −2 2 3 x=− 2
⎛ 3 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 2 ⎠ Set x = 0 −2 x + 5 y = 3 −2 ( 0 ) + 5 y = 3 5y = 3 5 3 y= 5 5 3 y= 5
⎛ 3⎞ y-intercept: ⎜ 0, ⎟ ⎝ 5⎠ ⎛ 3 ⎞ ⎛ 3⎞ ANSWER: ⎜ − , 0 ⎟ , ⎜ 0, ⎟ ⎝ 2 ⎠ ⎝ 5⎠
13.
PROBLEM: 1 1 x − y =1 4 3 SOLUTION: Set y = 0 1 1 x − y =1 4 3 1 1 x − (0) = 1 4 3 1 x =1 4 x = 1⋅ 4 x=4 x-intercept: (4, 0)
Set x = 0 1 1 x − y =1 4 3 1 1 ( 0) − y = 1 4 3 1 − y =1 3 y = −1 ⋅ 3 y = −3 y-intercept: (0, – 3) ANSWER: (4, 0), (0, – 3)
14.
PROBLEM: 2 3 − x+ y = 2 5 4 SOLUTION: Set y = 0 2 3 − x+ y =2 5 4 2 3 − x + (0) = 2 5 4 2 − x=2 5 ⎛ 5⎞ ⎛ 2 ⎞ ⎛ 5⎞ ⎜ − ⎟⋅⎜ − x ⎟ = 2⋅⎜ − ⎟ ⎝ 2⎠ ⎝ 5 ⎠ ⎝ 2⎠ x = −5 x-intercept: (– 5, 0)
Set x = 0 2 3 − x+ y = 2 5 4 2 3 − (0) + y = 2 5 4 4 3 4 ⋅ y = 2⋅ 3 4 3 8 y= 3 ⎛ 8⎞ y-intercept: ⎜ 0, ⎟ ⎝ 3⎠ ⎛ 8⎞ ANSWER: (– 5, 0), ⎜ 0, ⎟ ⎝ 3⎠ 15.
PROBLEM: y=6 SOLUTION: x-intercept: none
y=6 y-intercept: (0, 6) ANSWER: none, (0, 6)
16.
PROBLEM: y = −3 SOLUTION: x-intercept: none
y = –3 y-intercept: (0, –3) ANSWER: x-intercept: none, y-intercept: (0, –3)
17.
PROBLEM: x=2 SOLUTION: x = 2 x-intercept: (2, 0)
y-intercept: none ANSWER: (2, 0), none
18.
PROBLEM: x = −1 SOLUTION: x = –1 x-intercept: (–1, 0)
y-intercept: none ANSWER: x-intercept: (–1, 0), y-intercept: none
19.
PROBLEM: y = mx + b SOLUTION: Set y = 0 y = mx + b
( 0 ) = mx + b mx + b = 0 mx + b − b = −b mx = −b m b x=− m m b x=− m
⎛ b ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ m ⎠ Set x = 0 y = mx + b y = m ( 0) + b y =b y-intercept: (0, b)
⎛ b ⎞ ANSWER: ⎜ − , 0 ⎟ , (0, b) ⎝ m ⎠ 20.
PROBLEM: ax + by = c SOLUTION: Set y = 0 ax + by = c ax + b ( 0 ) = c ax = c a c x= a a c x= a
⎛c ⎞ x-intercept: ⎜ , 0 ⎟ ⎝a ⎠ Set x = 0 ax + by = c a ( 0 ) + by = c by = c b c y= b b c y= b
⎛ c⎞ y-intercept: ⎜ 0, ⎟ ⎝ b⎠ ⎛c ⎞ ANSWER: ⎜ , 0 ⎟ , ⎝a ⎠
⎛ c⎞ ⎜ 0, ⎟ ⎝ b⎠
Part B: Graph using Intercepts Find the intercepts and graph.
21.
PROBLEM: 3x + 4 y = 12 SOLUTION: Set y = 0 3x + 4 y = 12
3x + 4 ( 0 ) = 12 3x = 12 3 12 x= 3 3 x=4 x-intercept: (4, 0) Set x = 0 3 x + 4 y = 12 3 ( 0 ) + 4 y = 12 4 y = 12 4 12 y= 4 4 y=3 y-intercept: (0, 3)
ANSWER: (4, 0), (0, 3)
22.
PROBLEM: −2 x + 3 y = 6 SOLUTION: Set y = 0 −2 x + 3 y = 6
−2 x + 3 ( 0 ) = 6 −2 x = 6 −2 6 x=− −2 2 x = −3 x-intercept: (–3 , 0) Set x = 0 −2 x + 3 y = 6 −2 ( 0 ) + 3 y = 6 3y = 6 3 6 y= 3 3 y=2 y-intercept: (0, 2)
ANSWER: (–3 , 0), (0, 2)
23.
PROBLEM: 5 x − 2 y = 10 SOLUTION: Set y = 0 5 x − 2 y = 10
5 x − 2 ( 0 ) = 10 5 x = 10 5 10 x= 5 5 x=2 x-intercept: (2, 0) Set x = 0 5 x − 2 y = 10 5 ( 0 ) − 2 y = 10 −2 y = 10 10 −2 y=− 2 −2 y = −5 y-intercept: (0, –5)
ANSWER: (2, 0), (0, –5)
24.
PROBLEM: −4 x − 8 y = 16 SOLUTION: Set y = 0 −4 x − 8 y = 16
−4 x − 8 ( 0 ) = 16 −4 16 x=− −4 4 x = −4 x-intercept: (–4 , 0) Set x = 0 −4 x − 8 y = 16 −4 ( 0 ) − 8 y = 16 −8 y = 16 16 −8 y=− 8 −8 y = −2 y-intercept: (0, –2 )
ANSWER: (–4 , 0), (0, –2 )
25.
PROBLEM: 1 1 − x + y =1 2 3 SOLUTION: Set y = 0 1 1 − x + y =1 2 3 1 1 − x + ( 0) = 1 2 3 1 − x =1 2 x = −2 x-intercept: (–2 , 0)
Set x = 0 1 1 − x + y =1 2 3 1 1 − ( 0) + y = 1 2 3 1 y =1 3 y =3 y-intercept: (0, 3) ANSWER: (–2 , 0), (0, 3)
26.
PROBLEM: 3 1 x − y = −3 4 2 SOLUTION: Set y = 0 3 1 x − y = −3 4 2 3 1 x − ( 0 ) = −3 4 2 3 x = −3 4 4 3 4 ⋅ x = −3 ⋅ 3 4 3 x = −4 x-intercept: (–4 , 0)
Set x = 0 3 1 x − y = −3 4 2 3 1 ( 0 ) − y = −3 4 2 1 0 − y = −3 2 1 − y = −3 2 y=6 y-intercept: (0, 6) ANSWER: (–4 , 0), (0, 6)
27.
PROBLEM: 5 2 x − y = 10 2 SOLUTION: Set y = 0 5 2 x − y = 10 2 5 2 x − ( 0 ) = 10 2 2 x = 10
2 10 x= 2 2 x=5 x-intercept: (5, 0) Set x = 0 5 2 x − y = 10 2 5 2 ( 0 ) − y = 10 2 ⎛ 2 ⎞⎛ 5 ⎞ ⎛ 2⎞ ⎜ − ⎟⎜ − ⎟ y = 10 ⋅ ⎜ − ⎟ ⎝ 5 ⎠⎝ 2 ⎠ ⎝ 5⎠ y = −4 y-intercept: (0, –4) ANSWER: (5, 0), (0, –4)
28.
PROBLEM: 7 2 x − y = −14 3 SOLUTION: Set y = 0 7 2 x − y = −14 3 7 2 x − ( 0 ) = −14 3 2 14 x=− 2 2 x = −7 x-intercept: (–7 , 0)
Set x = 0 7 2 x − y = −14 3 7 2 ( 0 ) − y = −14 3 ⎛ 3⎞ ⎛ 7⎞ ⎛ 3⎞ ⎜ − ⎟ ⋅ ⎜ − ⎟ y = −14 ⋅ ⎜ − ⎟ ⎝ 7⎠ ⎝ 3⎠ ⎝ 7⎠ y=6 y-intercept: (0, 6) ANSWER: (–7 , 0), (0, 6)
29.
PROBLEM: 4 x − y = −8 SOLUTION: Set y = 0 4 x − y = −8 4 x − 0 = −8
4 8 x=− 4 4 x = −2 x-intercept: (–2, 0) Set x = 0 4 x − y = −8 4 ( 0 ) − y = −8 − y = −8 y =8 y-intercept: (0, 8) ANSWER: (–2, 0), (0, 8)
30.
PROBLEM: 6x − y = 6 SOLUTION: Set y = 0 6x − y = 6 6x − 0 = 6 6x = 6 6 6 x= 6 6 x =1 x-intercept: (1 , 0)
Set x = 0 6x − y = 6
6 ( 0) − y = 6 −y = 6 y = −6 y-intercept: (0, –6) ANSWER: (1 , 0), (0, –6)
31.
PROBLEM: –x + 2y =1 SOLUTION: Set y = 0 –x + 2y =1
− x + 2 ( 0) = 1 −x = 1 x = −1 x-intercept: (–1, 0) Set x = 0 –x + 2y =1 0 + 2y =1 2y =1 2 1 y= 2 2 1 y= 2 ⎛ 1⎞ y-intercept: ⎜ 0, ⎟ ⎝ 2⎠ ⎛ 1⎞ ANSWER: (–1, 0), ⎜ 0, ⎟ ⎝ 2⎠
32.
PROBLEM: 3x + 4 y = 6 SOLUTION: Set y = 0 3x + 4 y = 6
3x + 4 ( 0 ) = 6 3x = 6 3 6 x= 3 3 x=2 x-intercept: (2, 0) Set x = 0 3x + 4 y = 6 3 ( 0) + 4 y = 6 4y = 6 4 6 y= 4 4 3 y= 2
⎛ 3⎞ y-intercept: ⎜ 0, ⎟ ⎝ 2⎠ ⎛ 3⎞ ANSWER: (2, 0), ⎜ 0, ⎟ ⎝ 2⎠
33.
PROBLEM: 2 x + y = −1 SOLUTION: Set y = 0 2 x + y = −1 2 x + 0 = −1 2 x = −1 2 1 x=− 2 2 1 x=− 2 ⎛ 1 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 2 ⎠
Set x = 0 2 x + y = −1 2 ( 0 ) + y = −1 y = −1 y-intercept: (0, –1) ⎛ 1 ⎞ ANSWER: ⎜ − , 0 ⎟ , (0, –1) ⎝ 2 ⎠
34.
PROBLEM: −2 x + 6 y = 3 SOLUTION: Set y = 0 −2 x + 6 y = 3 −2 x + 6 ( 0 ) = 3 −2 x = 3 3 −2 x=− −2 2 3 x=− 2
⎛ 3 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 2 ⎠ Set x = 0 −2 x + 6 y = 3 −2 ( 0 ) + 6 y = 3 0 + 6y = 3 6y = 3 6 3 y= 6 6 1 y= 2 ⎛ 1⎞ y-intercept: ⎜ 0, ⎟ ⎝ 2⎠ ⎛ 3 ⎞ ANSWER: ⎜ − , 0 ⎟ , ⎝ 2 ⎠
⎛ 1⎞ ⎜ 0, ⎟ ⎝ 2⎠
35.
PROBLEM: 15 x + 4 y = −60 SOLUTION: Set y = 0 15 x + 4 y = −60
15 x + 4 ( 0 ) = −60 15 x = −60 15 60 x=− 15 15 x = −4 x-intercept: (–4, 0) Set x = 0 15 x + 4 y = −60 15 ( 0 ) + 4 y = −60 4 y = −60 4 60 y=− 4 4 y = −15 y-intercept: (0, –15)
ANSWER: (–4, 0), (0, –15)
36.
PROBLEM: −25 x + 3 y = 75 SOLUTION: Set y = 0 −25 x + 3 y = 75
−25 x + 3 ( 0 ) = 75 −25 x = 75 75 −25 x=− 25 −25 x = −3 x-intercept: (–3, 0) Set x = 0 −25 x + 3 y = 75 −25 ( 0 ) + 3 y = 75 3 y = 75 3 75 y= 3 3 y = 25 y-intercept: (0, 25)
ANSWER: (–3, 0), (0, 25)
37.
PROBLEM: 4x + 2 y = 0 SOLUTION: Set y = 0 4x + 2 y = 0
4x + 2 ( 0) = 0 x=0 x-intercept: (0, 0) Set x = 0 4x + 2 y = 0 4 ( 0) + 2 y = 0 2y = 0 y=0 y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
38.
PROBLEM: 3x − y = 0 SOLUTION: Set y = 0 3x − y = 0
3x − ( 0 ) = 0 x=0 x-intercept: (0, 0) Set x = 0 3x − y = 0 3( 0) − y = 0 y=0
y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
39.
PROBLEM: −12 x + 6 y = −4 SOLUTION: Set y = 0 −12 x + 6 y = −4 −12 x + 6 ( 0 ) = −4 −12 x = −4 −12 −4 x= −12 −12 1 x= 3 ⎛1 ⎞ x-intercept: ⎜ , 0 ⎟ ⎝3 ⎠
Set x = 0 −12 x + 6 y = −4 −12 ( 0 ) + 6 y = −4 6 y = −4 6 4 y=− 6 6 2 y=− 3
2⎞ ⎛ y-intercept: ⎜ 0, − ⎟ 3⎠ ⎝ 2⎞ ⎛1 ⎞ ⎛ ANSWER: ⎜ , 0 ⎟ , ⎜ 0, − ⎟ 3⎠ ⎝3 ⎠ ⎝
40.
PROBLEM: 3x + 12 y = −4 SOLUTION: Set y = 0 3x + 12 y = −4 3x + 12 ( 0 ) = −4 3x = −4 3 4 x=− 3 3 4 x=− 3
⎛ 4 ⎞ x-intercept: ⎜ − , 0 ⎟ ⎝ 3 ⎠ Set x = 0 3 x + 12 y = −4 3 ( 0 ) + 12 y = −4 12 y = −4 12 4 y=− 12 12 1 y=− 3
1⎞ ⎛ y-intercept: ⎜ 0, − ⎟ 3⎠ ⎝ ⎛ 4 ⎞ ANSWER: ⎜ − , 0 ⎟ , ⎝ 3 ⎠
1⎞ ⎛ ⎜ 0, − ⎟ 3⎠ ⎝
41.
PROBLEM: y = 2x + 4 SOLUTION: Set y = 0 y = 2x + 4
0 = 2x + 4 2x + 4 − 4 = 0 − 4 2 x = −4 2 4 x=− 2 2 x = −2 x-intercept: (–2, 0) Set x = 0 y = 2x + 4 y = 2 ( 0) + 4 y=4 y-intercept: (0, 4) ANSWER: (–2, 0), (0, 4)
42.
PROBLEM: y = −x + 3 SOLUTION: Set y = 0 y = −x + 3 0 = −x + 3 − x + 3 − 3 = −3 − x = −3 x=3 x-intercept: (3, 0)
Set x = 0 y = −x + 3 y = 0+3 y =3 y-intercept: (0, 3) ANSWER: (3, 0), (0, 3)
43.
PROBLEM: 1 y = x +1 2 SOLUTION: Set y = 0 1 y = x +1 2 1 0 = x +1 2 1 0 −1 = x + 1 −1 2 1 −1 = x 2 x = −2 x-intercept: (–2, 0)
Set x = 0 1 y = x +1 2 1 y = ( 0) + 1 2 y =1 y-intercept: (0, 1) ANSWER: (–2, 0), (0, 1)
44.
PROBLEM: 2 y = x −3 3 SOLUTION: Set y = 0 2 y = x−3 3 2 0 = x −3 3 2 0+3 = x −3+3 3 2 x=3 3 3 2 3 ⋅ x = 3⋅ 2 3 2 9 x= 2 ⎛ 1 ⎞ x-intercept: ⎜ 4 , 0 ⎟ ⎝ 2 ⎠ Set x = 0 2 y = x−3 3 2 y = ( 0) − 3 3 y = −3 y-intercept: (0, –3)
⎛ 1 ⎞ ANSWER: ⎜ 4 , 0 ⎟ , (0, –3) ⎝ 2 ⎠
45.
PROBLEM: 2 y = − x +1 5 SOLUTION: Set y = 0 2 y = − x +1 5 2 0 = − x +1 5 2 0 −1 = − x + 1 −1 5 2 −1 = − x 5 ⎛ 5⎞ ⎛ 2⎞ ⎛ 5⎞ ⎜ − ⎟ ⋅ ⎜ − ⎟ x = −1 ⋅ ⎜ − ⎟ ⎝ 2⎠ ⎝ 5⎠ ⎝ 2⎠ 5 x= 2 ⎛ 1 ⎞ x-intercept: ⎜ 2 , 0 ⎟ ⎝ 2 ⎠
Set x = 0 2 y = − x +1 5 2 y = − ( 0) + 1 5 y =1 y-intercept: (0, 1) ⎛ 1 ⎞ ANSWER: ⎜ 2 , 0 ⎟ , (0, 1) ⎝ 2 ⎠
46.
PROBLEM: 5 5 y =− x− 8 4 SOLUTION: Set y = 0 5 5 y =− x− 8 4 5 5 0=− x− 8 4 5 5 5 5 0+ = − x− + 4 8 4 4 5 ⎛ 8⎞ ⎛ 8⎞ ⎛ 5⎞ ⎜ − ⎟⋅⎜ − ⎟ x = ⋅⎜ − ⎟ 4 ⎝ 5⎠ ⎝ 5⎠ ⎝ 8⎠ x = −2 x-intercept: (–2, 0)
Set x = 0 5 5 y =− x− 8 4 5 5 y = − ( 0) − 8 4 5 y=− 4 1⎞ ⎛ y-intercept: ⎜ 0, −1 ⎟ 4⎠ ⎝ 1⎞ ⎛ ANSWER: (–2, 0), ⎜ 0, −1 ⎟ 4⎠ ⎝
47.
PROBLEM: 7 7 y =− x− 8 2 SOLUTION: Set y = 0 7 7 y =− x− 8 2 7 7 0=− x− 8 2 7 7 7 7 0+ = − x− + 2 8 2 2 7 7 − x= 8 2 ⎛ 8⎞ ⎛ 7⎞ ⎛7⎞ ⎛ 8⎞ ⎜ − ⎟⋅⎜ − ⎟ x = ⎜ ⎟⋅⎜ − ⎟ ⎝ 7⎠ ⎝ 8⎠ ⎝2⎠ ⎝ 7⎠ x = −4 x-intercept: (–4, 0)
Set x = 0 7 7 y =− x− 8 2 7 7 y = − ( 0) − 8 2 7 y=− 2 1⎞ ⎛ y-intercept: ⎜ 0, −3 ⎟ 2⎠ ⎝ 1⎞ ⎛ ANSWER: (–4, 0), ⎜ 0, −3 ⎟ 2⎠ ⎝
48.
PROBLEM: 3 y = −x + 2 SOLUTION: Set y = 0 3 y = −x + 2 3 0 = −x + 2 3 3 3 −x + − = 0 − 2 2 2 3 −x = − 2 3 x= 2 ⎛ 1 ⎞ x-intercept: ⎜1 , 0 ⎟ ⎝ 2 ⎠
Set x = 0
y = −x + y = 0+ y=
3 2
3 2
3 2
⎛ 1⎞ y-intercept: ⎜ 0,1 ⎟ 2⎠ ⎝ ⎛ 1 ⎞ ANSWER: ⎜1 , 0 ⎟ , ⎝ 2 ⎠
⎛ 1⎞ ⎜ 0,1 ⎟ 2⎠ ⎝
49.
PROBLEM: y=3 SOLUTION: No x-intercept
y =3 y-intercept: (0, 3) ANSWER: No x-intercept, (0, 3)
50.
PROBLEM: 3 y= 2 SOLUTION: No x-intercept
y=
3 2
⎛ 1⎞ y-intercept: ⎜ 0,1 ⎟ 2⎠ ⎝ ⎛ 1⎞ ANSWER: No x-intercept, ⎜ 0,1 ⎟ 2⎠ ⎝
51.
PROBLEM: x=5 SOLUTION: x = 5 x-intercept: (5, 0)
No y-intercept ANSWER: (5, 0), No y-intercept.
52.
PROBLEM: x = −2 SOLUTION: x = −2 x-intercept: (–2, 0)
No y-intercept ANSWER: (–2, 0), No y-intercept.
53.
PROBLEM: y = 5x SOLUTION: Set y = 0 y = 5x 0 = 5x x=0 x-intercept: (0, 0)
Set x = 0 y = 5x
y = 5 ( 0) y=0 y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
54.
PROBLEM: y = −x SOLUTION: Set y = 0 y = −x y=0 x-intercept: (0, 0)
Set x = 0 y = −x
0 = −x x=0 y-intercept: (0, 0) ANSWER: (0, 0), (0, 0)
Part C: Intercepts of Nonlinear Graphs. Given the graph find the x‐ and y‐intercepts.
55.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (–3, 0) and (3, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −3).
56.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (–3, 0) and (1, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −1).
57.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (–4, 0) and (0, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, 0).
58.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in one place. Therefore, this graph has only one x-intercept, namely (2, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only y-intercept is (0, −2).
59.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (−2, 0) and (2, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −1).
60.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (−1, 0) and (3, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, 3).
61.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in three places. Therefore, this graph has three x-intercepts, namely (−3, 0), (0, 0), and (2, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, 0).
62.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has two x-intercepts, namely (−4, 0) and (−1, 0). Furthermore, the graph intersects the y-axis in one place, hence, the only yintercept is (0, −2).
63.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has three x-intercepts, namely (−4, 0) and (4, 0). Furthermore, the graph intersects the y-axis in two places, hence, the y-intercept are (0, −4) and (0, 4).
64.
PROBLEM:
ANSWER: From the graph, we can see that it intersects the x-axis in one place. Therefore, this graph has one x-intercept, namely (0, 0). Furthermore, the graph intersects the y-axis in two places, hence, the y-intercept are (0, 0) and (0, −6).
Part D: Discussion Board Topics
65.
PROBLEM: What are the x-intercepts of the line y = 0 ? ANSWER: The line y = 0 does not have any x-intercepts.
66.
PROBLEM: What are the y-intercepts of the line x = 0 ? ANSWER: The line x = 0 does not have any y-intercepts.
67.
PROBLEM: Do all lines have intercepts? ANSWER: No. An example is a horizontal line like y = 3. This line never intersects the x-axis, therefore it does not have an x-intercept. (The value of y may vary)
68.
PROBLEM: How many intercepts can a circle have? Draw circles showing all possible numbers of intercepts. ANSWER: A circle can have a maximum of 4 intercepts.
From the graph, we can see that it intersects the x-axis in two places. Therefore, this graph has three x-intercepts, namely (−4, 0) and (4, 0). Furthermore, the graph intersects the y-axis in two places, hence, the y-intercept are (0, 4) and (0, −4). (The x- and y-intercepts may vary.)
69.
PROBLEM: Research and post the definitions of line segment, ray and line. Why are the arrows important? ANSWER: Line segment: A line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. (Source: http://en.wikipedia.org/wiki/Line_segment)
Ray: A ray is a part of a line that begins at a particular point (called the endpoint) and extends endlessly in one direction. A ray is also called half-line. (Source: http://www.icoachmath.com/SiteMap/Ray.html) Line: A straight line is the shortest distance between any two points on a plane. (Source: http://www.mathopenref.com/line.html) The arrows at the end of a line indicate the line can extend beyond the endpoints in either direction. 70.
PROBLEM: Can a function have more than one y-intercept? Explain ANSWER: Functions associate x values to no more than one y value as part of their definition, they can have at most one y-intercept. Some 2-dimensional mathematical relationships such as circles, ellipses, and hyperbolas can have more than one y-intercept. (Source: http://en.wikipedia.org/wiki/Y-intercept)
3.4 Graph using the yintercept and Slope
Part A: Slope Determine the slope and the y‐intercept of the given graph.
1.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, 3). ( x1 , y1 ) ( x2 , y2 )
(4, 0) (0,3) y − y 3−0 3 m= 2 1 = =− x2 − x1 0 − 4 4 3 ANSWER: − , (0, 3). 4 2.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, –2). ( x1 , y1 ) ( x2 , y2 )
(3, 0) (0, − 2) y − y −2 − 0 − 2 2 = = m= 2 1 = x2 − x1 0 − 3 −3 3 ANSWER:
2 , y-intercept is (0, –2). 3
3.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, 2). ( x1 , y1 ) ( x2 , y2 )
(1, 2) (0, 2) y − y 2−2 0 = =0 m= 2 1 = x2 − x1 1 − 0 1 ANSWER: 0, (0, 2).
4.
PROBLEM:
SOLUTION: From the graph, we can see that it does not intersect the y-axis. Hence, there is no y-intercept. ( x1 , y1 ) ( x2 , y2 )
(2, 0) (2, 2) y − y 2−0 2 = m= 2 1 = x2 − x1 2 − 2 0 ANSWER: The slope is not defined, no y-intercept.
5.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis at one place. Hence, the y-intercept is (0, 0). ( x1 , y1 ) ( x2 , y2 )
(0, 0) (1, 2) y − y 2−0 2 m= 2 1 = = =2 x2 − x1 1 − 0 1 ANSWER: 2, (0, 0).
6.
PROBLEM:
SOLUTION: From the graph, we can see that it intersects the y-axis. Hence, the y-intercept is (0, –1). ( x1 , y1 ) ( x2 , y2 )
(−2, 0) (0, − 1) y −y −1 − 0 1 m= 2 1 = =− x2 − x1 0 − ( −2 ) 2 1 ANSWER: − , y-intercept is (0, –1). 2
Determine the slope given two points.
7.
PROBLEM: (3, 2) and (5, 1) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(3, 2) (5,1) y − y 1− 2 1 m= 2 1 = =− x2 − x1 5 − 3 2 ANSWER: −
8.
1 2
PROBLEM: (7, 8) and (−3, 5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(7,8) (−3,5) y −y 5−8 3 −3 m= 2 1 = = = x2 − x1 −3 − 7 −10 10 ANSWER:
9.
3 10
PROBLEM: (2, −3) and (−3, 2) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(2, −3) (−3, 2) y − y 2 − ( −3 ) 5 m= 2 1 = = − = −1 x2 − x1 −3 − 2 5 ANSWER: −1
10.
PROBLEM: (−3, 5) and (7, −5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−3, 5) (7, −5) y −y −5 − 5 −10 m= 2 1 = = = −1 x2 − x1 7 − ( −3) 10 ANSWER: −1
11.
PROBLEM: (−1, −6) and (3, 2) SOLUTION: ( x1 , y1 ) ( x2 , y2 ) (−1, −6) m=
( 3, 2 )
y2 − y1 2 − ( −6 ) 8 = = =2 x2 − x1 3 − ( −1) 4
ANSWER: 2
12.
PROBLEM: (5, 3) and (4, 12) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( 5, 3) m=
( 4, 12 )
y2 − y1 12 − 3 9 = = = −9 x2 − x1 4 − 5 −1
ANSWER: −9
13.
PROBLEM: (−9, 3) and (−6, −5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−9, 3) (−6, −5) y −y −5 − 3 −8 8 = m= 2 1 = =− x2 − x1 −6 − ( −9 ) 3 3 ANSWER: −
14.
8 3
PROBLEM: (−22, 4) and (−8, −12) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−22, 4) (−8, −12) y −y −12 − 4 −16 8 m= 2 1 = = =− x2 − x1 −8 − ( −22 ) 14 7 ANSWER: −
15.
8 7
PROBLEM: ( 12 , − 13 ) and ( − 12 ,
2 3
)
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( 12 , − 13 ) ( − 12 , 23 )
2 ⎛ 1⎞ 2 1 3 − − + y2 − y1 3 ⎜⎝ 3 ⎟⎠ m= = = 3 3 = 3 = −1 1 1 1 1 2 x2 − x1 − − − − − 2 2 2 2 2 ANSWER: −1
16.
PROBLEM: ( − 34 , 32 ) and ( 14 , − 12 ) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( − 34 , 32 ) ( 14 , − 12 )
1 3 4 − − − y −y 2 2 = 2 = −2 m= 2 1 = 4 x2 − x1 1 ⎛ 3 ⎞ −⎜− ⎟ 4 4 ⎝ 4⎠ ANSWER: −2
17.
PROBLEM: ( − 13 , 85 ) and ( 12 , − 34 ) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( − 13 , 85 ) ( 12 , − 43 )
3 5 11 −6 − 5 − − − y2 − y1 4 8 = 8 = 8 = − 11 ⋅ 6 = − 33 m= = 3+ 2 5 8⋅5 20 x2 − x1 1 ⎛ 1 ⎞ −⎜− ⎟ 6 6 2 ⎝ 3⎠ 33 ANSWER: − 20
18.
PROBLEM: ( − 53 , − 23 ) and
( 101 , 54 )
SOLUTION: ( x1 , y1 ) ( x2 , y2 )
( − 53 , − 32 ) ( 101 , 54 ) 4 ⎛ 3 ⎞ 8 + 15 23 −⎜− ⎟ y2 − y1 23 5 ⎝ 2⎠ m= = = 10 = 10 = 7 1 ⎛ 3 ⎞ 1+ 6 x2 − x1 7 −⎜− ⎟ 10 10 10 ⎝ 5 ⎠
ANSWER:
23 7
19.
PROBLEM: (3, −5) and (5, −5) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(3, −5)
(5, −5) y − y − 5 − ( −5 ) 0 m= 2 1 = = =0 x2 − x1 5−3 2
ANSWER: 0
20.
PROBLEM: (−3, 1) and (−14, 1) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−3, 1) (−14, 1) y −y 1−1 0 m= 2 1 = = =0 x2 − x1 −14 − ( −3 ) −11 ANSWER: 0
21.
PROBLEM: (−2, 3) and (−2, −4) SOLUTION: ( x1 , y1 ) ( x2 , y2 )
(−2, 3) (−2, −4) y −y −4 − 3 −7 m= 2 1 = = x2 − x1 −2 − ( −2 ) 0 ANSWER: The slope is not defined.
22.
PROBLEM: (−4, −4) and (5, 5) SOLUTION: ( x1 , y1 ) ( x2 , y2 ) (−4, −4) m=
( 5, 5)
y2 − y1 5 − ( −4 ) 9 = = =1 x2 − x1 5 − ( −4 ) 9
ANSWER: 1
23.
PROBLEM: A roof drops 4 ft for every 12 ft forward. Determine the slope of the roof. SOLUTION: If the roof moves forward, the x-intercept is 12, i.e. (12, 0). If the roof drops down, the y-intercept is 4, i.e. (0, 4). ( x1 , y1 ) ( x2 , y2 )
(0, 4) (12, 0) y −y 0−4 4 1 =− =− m= 2 1 = x2 − x1 12 − 0 12 3 ANSWER: −
24.
1 3
PROBLEM: A road drops 300 ft for every 5,280 ft forward. Determine the slope of the road. SOLUTION: If the road moves forward, the x-intercept is 5280, i.e. (5280, 0). If the road drops down, the y-intercept is 300, i.e. (0, 300). ( x1 , y1 ) ( x2 , y2 )
( 5280, 0 ) ( 0, 300 ) m=
y2 − y1 300 − 0 300 5 = =− =− x2 − x1 0 − 5280 5280 88
ANSWER: −
5 88
25.
PROBLEM: The following graph gives the US population of persons 65 years old and over (Source: U.S. Census Bureau). At what rate was this population increasing from the year 2000 to 2008?
SOLUTION: The rate of population increasing from the year 2000 to 2008 can be found out by calculating the slope between the points corresponding to 2000 and 2008 on the x-axis and 33 and 37 corresponding to the y-axis. 37 − 33 4 1 m= = = 2008 − 2000 8 2 1 ANSWER: The population is increasing at a rate of million per year from 2 2000 to 2008.
26.
PROBLEM: The following graph gives total consumer credit outstanding in the US (Source: U.S. Census Bureau). At what rate was consumer credit increasing from the year 2002 to 2008?
SOLUTION: The rate of increasing consumer credit from the year 2002 to 2008 can be found out by calculating the slope between the points corresponding to 2002 and 2008 on the x-axis and 2.0 and 2.6 corresponding to the y-axis. 2.6 − 2.0 0.6 m= = = 0.1 2008 − 2002 6 ANSWER: The consumer credit increases at a rate of $0.1 billion per year from 2002 to 2008.
27.
PROBLEM: A commercial van was purchased new for $20,000 and is expected to be worth $4,000 in 8 years. Determine the rate at which the van will depreciate in value. SOLUTION: The rate of depreciation can be found out by calculating the slope, which helps in determining the rate at which the van will depreciate in value. Price is plotted in the y-axis and the number of years in the x-axis. $20, 000 − $4000 $16, 000 m= = = $2, 000 8−0 8 ANSWER: The van will depreciate by $2,000 per year.
28.
PROBLEM: A commercial grade copy machine was purchased new for $4,800 and will be considered worthless in 6 years. Determine the rate at which the copy machine will depreciate in value. SOLUTION: The rate of depreciation can be found out by calculating the slope, which helps in determining the rate at which the commercial grade copy machine will depreciate in value. Price is plotted in the y-axis and the number of years in the x-axis. $4800 − 0 $4800 m= = = $800 6−0 6 ANSWER: The commercial grade copy machine will depreciate by $800 per year.
29.
PROBLEM: Find y if the slope of the line passing through (−2, 3) and (4, y) is 12. SOLUTION: m = 12 ( x1 , y1 ) ( x2 , y2 )
(−2, 3) (4, y ) y −y y −3 y −3 12 = 2 1 = = 6 x2 − x1 4 − ( −2 ) y − 3 + 3 = 72 + 3 y = 75
ANSWER: y = 75
30.
PROBLEM: Find y if the slope of the line passing through (5, y) and (6, −1) is 10. SOLUTION: m = 10 ( x1 , y1 ) ( x2 , y2 )
( 5, y )
10 =
(6, −1)
y2 − y1 −1 − y −1 − y = = 6−5 1 x2 − x1
−1 − y = 10 −1 + 1 − y = 10 + 1 − y = 11 y = −11
ANSWER: y = −11
31.
PROBLEM: Find y if the slope of the line passing through (5, y) and (−4, 2) is 0. SOLUTION: m=0 ( x1 , y1 ) ( x2 , y2 )
( 5, y ) 0=
(−4, 2)
y2 − y1 2 − y 2 − y = = x2 − x1 −4 − 5 −9
2−2− y = 0−2 − y = −2 y=2
ANSWER: y = 2
32.
PROBLEM: Find x if the slope of the line passing through (−3, 2) and (x, 5) is undefined. SOLUTION: m = undefined ( x1 , y1 ) ( x2 , y2 ) (−3, 2)
( x, 5 )
3 y2 − y1 5−2 3 = = = 0 x2 − x1 x − ( −3) x + 3 x+3= 0 x + 3 − 3 = −3 x = −3
ANSWER: x = −3
Part B: Slope‐Intercept Form Express the given linear equation in slope‐intercept form and identify the slope and y‐intercept.
33.
PROBLEM: 6 x − 5 y = 30 SOLUTION: 6 x − 5 y = 30 6 x − 6 x − 5 y = 30 − 6 x 5 y = 6 x − 30 5 6 30 y = x− 5 5 5 6 y = x−6 5
6 ANSWER: The y-intercept is (0, – 6) and the slope is m = . 5 34.
PROBLEM: −2 x + 7 y = 28 SOLUTION: −2 x + 7 y = 28 −2 x + 2 x + 7 y = 28 + 2 x 7 y = 2 x + 28 7 2 28 y = x+ 7 7 7 2 y = x+4 7 ANSWER: The y-intercept is (0, 4) and the slope is m =
35.
2 . 7
PROBLEM: 9 x − y = 17 SOLUTION: 9 x − y = 17 9 x − 9 x − y = 17 − 9 x
y = 9 x − 17 ANSWER: The y-intercept is (0, – 17) and the slope is m = 9.
36.
PROBLEM: x − 3 y = 18 SOLUTION: x − 3 y = 18 x − x − 3 y = 18 − x
3 y = x − 18 3 1 18 y = x− 3 3 3 1 y = x−6 3
1 ANSWER: The y-intercept is (0, –6) and the slope is m = . 3 37.
PROBLEM: 2x − 3 y = 0 SOLUTION: 2x − 3y = 0 2x − 2x − 3y = 0 − 2x
3 y = 2x 3 2 y= x 3 3 2 y= x 3 2 ANSWER: The y-intercept is (0, 0) and the slope is m = . 3 38.
PROBLEM: −6 x + 3 y = 0 SOLUTION: −6 x + 3 y = 0 −6 x + 6 x + 3 y = 0 + 6 x 3y = 6x 3 6 y= x 3 3 y = 2x ANSWER: The y-intercept is (0, 0) and the slope is m = 2.
39.
PROBLEM: 5 2 3 x − 4 y = 10 SOLUTION: 5 2 3 x − 4 y = 10 2 3
x − 23 x − 54 y = 10 − 32 x
4 5
⋅ 54 y = 54 ⋅ 23 x − 54 ⋅10
y = 158 x − 8 ANSWER: The y-intercept is (0, –8) and the slope is m =
40.
8 . 15
PROBLEM: − 43 x + 15 y = −5 SOLUTION: − 43 x + 15 y = −5
− 43 x + 43 x + 15 y = −5 + 34 x 1 5
y = 43 x − 5
5 1
⋅ 15 y = 15 ⋅ 34 x − 15 ⋅ 5
y=
20 3
x − 25
ANSWER: The y-intercept is (0, –25) and the slope is m =
20 . 3
Graph the line given the slope and the y‐intercept.
41.
PROBLEM: 1 m = and (0, −2) 3 SOLUTION: 1 rise m= = 3 run Starting from the point (0, −2) we will use the slope to mark another point up 1 unit and to the right 3 units.
ANSWER:
42.
PROBLEM: 2 m = − and (0, 4) 3 SOLUTION:
2 rise = 3 run Starting from the point (0, 4) we will use the slope to mark another point down 2 units and to the right 3 units. m=−
ANSWER:
43.
PROBLEM: m = 3 and (0, 1) SOLUTION: 3 rise m= = 1 run Starting from the point (0, 1) we will use the slope to mark another point up 3 units and to the right 1 unit.
ANSWER:
44.
PROBLEM: m = −2 and (0, −1) SOLUTION: 2 rise m=− = 1 run Starting from the point (0, −1) we will use the slope to mark another point down 2 units and to the right 1 unit. ANSWER:
45.
PROBLEM: m = 0 and (0, 5) SOLUTION: rise m=0= run Starting from the point (0, 5) the line is a horizontal line with no slope.
ANSWER:
46.
PROBLEM: m undefined and (0, 0) SOLUTION:
rise run Starting from the point (0, 0) the line is along the y-axis since the slope is undefined. m = undefined =
ANSWER:
47.
PROBLEM: m = 1 and (0, 0) SOLUTION: 1 rise m= = 1 run Starting from the point (0, 0) we will use the slope to mark another point up 1 unit and to the right 1 unit.
ANSWER:
48.
PROBLEM: m = −1 and (0, 0) SOLUTION: 1 rise m=− = 1 run Starting from the point (0, 0) we will use the slope to mark another point down 1 unit and to the right 1 unit. ANSWER:
49.
PROBLEM: 15 m=− and (0, 20) 3 SOLUTION: 15 rise m=− = 3 run Starting from the point (0, 20) we will use the slope to mark another point down 15 units and to the right 3 units.
ANSWER:
50.
PROBLEM: m = −10 and (0, −5) SOLUTION: 10 rise m=− = 1 run Starting from the point (0, −5) we will use the slope to mark another point down 10 units and to the right 1 unit. ANSWER:
Graph using the slope and y‐intercept.
51.
PROBLEM: 2 y = x−2 3 SOLUTION:
2 The y-intercept is (0, –2) and the slope is m = . 3
2 rise = 3 run Starting from the point (0, –2) we will use the slope to mark another point up 2 units and to the right 3 units. m=
ANSWER:
52.
PROBLEM: 1 y = − x +1 3 SOLUTION:
1 The y-intercept is (0, 1) and the slope is m = − . 3 1 rise m=− = 3 run Starting from the point (0, 1) we will use the slope to mark another point down 1 unit and to the right 3 units. ANSWER:
53.
PROBLEM: y = −3 x + 6 SOLUTION: The y-intercept is (0, 6) and the slope is m = –3. 3 rise m=− = 1 run Starting from the point (0, 6) we will use the slope to mark another point down 3 units and to the right 1 unit. ANSWER:
54.
PROBLEM: y = 3x + 1 SOLUTION: The y-intercept is (0, 1) and the slope is m = 3. 3 rise m= = 1 run Starting from the point (0, 1) we will use the slope to mark another point down 3 units and to the right 1 unit. ANSWER:
55.
PROBLEM: 3 y= x 5 SOLUTION:
3 The y-intercept is (0, 0) and the slope is m = . 5 3 rise m= = 5 run Starting from the point (0, 0) we will use the slope to mark another point up 3 units and to the right 5 units. ANSWER:
56.
PROBLEM: 3 y=− x 7 SOLUTION:
3 The y-intercept is (0, 0) and the slope is m = − . 7 3 rise m=− = 7 run Starting from the point (0, 0) we will use the slope to mark another point down 3 units and to the right 7 units.
ANSWER:
57.
PROBLEM: y = −8 SOLUTION: The y-intercept is (0, –8) and the slope is m = 0. rise m=0= run Starting from the point (0, –8) the line is a horizontal line with no slope. ANSWER:
58.
PROBLEM: y=7 SOLUTION: The y-intercept is (0, 7) and the slope is m = 0. rise m=0= run Starting from the point (0, 7) the line is a horizontal line with no slope.
ANSWER:
59.
PROBLEM: y = −x + 2 SOLUTION: The y-intercept is (0, 2) and the slope is m = –1. 1 rise m=− = 1 run Starting from the point (0, 2) we will use the slope to mark another point down 1 unit and to the right 1 unit. ANSWER:
60.
PROBLEM: y = x +1 SOLUTION: The y-intercept is (0, 1) and the slope is m = 1. 1 rise m= = 1 run Starting from the point (0, 1) we will use the slope to mark another point up 1 unit and to the right 1 unit.
ANSWER:
61.
PROBLEM: 1 3 y = x+ 2 2 SOLUTION:
1 ⎛ 1⎞ The y-intercept is ⎜ 0,1 ⎟ and the slope is m = . 2 2⎠ ⎝ 1 rise m= = 2 run ⎛ 1⎞ Starting from the point ⎜ 0,1 ⎟ we will use the slope to mark another point up 1 2⎠ ⎝ unit and to the right 2 units. ANSWER:
62.
PROBLEM: 3 5 y =− x+ 4 2 SOLUTION:
1⎞ 3 ⎛ The y-intercept is ⎜ 0, 2 ⎟ and the slope is m = − . 2⎠ 4 ⎝ 3 rise m=− = 4 run 1⎞ ⎛ Starting from the point ⎜ 0, 2 ⎟ we will use the slope to mark another point down 2⎠ ⎝ 3 units and to the right 4 units. ANSWER:
63.
PROBLEM: 4x + y = 7 SOLUTION: 4x + y = 7 4x − 4x + y = 7 − 4x
y = −4 x + 7 The y-intercept is (0, 7) and the slope is m = –4. 4 rise m=− = 1 run Starting from the point (0, 7) we will use the slope to mark another point down 4 units and to the right 1 unit.
ANSWER:
64.
PROBLEM: 3x − y = 5 SOLUTION:
3x − y = 5 3x − 3x − y = 5 − 3x
y = 3x − 5 The y-intercept is (0, –5) and the slope is m = 3. 3 rise m= = 1 run Starting from the point (0, –5) we will use the slope to mark another point up 3 units and to the right 1 unit. ANSWER:
65.
PROBLEM: 5 x − 2 y = 10 SOLUTION: 5 x − 2 y = 10 5 x − 5 x − 2 y = 10 − 5 x 2 y = 5 x − 10 2 5 10 y = x− 2 2 2 5 y = x −5 2
5 The y-intercept is (0, –5) and the slope is m = . 2 5 rise m= = 2 run Starting from the point (0, –5) we will use the slope to mark another point up 5 units and to the right 2 units. ANSWER:
66.
PROBLEM: −2 x + 3 y = 18 SOLUTION: −2 x + 3 y = 18 −2 x + 2 x + 3 y = 18 + 2 x 3 y = 2 x + 18 3 2 18 y = x+ 3 3 3 2 y = x+6 3
2 The y-intercept is (0, 6) and the slope is m = . 3 2 rise m= = 3 run Starting from the point (0, 6) we will use the slope to mark another point up 2 units and to the right 3 units. ANSWER:
67.
PROBLEM: x− y =0 SOLUTION: x− y =0 x− x− y = 0− x
y=x The y-intercept is (0, 0) and the slope is m = 1. 1 rise m= = 1 run Starting from the point (0, 0) we will use the slope to mark another point up 1 unit and to the right 1 unit.
ANSWER:
68.
PROBLEM: x+ y =0 SOLUTION:
x+ y =0 x− x+ y = 0− x
y = −x The y-intercept is (0, 0) and the slope is m = –1. 1 rise m=− = 1 run Starting from the point (0, 0) we will use the slope to mark another point down 1 unit and to the right 1 unit. ANSWER:
69.
PROBLEM: 1 1 x − y =1 2 3 SOLUTION: 1 1 x − y =1 2 3 1 1 1 1 x − x − y = 1− x 2 2 3 2 1 1 y = x −1 3 2 1 1 3 ⋅ y = 3 ⋅ x − 3 ⋅1 3 2 3 y = x−3 2
3 The y-intercept is (0, –3) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, –3) we will use the slope to mark another point up 3 units and to the right 2 units. ANSWER:
70.
PROBLEM: 2 1 − x+ y =2 3 2 SOLUTION:
2 1 − x+ y =2 3 2 2 2 1 2 − x+ x+ y = 2+ x 3 3 2 3 1 2 y = x+2 2 3 1 2 2⋅ y = 2⋅ x + 2⋅2 2 3 4 y = x+4 3 4 The y-intercept is (0, 4) and the slope is m = . 3 4 rise m= = 3 run Starting from the point (0, 4) we will use the slope to mark another point up 4 units and to the right 3 units. ANSWER:
71.
PROBLEM: 3x + 2 y = 1 SOLUTION: 3x + 2 y = 1 3x − 3x + 2 y = 1 − 3x 2 y = −3 x + 1 2 3 1 y =− x+ 2 2 2 3 1 y =− x+ 2 2 3 ⎛ 1⎞ The y-intercept is ⎜ 0, ⎟ and the slope is m = − . 2 ⎝ 2⎠ 3 rise m=− = 2 run ⎛ 1⎞ Starting from the point ⎜ 0, ⎟ we will use the slope to mark another point down 3 ⎝ 2⎠ units and to the right 2 units. ANSWER:
72.
PROBLEM: −5 x + 3 y = −1 SOLUTION; −5 x + 3 y = −1 −5 x + 5 x + 3 y = −1 + 5 x 3 y = 5x −1 3 5 1 y = x− 3 3 3 5 1 y = x− 3 3 1⎞ 5 ⎛ The y-intercept is ⎜ 0, − ⎟ and the slope is m = . 3 3⎠ ⎝ 5 rise m= = 3 run Starting from the point (0, 4) we will use the slope to mark another point up 5 units and to the right 3 units. ANSWER:
73.
PROBLEM: On the same set of axes graph the three lines where y = 32 x + b where b = {−2, 0, 2}. SOLUTION: y = 32 x + b When b = −2 y = 32 x − 2
3 The y-intercept is (0, –2) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, –2) we will use the slope to mark another point up 3 units and to the right 2 units. When b = 0 y = 32 x + 0 y = 32 x 3 The y-intercept is (0, 0) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, 0) we will use the slope to mark another point up 3 units and to the right 2 units. When b = 2 y = 32 x + 2 3 The y-intercept is (0, 2) and the slope is m = . 2 3 rise m= = 2 run Starting from the point (0, 2) we will use the slope to mark another point up 3 units and to the right 2 units.
ANSWER:
74.
PROBLEM: On the same set of axes graph the three lines where y = mx + 1 where m = {−1/2, 0, 1/2}. SOLUTION:
y = mx + 1
When m = −1/2
1 y = − x +1 2 1 The y-intercept is (0, 1) and the slope is m = − . 2 1 rise m=− = 2 run Starting from the point (0, 1) we will use the slope to mark another point down 1 unit and to the right 2 units. When m = 0
y = ( 0) x + 1
y =1 The y-intercept is (0, 1) and the slope is m = 0. rise m=0= run Starting from the point (0, 1) the line will be a horizontal line with no slope.
When m = 1/2
y=
1 x +1 2
1 The y-intercept is (0, 1) and the slope is m = . 2
1 rise = 2 run Starting from the point (0, 1) we will use the slope to mark another point up 1 unit and to the right 2 units. m=
ANSWER:
Part C: Discussion Board Topics
75.
PROBLEM: Name three methods for graphing lines. Discuss the pros and cons of each method? ANSWER: The three methods of graphing lines are by using the x-intercepts and y-intercepts, slope intercept form, and point slope form. Pros of the x-intercept and y-intercept method Identifies the y intercept Identifies x intercept Cons of the x-intercept and y-intercept method Does not involve understanding of gradient or rate of change. Pros of the slope intercept form Identifies y intercept. Shows understanding of gradient or rate of change. Cons of the slope intercept form Rule must be in y = mx + b form. Does not find x intercept
Pros of the point-slope form The beauty of the slope formula is that we can obtain the slope, given two points, using only algebra. Cons of the point-slope formula When using the slope formula, take care to be consistent since order does matter. We must subtract the coordinates of the first point from the second point for both the numerator and denominator in the same order. When considering the slope as a rate of change it is important to include the correct units. (Source: henchmaths.wikispaces.com/.../Sketching+Linear+Graphs+4+ways.ppt)
76.
PROBLEM: Choose a linear equation and graph it 3 different ways. Scan the work and share it on the discussion board. ANSWER: Let the equation be x – y = 0 x-intercept and the y-intercept form Set x = 0 y=0 y-intercept: (0, 0) Set y = 0 x=0 x-intercept: (0, 0) Slope intercept form x–y–x=0–x y=x m = 1; y intercept: (0, 0) Point slope form Let the two coordinates be (0, 0) and (1, 1). y −y y − y1 m= 2 1 = x2 − x1 x − x1
1− 0 y − 0 = 1 −1 x − 0 y=x
m=
m = 1; y intercept: (0, 0) (Answers may vary)
77.
PROBLEM: Why do we use the letter m for slope? ANSWER: A third theory distinguishes between variables in the beginning of the alphabet--a, b, c--from variables in the middle of the alphabet--l, m, n--from variables at the end of the alphabet--x, y, z. Variables in the beginning of the alphabet represent constants, such as Avogadro's number. Variables in the middle of the alphabet represent parameters. Variables at the end of the alphabet are, well, variable.
78.
PROBLEM: How are equivalent fractions useful when working with slopes? ANSWER: Equivalent fractions help in simplifying the graphing. The rise and the run of the slope are equal which simplifies the graphing method.
79.
PROBLEM: Can we graph a line knowing only its slope? ANSWER: Yes. If the coordinates of two points in a line are known then the line can be graphed.
80.
PROBLEM:
Research and discuss the alternative notation m =
Δy . Δx
ANSWER: Δy y2 − y1 m= = Δx x2 − x1
Δy shows the change in the y-coordinate or y value. Δx shows the change in the x-coordinate or the x value. 81.
PROBLEM: What strategies for graphing lines should be brought to an exam? Explain. ANSWER: A clear straight edge, a mechanical pencil (preferably with a lead of 0.5 mm thickness), and graph paper. The three methods to graph lines should be studied thoroughly. Mathematical calculations should be done accurately.
3.5 Finding Linear Equations Part A: Slope‐Intercept Form Determine the slope and y‐intercept
1.
PROBLEM: 5x − 3 y = 18 SOLUTION: 5x − 3 y = 18
5x − 5x − 3 y = 18 − 5 x 3 y = 5 x − 18 3 5 18 y = x− 3 3 3 5 y = x−6 3 ANSWER: The y-intercept is (0, –6) and the slope is m =
2.
5 . 3
PROBLEM: −6 x + 2 y = 12 SOLUTION: −6 x + 2 y = 12
−6 x + 6 x + 2 y = 12 + 6 x 2 y = 6 x + 12 2 6 12 y = x+ 2 2 2 y = 3x + 6 ANSWER: The y-intercept is (0, 6) and the slope is m = 3 .
3.
PROBLEM: x− y =5 SOLUTION: x− y =5
x + y − x = 5− x y = x −5 ANSWER: The y-intercept is (0, –5) and the slope is m = 1 .
4.
PROBLEM: −x + y = 0 SOLUTION: −x + y = 0
−x + x + y = 0 + x y=x ANSWER: The y-intercept is (0, 0) and the slope is m = 1 .
5.
PROBLEM: 4 x − 5 y = 15 SOLUTION: 4 x − 5 y = 15
4 x − 4 x − 5 y = 15 − 4 x 5 y = 4 x − 15 5 4 15 y = x− 5 5 5 4 y = x −3 5 ANSWER: The y-intercept is (0, –3) and the slope is m =
4 . 5
6.
PROBLEM: −7 x + 2 y = 3 SOLUTION: −7 x + 2 y = 3
−7 x + 7 x + 2 y = 3 + 7 x 2 y = 7x + 3 2 7 3 y = x+ 2 2 2 7 3 y = x+ 2 2 7 ⎛ 3⎞ ANSWER: The y-intercept is ⎜ 0, ⎟ and the slope is m = . 2 ⎝ 2⎠ 7.
PROBLEM: y =3 SOLUTION: y =3 ANSWER: The y-intercept is (0,3) and the slope is m = 0 .
8.
PROBLEM: 3 y=− 4 SOLUTION: 3 y=− 4
3⎞ ⎛ ANSWER: The y-intercept is ⎜ 0, − ⎟ and the slope is m = 0 . 4⎠ ⎝
9.
PROBLEM: 1 1 x − y = −1 5 3 SOLUTION: 1 1 x − y = −1 5 3 1 1 1 x − x − y = −1 − x 5 3 5 1 1 y = x +1 3 5 1 1 3 ⋅ y = 3 ⋅ x + 3 ⋅1 3 5 3 y = x+3 5 ANSWER: The y-intercept is (0, 3) and the slope is m =
10.
3 . 5
PROBLEM: 5 3 x+ y =9 16 8 SOLUTION: 5 3 x+ y =9 16 8 5 5 3 5 x− x+ y =9− x 16 16 8 16 3 5 y = − x+9 8 16 8 3 8 ⎛ 5 ⎞ 8 ⋅ y = ⋅⎜ − x⎟ + ⋅9 3 8 3 ⎝ 16 ⎠ 3 5 y = − x + 24 6
5 ANSWER: The y-intercept is (0, 24) and the slope is m = − . 6
11.
PROBLEM: 2 5 5 − x+ y = 3 2 4 SOLUTION: 2 5 5 − x+ y = 3 2 4 2 2 5 5 2 − x+ x+ y = + x 3 3 2 4 3 5 2 5 y = x+ 2 3 4 2 5 2 2 2 5 ⋅ y = ⋅ x+ ⋅ 5 2 5 3 5 4 4 1 y = x+ 15 2
4 ⎛ 1⎞ ANSWER: The y-intercept is ⎜ 0, ⎟ and the slope is m = . 15 ⎝ 2⎠ 12.
PROBLEM: 1 3 1 x− y = − 2 4 2 SOLUTION: 1 3 1 x− y = − 2 4 2 1 1 3 1 1 x− x− y =− − x 2 2 4 2 2 3 1 1 y = x+ 4 2 2 4 3 4 1 4 1 ⋅ y = ⋅ x+ ⋅ 3 4 3 2 3 2 2 2 y = x+ 3 3
2 ⎛ 2⎞ ANSWER: The y-intercept is ⎜ 0, ⎟ and the slope is m = . 3 ⎝ 3⎠
Part B: Finding Equations in Slope‐Intercept Form Given the slope and y‐intercept determine the equation of the line.
13.
PROBLEM: m = 1/2; (0, 5) SOLUTION: y = mx + b 1 b = 5; m = 2 ANSWER: y =
14.
1 x+5 2
PROBLEM: m = 4; (0, −1) SOLUTION: y = mx + b b = −1; m = 4 ANSWER: y = 4x – 1
15.
PROBLEM: m = −2/3; (0, −4) SOLUTION: y = mx + b
b=–4;m= −
2 3
2 ANSWER: y = − x – 4 3 16.
PROBLEM: m = −3; (0, 9) SOLUTION: y = mx + b b = 9; m = −3 ANSWER: y = −3x + 9
17.
PROBLEM: m = 0; (0, −1) SOLUTION: y = mx + b b=–1;m=0 y = (0)x – 1 ANSWER: y = –1
18.
PROBLEM: m = 5; (0, 0) SOLUTION: y = mx + b b = 0; m = 5 y = 5x + 0 ANSWER: y = 5x
Given the graph find the equation in slope‐intercept form.
19.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 3). Hence the b = 3.
rise 2 = − = −1 run 2 y = mx + b
m=
ANSWER: y = –x + 3
20.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, –2). Hence the b = –2.
rise 4 = run 3 y = mx + b
m=
ANSWER: y =
21.
4 x−2 3
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 2). Hence the b = 2.
rise 0 = =0 run 1 y = mx + b y = (0)x + 2
m=
ANSWER: y = 2
22.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 2). Hence the b = 2.
rise −4 = =1 run −4 y = mx + b m=
ANSWER: y = x + 2
23.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 0). Hence the b = 0.
rise 4 1 = = run 8 2 y = mx + b 1 y= x+0 2 1 ANSWER: y = x 2 m=
24.
PROBLEM:
SOLUTION: The line intersects the y axis at (0, 0). Hence the b = 0.
rise −2 2 = =− run 5 5 y = mx + b
m=
ANSWER: 2 y = − x+0 5 2 y=− x 5 Find the equation given the slope and a point.
25.
PROBLEM: m = 2/3; (−9, 2) SOLUTION: y = mx + b 2 y = x+b 3 Substituting the point (−9, 2) in the equation, 2 y = x+b 3 2 2 = ( −9 ) + b 3 2 = −6 + b 8=b Completing the equation y = mx + b,
y=
2 x+b 3
ANSWER: y =
26.
2 x +8 3
PROBLEM: m = −1/5; (5, −5) SOLUTION: y = mx + b 1 y = − x+b 5 Substituting the point (5, −5) in the equation, 1 y = − x+b 5 1 ( −5 ) = − ( 5 ) + b 5 −5 = −1 + b −4 = b Completing the equation y = mx + b, 1 y = − x+b 5 1 ANSWER: y = − x − 4 5
27.
PROBLEM: m = 0; (−4,3) SOLUTION: y = mx + b y = (0) x + b
y=b Substituting the point (−4,3) in the equation, y=b 3=b Completing the equation y = mx + b, y=b ANSWER: y = 3
28.
PROBLEM: m = 3; (−2,1) SOLUTION: y = mx + b y = 3x + b Substituting the point (−2, 1) in the equation, y = 3x + b 1 = 3 ( −2 ) + b 1 = −6 + b 7=b Completing the equation y = mx + b, y = 3x + b
ANSWER: y = 3 x + 7
29.
PROBLEM: m = −5; (−2, 8) SOLUTION: y = mx + b y = ( −5 ) x + b
y = −5x + b Substituting the point (−2, 8) in the equation, y = −5 x + b 8 = −5 ( −2 ) + b 8 = 10 + b –2 = b Completing the equation y = mx + b, ANSWER: y = –5x – 2
30.
PROBLEM: m = −4; (1/2, −3/2) SOLUTION: y = mx + b y = −4 x + b Substituting the point (1/2, −3/2) in the equation, y = −4 x + b 3 ⎛1⎞ = −4 ⎜ ⎟ + b 2 ⎝2⎠ 3 4 − = − +b 2 2 1 =b 2 Completing the equation y = mx + b, 1 ANSWER: y = −4 x + 2 −
31.
PROBLEM: m = −1/2; (3, 2) SOLUTION: y = mx + b 1 y = − x+b 2 Substituting the point (3, 2) in the equation, 1 y = − x+b 2 1 2 = − ( 3) + b 2 3 2 = − +b 2 7 =b 2 Completing the equation y = mx + b, 1 7 ANSWER: y = − x + 2 2
32.
PROBLEM: m = 3/4; (1/3, 5/4) SOLUTION: y = mx + b 3 y = x+b 4 Substituting the point (1/3, 5/4) in the equation, 3 y = x+b 4 5 3 1 = ⋅ +b 4 4 3 5 1 = +b 4 4 1= b Completing the equation y = mx + b, 3 ANSWER: y = x + 1 4
33.
PROBLEM: m = 0; (3,0) SOLUTION: y = mx + b
y = ( 0) x + b y =b Substituting the point (3,0) in the equation, y=b 0=b Completing the equation y = mx + b, ANSWER: y = 0
34.
PROBLEM: m undefined; (3,0) ANSWER: If the slope is undefined, the line is vertical line x = 3.
Given two points find the equation of the line.
35.
PROBLEM: (−6, 6), (2, 2) SOLUTION: y −y 2−6 −4 1 m= 2 1 = = =− x2 − x1 2 − ( −6 ) 8 2 y = mx + b 1 y = − x+b 2 Substitute the point (2, 2) in the equation, ⎛ 1⎞ y = ⎜− ⎟x +b ⎝ 3⎠ ⎛ 1⎞ 2 = ⎜− ⎟⋅2 + b ⎝ 2⎠ 3=b Completing the equation y = mx + b, 1 ANSWER: y = − x + 3 2
36.
PROBLEM: (−10, −3), (5, 0) SOLUTION: 0 − ( −3 ) y −y 3 1 m= 2 1 = = = x2 − x1 5 − ( −10 ) 15 5 y = mx + b 1 y = x+b 5 Substitute the point (5, 0) in the equation, 1 y = x+b 5 1 0 = ( 5) + b 5 −1 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 1 5
37.
PROBLEM: (0, 1/2), (1/2, −1) SOLUTION: 1 3 −1 − − y2 − y1 2 = 2 = −3 = m= 1 1 x2 − x1 −0 2 2 y = mx + b
y = −3 x + b Substitute the point (1/2, −1) in the equation, y = −3 x + b ⎛1⎞ −1 = −3 ⎜ ⎟ + b ⎝2⎠ 3 −1 = − + b 2 1 =b 2 Completing the equation y = mx + b, 1 ANSWER: y = −3 x + 2
38.
PROBLEM: (1/3, 1/3), (2/3, 1) SOLUTION: 1 2 y −y 3 = 3 =2 m= 2 1 = 2 1 1 x2 − x1 − 3 3 3 y = mx + b 1−
y = 2x + b Substitute the point (2/3, 1) in the equation, y = 2x + b ⎛2⎞ 1 = 2⎜ ⎟ + b ⎝3⎠ 4 1− = b 3 1 − =b 3 Completing the equation y = mx + b, 1 ANSWER: y = 2 x − 3
39.
PROBLEM: (3, −4), (−6, −7) SOLUTION: y − y −7 − ( − 4 ) −3 1 m= 2 1 = = = x2 − x1 −6 − 3 −9 3 y = mx + b 1 y = x+b 3 Substitute the point (−6, −7) in the equation, 1 y = x+b 3 1 −7 = ( −6 ) + b 3 −5 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 5 3
40.
PROBLEM: (−5, 2), (3, 2) SOLUTION: y −y 2−2 0 m= 2 1 = = =0 x2 − x1 3 − ( −5 ) 8 y = mx + b
y = ( 0) x + b y =b Substitute the point (3, 2) in the equation, y=b 2=b Completing the equation y = mx + b, ANSWER: y = 2
41.
PROBLEM: (−6, 4), (−6,−3) SOLUTION: y −y −3 − 4 −7 m= 2 1 = = = undefined x2 − x1 −6 − ( −6 ) 0 ANSWER: Since the slope is undefined, the line is a vertical line x = –6.
42.
PROBLEM: (−4, −4), (−1, −1) SOLUTION: y − y −1 − ( −4 ) 3 m= 2 1 = = =1 x2 − x1 −1 − ( −4 ) 3 y = mx + b y = x+b Substitute the point (−1, −1) in the equation, y = x+b − 1 = −1 + b 0=b Completing the equation y = mx + b,
ANSWER: y = x
43.
PROBLEM: (3, −3), (−5, 5) SOLUTION: y − y 5 − ( −3 ) 8 m= 2 1 = = = −1 x2 − x1 −5 − 3 −8 y = mx + b
y = −x + b Substitute the point (−5, 5) in the equation, y = −x + b
5 = − ( −5 ) + b 0=b Completing the equation y = mx + b, ANSWER: y = − x
44.
PROBLEM: (0, 8), (−4, 0) SOLUTION: y −y 0−8 m= 2 1 = =2 x2 − x1 −4 − 0 y = mx + b
y = 2x + b Substitute the point (0, 8) in the equation, y = 2x + b
8 = 2 ( 0) + b 8=b Completing the equation y = mx + b, ANSWER: y = 2 x + 8 Part C: Equations using Point‐Slope Form Find the equation given the slope and a point.
45.
PROBLEM: m = 1/2; (4, 3) SOLUTION: y = mx + b 1 y = x+b 2 Substituting the point (4, 3) in the equation, 1 y = x+b 2 1 3 = ( 4) + b 2 1= b Completing the equation y = mx + b, 1 ANSWER: y = x + 1 2
46.
PROBLEM: m = −1/3; (9, −2) SOLUTION: y = mx + b
1 y = − x+b 3 Substituting the point (9, −2) in the equation, 1 y = − x+b 3 1 −2 = − ( 9 ) + b 3 1= b Completing the equation y = mx + b, 1 ANSWER: y = − x + 1 3 47.
PROBLEM: m = 6; (1, −5) SOLUTION: y = mx + b
y = 6x + b Substituting the point (1, −5) in the equation, y = 6x + b
−5 = 6 (1) + b −11 = b Completing the equation y = mx + b, ANSWER: y = 6 x − 11
48.
PROBLEM: m = −10; (1, −20) SOLUTION: y = mx + b y = −10 x + b Substituting the point (1, −20) in the equation, −20 = −10 (1) + b
−10 = b Completing the equation y = mx + b, ANSWER: y = −10 x − 10
49.
PROBLEM: m = −3; (2, 3) SOLUTION: y = mx + b y = −3 x + b Substituting the point (2, 3) in the equation, y = −3x + b
3 = −3 ( 2 ) + b 9=b Completing the equation y = mx + b, ANSWER: y = −3 x + 9
50.
PROBLEM: m = 2/3; (−3, −5) SOLUTION: y = mx + b 2 y = x+b 3 Substituting the point (−3, −5) in the equation, 2 y = x+b 3 2 −5 = ( −3) + b 3 −3 = b Completing the equation y = mx + b, 2 ANSWER: y = x − 3 3
51.
PROBLEM: m = −3/4; (−8,3) SOLUTION: y = mx + b 3 y = − x+b 4 Substituting the point (−8,3) in the equation, 3 y = − x+b 4 3 3 = − ( −8 ) + b 4 −3 = b Completing the equation y = mx + b, 3 ANSWER: y = − x − 3 4
52.
PROBLEM: m = 5; (1/5, −3) SOLUTION: y = mx + b
y = 5x + b Substituting the point (1/5, −3) in the equation, y = 5x + b
⎛1⎞ −3 = 5 ⎜ ⎟ + b ⎝5⎠ −4 = b Completing the equation y = mx + b, ANSWER: y = 5 x − 4
53.
PROBLEM: m = −3; (−1/9, 2) SOLUTION: y = mx + b
y = −3 x + b Substituting the point (−1/9, 2) in the equation, y = −3x + b ⎛ 1⎞ 2 = −3 ⎜ − ⎟ + b ⎝ 9⎠ 5 =b 3 Completing the equation y = mx + b, 5 ANSWER: y = −3 x + 3 54.
PROBLEM: m = 0; (4, −6) SOLUTION: y = mx + b
y = ( 0) x + b y =b Substituting the point (4, −6) in the equation, y=b −6 = b Completing the equation y = mx + b, ANSWER: y = −6
55.
PROBLEM: m = 0; (−5, 10) SOLUTION: y = mx + b
y = ( 0) x + b y =b Substituting the point (−5, 10) in the equation, y=b 10 = b Completing the equation y = mx + b, ANSWER: y = 10
56.
PROBLEM: m = 5/8; (4,3) SOLUTION: y = mx + b 5 y = x+b 8 Substituting the point (4,3) in the equation, 5 y = x+b 8 5 3 = ( 4) + b 8 1 =b 2 Completing the equation y = mx + b, 5 1 ANSWER: y = x + 8 2
57.
PROBLEM: m = −3/5; (−2, −1) SOLUTION: y = mx + b 3 y = − x+b 5 Substituting the point (−2, −1) in the equation, 3 y = − x+b 5 ⎛ 3⎞ −1 = ⎜ − ⎟ ( −2 ) + b ⎝ 5⎠ 11 − =b 5 Completing the equation y = mx + b, 3 11 ANSWER: y = − x − 5 5
58.
PROBLEM: m = 1/4; (12, −2) SOLUTION: y = mx + b 1 y = x+b 4 Substituting the point (12, –2) in the equation, 1 y = x+b 4 1 −2 = (12 ) + b 4 −5 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 5 4
59.
PROBLEM: m = 1; (0, 0) SOLUTION: y = mx + b
y = x+b Substituting the point (0, 0) in the equation, y = x+b 0 = 0+b 0=b Completing the equation y = mx + b, ANSWER: y = x
60.
PROBLEM: m = −3/4; (0, 0) SOLUTION: y = mx + b 3 y = − x+b 4 Substituting the point (0, 0) in the equation, 3 y = − x+b 4 3 0 = − ( 0) + b 4 0=b Completing the equation y = mx + b, 3 ANSWER: y = − x 4
Use the point‐slope formula to find the equation given the Graph.
61.
PROBLEM:
SOLUTION: The two points seen in the graph are (2, 1) and (1, 3) y − y 3 −1 2 m= 2 1 = = − = −2 x2 − x1 1 − 2 1 y = mx + b y = −2 x + b Substitute the point (2, 1) in the equation, y = −2 x + b
1 = −2 ( 2 ) + b 5=b Completing the equation y = mx + b, ANSWER: y = −2 x + 5
62.
PROBLEM:
SOLUTION: The two points seen in the graph are (–1, –2) and (4, 1) y − y 1 − ( −2 ) 3 = m= 2 1 = x2 − x1 4 − ( −1) 5 y = mx + b 3 y = x+b 5 Substitute the point (4, 1) in the equation, 3 y = x+b 5 3 1 = ( 4) + b 5 7 − =b 5 Completing the equation y = mx + b, 3 7 ANSWER: y = x − 5 5
63.
PROBLEM:
SOLUTION: The two points seen in the graph are (–7, 1) and (–1, 5) y −y 5 −1 4 2 m= 2 1 = = = x2 − x1 −1 − ( −7 ) 6 3 y = mx + b 2 y = x+b 3 Substitute the point (–1, 5) in the equation, 2 y = x+b 3 2 5 = ( −1) + b 3 17 =b 3 Completing the equation y = mx + b, 2 17 ANSWER: y = x + 3 3
64.
PROBLEM:
SOLUTION: The two points seen in the graph are (–6, 5) and (–1, 5) y −y 5−5 0 m= 2 1 = = =0 x2 − x1 1 − ( −6 ) 7 y = mx + b
y = ( 0) x + b y =b Substitute the point (–1, 5) in the equation, y=b 5=b Completing the equation y = mx + b, ANSWER: y = 5
65.
PROBLEM:
SOLUTION: The two points seen in the graph are (1, –1) and (–9, 5) y − y 5 − ( −1) 6 3 m= 2 1 = =− =− x2 − x1 −9 − 1 10 5 y = mx + b 3 y = − x+b 5 Substitute the point (1, –1) in the equation, 3 y = − x+b 5 3 −1 = − (1) + b 5 2 − =b 5 Completing the equation y = mx + b, 3 2 ANSWER: y = − x − 5 5
66.
PROBLEM:
SOLUTION: The two points seen in the graph are (–7, 3) and (2, 2) y −y 2−3 1 m= 2 1 = =− x2 − x1 2 − ( −7 ) 9 y = mx + b 1 y = − x+b 9 Substitute the point (2, 2) in the equation, 1 y = − x+b 9 1 2 = − ( 2) + b 9 20 =b 9 Completing the equation y = mx + b, 1 20 ANSWER: y = − x + 9 9
Use the point‐slope formula to find the equation passing through the two points.
67.
PROBLEM: (−4, 0), (0, 5) SOLUTION: y −y 5−0 5 m= 2 1 = = x2 − x1 0 − ( −4 ) 4 y = mx + b 5 y = x+b 4 Substitute the point (0, 5) in the equation, 5 y = x+b 4 5 5 = ( 0) + b 4 5=b Completing the equation y = mx + b, 5 ANSWER: y = x + 5 4
68.
PROBLEM: (−1, 2), (0, 3) SOLUTION: y −y 3− 2 1 m= 2 1 = = =1 x2 − x1 0 − ( −1) 1 y = mx + b
y = x+b Substitute the point (0, 3) in the equation, y = x+b 3= 0+b 3=b Completing the equation y = mx + b, ANSWER: y = x + 3
69.
PROBLEM: (−3, −2), (3, 2) SOLUTION: y − y 2 − ( −2 ) 4 2 = = m= 2 1 = x2 − x1 3 − ( −3) 6 3 y = mx + b 2 y = x+b 3 Substitute the point (3, 2) in the equation, 2 y = x+b 3 2 2 = ( 3) + b 3 0=b Completing the equation y = mx + b, 2 ANSWER: y = x 3
70.
PROBLEM: (3, −1), (2, −3) SOLUTION: y − y −3 − ( −1) −2 m= 2 1 = = =2 x2 − x1 2−3 −1 y = mx + b
y = 2x + b Substitute the point (2, −3) in the equation, y = 2x + b −3 = 2 ( 2 ) + b −7 = b Completing the equation y = mx + b, ANSWER: y = 2 x − 7
71.
PROBLEM: (−2, 4), (2, −4) SOLUTION: y −y −4 − 4 −8 m= 2 1 = = = −2 x2 − x1 2 − ( −2 ) 4 y = mx + b
y = −2 x + b Substitute the point (2, −4) in the equation, y = −2 x + b − 4 = −2 ( 2 ) + b 0=b Completing the equation y = mx + b, ANSWER: y = −2 x 72.
PROBLEM: (−5, −2), (5, 2) SOLUTION: y − y 2 − ( −2 ) 4 2 = = m= 2 1 = x2 − x1 5 − ( −5 ) 10 5 y = mx + b 2 y = x+b 5 Substitute the point (5, 2) in the equation, 4 y = x+b 5 2 2 = ( 5) + b 5 0=b Completing the equation y = mx + b, 2 ANSWER: y = x 5
73.
PROBLEM: (−3, −1), (3, 3) SOLUTION: y − y 3 − ( −1) 4 2 = = m= 2 1 = x2 − x1 3 − ( −3) 6 3 y = mx + b 2 y = x+b 3 Substitute the point (3, 3) in the equation, 2 y = x+b 3 2 3 = ( 3) + b 3 1= b Completing the equation y = mx + b, 2 ANSWER: y = x + 1 3
74.
PROBLEM: (1, 5), (0, 5) SOLUTION: y − y 5−5 0 =0 m= 2 1 = = x2 − x1 0 − 1 −1 y = mx + b
y = ( 0) x + b y =b Substitute the point (0, 5) in the equation, y=b 5=b Completing the equation y = mx + b, ANSWER: y = 5
75.
PROBLEM: (1, 2), (2, 4) SOLUTION: y − y 4−2 2 m= 2 1 = = =2 x2 − x1 2 − 1 1 y = mx + b
y = 2x + b Substitute the point (2, 4) in the equation, y = 2x + b 4 = 2 ( 2) + b 0=b Completing the equation y = mx + b, ANSWER: y = 2 x
76.
PROBLEM: (6, 3), (2, −3) SOLUTION: y − y −3 − 3 −6 3 m= 2 1 = = = x2 − x1 2 − 6 −4 2 y = mx + b 3 y = x+b 2 Substitute the point (2, −3) in the equation, 3 y = x+b 2 3 −3 = ( 2 ) + b 2 −6 = b Completing the equation y = mx + b, 3 ANSWER: y = x − 6 2
77.
PROBLEM: (10, −3), (5, −4) SOLUTION: y − y −4 − ( −3) −1 1 m= 2 1 = = = x2 − x1 5 − 10 −5 5 y = mx + b 1 y = x+b 5 Substitute the point (5, −4) in the equation, 1 −4 = ( 5 ) + b 5 −5 = b Completing the equation y = mx + b, 1 ANSWER: y = x − 5 5
78.
PROBLEM: (−3, 3), (−1,12) SOLUTION: y −y 12 − 3 9 m= 2 1 = = x2 − x1 −1 − ( −3) 2 y = mx + b 9 y = x+b 2 Substitute the point (−1,12) in the equation, 9 y = x+b 2 9 12 = ( −1) + b 2 33 =b 2 Completing the equation y = mx + b, 9 33 ANSWER: y = x + 2 2
79.
PROBLEM: (4/5, −1/3), (−1/5, 2/3) SOLUTION: 2 ⎛ 1⎞ 3 −⎜− ⎟ y −y 3 ⎝ 3⎠ m= 2 1 = = 3 = −1 1 4 5 x2 − x1 − − − 5 5 5 y = mx + b
y = −x + b Substitute the point (−1/5, 2/3) in the equation, y = −x + b 2 ⎛ 1⎞ = −⎜− ⎟ + b 3 ⎝ 5⎠ 7 =b 15 Completing the equation y = mx + b, 7 ANSWER: y = − x + 15 80.
PROBLEM: (5/3, 1/3), (−10/3, −5/3) SOLUTION: 5 1 6 − − − y2 − y1 6 2 m= = 3 3 = 3 = = x2 − x1 − 10 − 5 − 15 15 5 3 3 3 y = mx + b
2 x+b 5 Substitute the point (−10/3, −5/3) in the equation, 2 y = x+b 5 5 2 ⎛ 10 ⎞ − = ⎜− ⎟+b 3 5⎝ 3 ⎠ 1 − =b 3 Completing the equation y = mx + b, 2 1 ANSWER: y = x − 5 3 y=
81.
PROBLEM: (3, −1/4), (4, −1/2) SOLUTION: 1 ⎛ 1⎞ − −⎜− ⎟ − 1 y −y 1 2 ⎝ 4⎠ m= 2 1 = = 4 =− x2 − x1 4−3 1 4 y = mx + b
1 y = − x+b 4 Substitute the point (4, −1/2) in the equation, 1 y = − x+b 4 1 1 − = − ( 4) + b 2 4 1 =b 2 Completing the equation y = mx + b, 1 1 ANSWER: y = − x + 4 2 82.
PROBLEM: (0, 0), (−5, 1) SOLUTION: y −y 1− 0 1 =− m= 2 1 = x2 − x1 −5 − 0 5 y = mx + b
1 y = − x+b 5 Substitute the point (−5, 1) in the equation, 1 y = − x+b 5 1 1 = − ( −5 ) + b 5 0=b Completing the equation y = mx + b, 1 ANSWER: y = − x 5
83.
PROBLEM: (2, −4), (0, 0) SOLUTION: y − y 0 − ( −4 ) 4 m= 2 1 = = = −2 x2 − x1 0−2 −2 y = mx + b y = −2 x + b Substitute the point (0, 0) in the equation, y = −2 x + b
0 = −2 ( 0 ) + b 0=b Completing the equation y = mx + b, ANSWER: y = −2 x
84.
PROBLEM: (3, 5), (3, −2) SOLUTION: y − y − 2 − 5 −7 m= 2 1 = = = undefined x2 − x1 3−3 0 ANSWER: Since slope is not defined, the line is a vertical line x = 3.
85.
PROBLEM: (−4, 7), (−1, 7) SOLUTION: y −y 7−7 0 = =0 m= 2 1 = x2 − x1 −1 − ( −4 ) 3 y = mx + b
y = ( 0) x + b y =b Substitute the point (−1, 7) in the equation, y=b 7=b Completing the equation y = mx + b, ANSWER: y = 7
86.
PROBLEM: (−8, 0), (6, 0) SOLUTION: y −y 0−0 0 m= 2 1 = = =0 x2 − x1 6 − ( −8 ) 14 y = mx + b
y = ( 0) x + b y =b Substitute the point (6, 0) in the equation, y=b 0=b Completing the equation y = mx + b, ANSWER: y = 0 Part D: Applications
87.
PROBLEM: Joe has been keeping track of his cellular phone bills for the last two months. The bill for the first month was $38 for 100 minutes of usage and the bill for the second month was $45.50 for 150 minutes of usage. Find a linear equation that gives the total monthly bill based on the minutes of usage. SOLUTION: Bill amount is plotted on the y-axis; minutes of usage on the x-axis. Bill for the first month is $38 for 100 minutes of usage: (100, 38) Bill for the first month is $45.50 for 150 minutes of usage: (150, 45.50) y − y 45.50 − 38 3 Slope, m = 2 1 = = x2 − x1 150 − 100 20 y = mx + b 3 x+b y= 20 Substitute the point (100, 38) in the equation, 3 y= x+b 20 3 38 = (100 ) + b 20 23 = b Completing the equation y = mx + b,
3 x + 23 20 y = 0.15 x + 23 y=
ANSWER: y = 0.15 x + 23
88.
PROBLEM: A company in its first year of business produced 150 training manuals for a total cost of $2,350. The following year the company produced 50 more manuals at a cost of $1,450. Use this information to find a linear equation that gives the total cost of producing training manuals. SOLUTION: Cost of the training manuals produced is plotted on the y-axis; number of training manuals is plotted on the x-axis. 150 training manuals for a total cost of $2,350: (150, 2350) 50 training manuals for a total cost of $1,450: (50, 1450) y − y 1450 − 2350 Slope, m = 2 1 = =9 x2 − x1 50 − 150 y = mx + b y = 9x + b Substitute the point (50, 1450) in the equation, y = 9x + b
1450 = 9 ( 50 ) + b 1000 = b Completing the equation y = mx + b, ANSWER: y = 9 x + 1000
89.
PROBLEM: A corn farmer in California was able to produce 154 bushels of corn per acre 2 years after starting his operation. Currently, after 7 years of operation, he has increased his yield to 164 bushels per acre. Use this information to write a linear equation that gives the total yield per acre based on number of years of operation and use it to predict the yield for next year. SOLUTION: Number of bushels produced in a year is plotted on the y-axis; number of years of operation is plotted on the x-axis. 154 bushels of corn per acre for 3 years: (2, 154) 164 bushels of corn per acre for 7 years: (7, 164) y − y 164 − 154 10 Slope, m = 2 1 = = =2 x2 − x1 7−2 5
y = mx + b y = 2x + b Substitute the point (7, 164) in the equation, y = 2x + b 164 = 2 ( 7 ) + b 150 = b Completing the equation y = mx + b, y = 2 x + 150 To find the yield for the next year; put x = 8 in the above equation. y = 2 x + 150 y = 2 ( 8 ) + 150 y = 166 ANSWER: y = 2 x + 150 . Yield for the 8th year is 166 bushels.
90.
PROBLEM: A webmaster has noticed that the number of registered users has been steadily increasing since beginning an advertising campaign. Before starting to advertise he had 1,200 registered users and after 3 months of advertising he now has 1,590 registered users. Use this data to write a linear equation that gives the total number of registered users given the number of months after starting to advertise. Use the equation to predict the number of users 7 months into the advertising campaign. SOLUTION: Number of registered users is plotted on the y-axis; number of months of advertising is plotted on the x-axis. Number of registered users before advertising 1,200: (0, 1200) Number of registered users after 3 months of advertising: (3, 1590) y − y 1590 − 1200 Slope, m = 2 1 = = 130 x2 − x1 3−0 y = mx + b y = 130 x + b Substitute the point (0, 1200) in the equation, y = 130 x + b
1200 = 130 ( 0 ) + b 1200 = b Completing the equation y = mx + b, y = 130 x + 1200
To find number of registered users after 7 months of advertising, put x = 7 in the equation above. y = 130x + 1200 y = 130 ( 7 ) + 1200 y = 2110 ANSWER: Number of registered users after 7 months of advertising is 2,110.
91.
PROBLEM: A car purchased new cost $22,000 and was sold 10 years later for $7,000. Write a linear equation that gives the value of the car in terms of its age in years. SOLUTION: Value of the car is plotted on the y-axis; number of years is plotted on the x-axis. Initial cost of the car is $22,000: (0, 22000) Cost of the car after 10 years is $7,000: (10, 7000) y − y 7000 − 22000 −15000 Slope, m = 2 1 = = = −1500 x2 − x1 10 − 0 10 y = mx + b y = 1500 x + b Substitute the point (0, 22000) in the equation, y = −1500 x + b
22000 = −1500 ( 0 ) + b 22000 = b Completing the equation y = mx + b, ANSWER: y = −1500 x + 22000 92.
PROBLEM: An antique clock was purchased in 1985 for $1,500 and sold at auction in 1997 for $5,700. Determine a linear equation that models the value of the clock in terms of years since 1985. SOLUTION: Value of the clock is plotted on the y-axis; years are plotted on the x-axis. Assume the year 1985 represents the origin. Value of the clock in the year 1985 is $1,500: (0, 1500) Value of the clock in the year 1997 is $5,700: (12, 5700) y − y 5700 − 1500 4200 Slope, m = 2 1 = = = 350 x2 − x1 12 − 0 12 y = mx + b y = 350 x + b Substitute the point (0, 1500) in the equation,
y = 350 x + b 1500 = 350 ( 0 ) + b 1500 = b Completing the equation y = mx + b, ANSWER: y = 350 x + 1500 Part E: Discussion Board Topics
93.
PROBLEM: Discuss the merits and drawbacks of point-slope form and y-intercept form. ANSWER: Merits of the point-slope form The beauty of the slope formula is that we can obtain the slope, given two points, using only algebra. Drawbacks of the point-slope formula When using the slope formula, take care to be consistent since order does matter. We must subtract the coordinates of the first point from the second point for both the numerator and denominator in the same order. When considering the slope as a rate of change it is important to include the correct units. Merits of the y-intercept form Identifies y intercept. Shows understanding of gradient or rate of change. Drawbacks of the y-intercept form Rule must be in y = mx + b form. Does not find x intercept (Source: henchmaths.wikispaces.com/.../Sketching+Linear+Graphs+4+ways.ppt)
94.
PROBLEM: Research and discuss linear depreciation. In a linear depreciation model, what does the slope and y-intercept represent? ANSWER: The simplest and most commonly used depreciation method, linear depreciation is calculated by taking the purchase or acquisition price of an asset subtracted by the salvage value divided by the total productive years the asset can be reasonably expected to benefit the company (called “useful life” in accounting jargon).
Purchase cost = (Depreciation expense/year)(Number of years) + Salvage value Slope represents the depreciation expense/year. The y-intercept represents the salvage value of the asset. (Source: http://beginnersinvest.about.com/od/incomestatementanalysis/a/straightline-depreciation.htm)
3.6 Parallel and Perpendicular Lines Part A: Parallel and Perpendicular Lines Determine the slope of a parallel and perpendicular line.
1.
PROBLEM: 3 y = − x +8 4 SOLUTION: 3 y = − x +8 4
2.
3 The slope of the line is m = − . 4 ANSWER: Slope of the parallel line m|| = m 3 m|| = − 4 1 1 4 Slope of the perpendicular line m⊥ = − = − = 3 m 3 − 4 PROBLEM: 1 y = x −3 2 SOLUTION: 1 y = x −3 2
1 . 2 ANSWER: Slope of the parallel line m|| = m 1 m|| = 2 1 1 Slope of the perpendicular line m⊥ = − = − = −2 1 m 2
The slope of the line is m =
3.
PROBLEM: y = 4x + 4 SOLUTION: y = 4x + 4
The slope of the line is m = 4 . ANSWER: Slope of the parallel line m|| = m m|| = 4 1 1 Slope of the perpendicular line m⊥ = − = − m 4 4.
PROBLEM: y = −3x + 7 SOLUTION: y = −3x + 7
The slope of the line is m = −3 . ANSWER: Slope of the parallel line m|| = m m|| = −3 1 1 1 Slope of the perpendicular line m⊥ = − = − = m −3 3
5.
PROBLEM: 5 1 y =− x− 8 2 SOLUTION: 5 1 y =− x− 8 2
5 The slope of the line is m = − . 8 ANSWER: Slope of the parallel line m|| = m 5 m|| = − 8 1 1 8 Slope of the perpendicular line m⊥ = − = − = 5 m 5 − 8
6.
PROBLEM: 7 3 y = x+ 3 2 SOLUTION: 7 3 y = x+ 3 2
7 . 3 ANSWER: Slope of the parallel line m|| = m 7 m|| = 3 1 1 3 Slope of the perpendicular line m⊥ = − = − = − 7 m 7 3 The slope of the line is m =
7.
PROBLEM: 2 y = 9x − 5 SOLUTION: 2 y = 9x − 5 The slope of the line is m = 9 . ANSWER: Slope of the parallel line m|| = m m|| = 9 1 1 Slope of the perpendicular line m⊥ = − = − m 9
8.
PROBLEM: 1 y = −10 x + 5 SOLUTION: 1 y = −10 x + 5 The slope of the line is m = −10 . ANSWER: Slope of the parallel line m|| = m m|| = −10 1 1 1 Slope of the perpendicular line m⊥ = − = − = m −10 10
9.
PROBLEM: y =5 SOLUTION: y =5
The slope of the line is m = 0 . ANSWER: Slope of the parallel line m|| = m m|| = 0 1 1 Slope of the perpendicular line m⊥ = − = − = undefined m 0 10.
PROBLEM: 1 x=− 2 SOLUTION: 1 x=− 2 The slope of the line is m = undefined ANSWER: Slope of the parallel line m|| = m m|| = undefined
Slope of the perpendicular line m⊥ = −
1 1 =− =0 m undefined
11.
PROBLEM: x− y =0 SOLUTION: x− y =0
x − x − y = −x y=x The slope of the line is m =1. ANSWER: Slope of the parallel line m|| = m m|| = 1 1 1 Slope of the perpendicular line m⊥ = − = − = −1 m 1 12.
PROBLEM: x+ y =0 SOLUTION: x+ y =0
x − x + y = −x y = −x The slope of the line is m = −1 ANSWER: Slope of the parallel line m|| = m m|| = −1 1 1 Slope of the perpendicular line m⊥ = − = − = 1 m −1
13.
PROBLEM: 4x + 3y = 0 SOLUTION: 4x + 3y = 0
4 x − 4 x + 3 y = −4 x 3 y = −4 x 3 4 y=− x 3 3 4 y=− x 3
4 The slope of the line is m = − . 3 ANSWER: Slope of the parallel line m|| = m 4 m|| = − 3 1 1 3 = Slope of the perpendicular line m⊥ = − = − 4 4 m − 3 14.
PROBLEM: 3x − 5 y = 10 SOLUTION: 3x − 5 y = 10
3x − 3x − 5 y = 10 − 3x 5 y = 3x − 10 5 3 10 y = x− 5 5 5 3 y = x−2 5 3 5 ANSWER: Slope of the parallel line m|| = m 3 m|| = 5 1 1 5 Slope of the perpendicular line m⊥ = − = − = − 3 m 3 5 The slope of the line is m =
15.
PROBLEM: −2 x + 7 y = 14 SOLUTION: −2 x + 7 y = 14 −2 x + 2 x + 7 y = 14 + 2 x
7 y = 2 x + 14 7 2 14 y = x+ 7 7 7 2 y = x+2 7 2 The slope of the line is m = . 7 ANSWER: Slope of the parallel line m|| = m 2 m|| = 7 1 1 7 Slope of the perpendicular line m⊥ = − = − = − 2 m 2 7 16.
PROBLEM: − x − y = 15 SOLUTION: − x − y = 15
− x + x − y = 15 + x y = − x − 15 The slope of the line is m = −1 ANSWER: Slope of the parallel line m|| = m m|| = −1 1 1 Slope of the perpendicular line m⊥ = − = − = 1 m −1
17.
PROBLEM: 1 1 x − y = −1 2 3 SOLUTION: 1 1 x − y = −1 2 3 1 1 1 1 x − x − y = −1 − x 2 2 3 2 1 1 y = x +1 3 2 1 1 3 ⋅ y = 3 ⋅ x + 3 ⋅1 3 2 3 y = x+3 2
3 . 2 ANSWER: Slope of the parallel line m|| = m 3 m|| = 2 1 1 2 Slope of the perpendicular line m⊥ = − = − = − 3 m 3 2 The slope of the line is m =
18.
PROBLEM: 2 4 − x+ y =8 3 5 SOLUTION: 2 4 − x+ y =8 3 5 2 2 4 2 − x + x + y = 8+ x 3 3 5 3 4 2 y = x +8 5 3 5 4 5 2 5 ⋅ y = ⋅ x + ⋅8 4 5 4 3 4 5 y = x + 10 6
5 6 ANSWER: Slope of the parallel line m|| = m 5 m|| = 6 1 1 6 Slope of the perpendicular line m⊥ = − = − = − 5 m 5 6 The slope of the line is m =
19.
PROBLEM: 1 1 2x − y = 5 10 SOLUTION: 1 1 2x − y = 5 10 1 1 2x − 2x − y = − 2x 5 10 1 1 y = 2x − 5 10 1 1 5 ⋅ y = 5 ⋅ 2x − 5 ⋅ 5 10 1 y = 10 x − 2 The slope of the line is m = 10 . ANSWER: Slope of the parallel line m|| = m m|| = 10 1 1 Slope of the perpendicular line m⊥ = − = − m 10
20.
PROBLEM: 4 − x − 2y = 7 5 SOLUTION: 4 − x − 2y = 7 5 4 4 4 − x + x − 2y = 7 + x 5 5 5 4 2y = − x − 7 5 2 4 7 y=− x− 2 5⋅ 2 2 2 7 y =− x− 5 2
2 5 ANSWER: Slope of the parallel line m|| = m 2 m|| = − 5 1 1 5 Slope of the perpendicular line m⊥ = − = − = 2 2 m − 5 The slope of the line is m = −
Determine if the lines are parallel, perpendicular, or neither.
21.
PROBLEM: 2 ⎧ ⎪⎪ y = 3 x + 3 ⎨ ⎪y = 2 x −3 ⎪⎩ 3 SOLUTION: 2 y = x+3 3 2 m= 3
2 x −3 3 2 m= 3 y=
ANSWER: Slopes of the two lines are equal. Hence lines are parallel to each other.
22.
PROBLEM: 3 ⎧ ⎪⎪ y = 4 x − 1 ⎨ ⎪y = 4 x +3 ⎪⎩ 3 SOLUTION: 3 y = x −1 4 3 m= 4
4 x+3 3 4 m= 3 y=
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
23.
PROBLEM: ⎧ y = −2 x + 1 ⎪ ⎨ 1 ⎪⎩ y = 2 x + 8 SOLUTION:
1 x +8 y = −2 x + 1 2 1 m = −2 m= 2 ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other. y=
24.
PROBLEM: 1 ⎧ ⎪ y = 3x − 2 ⎨ ⎪⎩ y = 3x + 2 SOLUTION: 1 y = 3x − 2 m=3
y = 3x + 2 m=3
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
25.
PROBLEM: ⎧ y =5 ⎨ ⎩ x = −2 SOLUTION: x = −2 y=5 m = undefined m=0 ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
26.
PROBLEM: ⎧y = 7 ⎪ 1 ⎨ ⎪⎩ y = − 7 SOLUTION: y=7 m=0
y=−
1 7
m=0 ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other. 27.
PROBLEM: ⎧3x − 5 y = 15 ⎨ ⎩ 5x + 3 y = 9 SOLUTION: 3x − 5 y = 15 3x − 3 x − 5 y = 15 − 3 x 5 y = 3 x − 15 5 3 15 y = x− 5 5 5 3 y = x−3 5 3 m= 5
5x + 3 y = 9 5x − 5x + 3 y = 9 − 5x 3 y = −5 x + 9 3 5 9 y = − x+ 3 3 3 5 y = − x+3 3 5 m=− 3
ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
28.
PROBLEM: ⎧ x− y =7 ⎨ ⎩3x + 3 y = 2 SOLUTION:
3x + 3 y = 2 3x − 3x + 3 y = 2 − 3x x− y =7 3 y = −3 x + 2 x−x− y =7−x 3 3 2 y = − x+ y = x−7 3 3 3 m =1 2 y = −x + 3 m = −1 ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other. 29.
PROBLEM: ⎧ 2x − 6 y = 4 ⎨ ⎩ − x + 3 y = −2 SOLUTION: 2x − 6 y = 4 2x − 2x − 6 y = 4 − 2x 6 y = 2x − 4 6 2 4 y = x− 6 6 6 1 2 y = x− 3 3 1 m= 3
− x + 3 y = −2 − x + x + 3 y = −2 + x 3y = x − 2 3 1 2 y = x− 3 3 3 1 2 y = x− 3 3 1 m= 3
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
30.
PROBLEM: ⎧−4 x + 2 y = 3 ⎨ ⎩ 6 x − 3 y = −3 SOLUTION: −4 x + 2 y = 3 −4 x + 4 x + 2 y = 3 + 4 x 2 y = 3 + 4x 2 4 3 y = x+ 2 2 2 3 y = 2x + 2 m=2
6 x − 3 y = −3 6 x − 6 x − 3 y = −3 − 6 x 3y = 6 y + 3 3 6 3 y = y+ 3 3 3 y = 2 y +1 m=2
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
31.
PROBLEM: ⎧ x + 3y = 9 ⎨ ⎩2 x + 3 y = 6 SOLUTION: x + 3y = 9 x − x + 3y = 9 − x 3y = −x + 9 3 1 9 y = − x+ 3 3 3 1 y = − x+3 3 1 m=− 3
2x + 3y = 6 2x − 2x + 3y = 6 − 2x 3 y = −2 x + 6 3 2 6 y =− x+ 3 3 3 2 y =− x+2 3 2 m=− 3
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
32.
PROBLEM: ⎧ y − 10 = 0 ⎨ ⎩ x − 10 = 0 SOLUTION: y − 10 = 0 m=0
x − 10 = 0 m = undefined
ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
33.
PROBLEM: ⎧ y+2=0 ⎨ ⎩2 y − 10 = 0 SOLUTION: y+2=0 m=0
2 y − 10 = 0 m=0
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
34.
PROBLEM: ⎧3x + 2 y = 6 ⎨ ⎩2 x + 3 y = 6 SOLUTION: 3x + 2 y = 6 3x − 3x + 2 y = 6 − 3x 2 y = −3x + 6 2 3 6 y =− x+ 2 2 2 3 y = − x+3 2 3 m=− 2
2x + 3y = 6 2x − 2x + 3y = 6 − 2x 3 y = −2 x + 6 3 2 6 y =− x+ 3 3 3 2 y =− x+2 3 2 m=− 3
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
35.
PROBLEM: ⎧−5 x + 4 y = 20 ⎨ ⎩ 10 x − 8 y = 16 SOLUTION: −5 x + 4 y = 20 −5 x + 5 x + 4 y = 20 + 5 x 4 y = 5 x + 20 4 5 20 y = x+ 4 4 4 5 y = x+5 4 5 m= 4
10 x − 8 y = 16 10 x − 10 x − 8 y = 16 − 10 x 8 y = 10 x − 16 8 10 16 y = x− 8 8 8 5 y = x−2 4 5 m= 4
ANSWER: Slopes of the two lines are equal. Hence the lines are parallel to each other.
36.
PROBLEM: 1 ⎧ 1 ⎪⎪ 2 x − 3 y = 1 ⎨ ⎪ 1 x + 1 y = −2 ⎪⎩ 6 4 SOLUTION: 1 1 x − y =1 2 3 1 1 1 1 x − x − y = 1− x 2 2 3 2 1 1 y = x −1 3 2 1 1 3 ⋅ y = 3 ⋅ x − 3 ⋅1 3 2 3 y = x −3 2 3 m= 2
1 1 x + y = −2 6 4 1 1 1 1 x − x + y = −2 − x 6 6 4 6 1 1 y = − x−2 4 6 1 1 4 ⋅ y = −4 ⋅ x − 4 ⋅ 2 4 6 2 y = − x −8 3 2 m=− 3
ANSWER: Slopes of the two lines are negative reciprocals. Hence the lines are perpendicular to each other.
Part B: Equations in Point‐Slope Form Find the equation of the line.
37.
PROBLEM:
Parallel to y =
1 x + 2 and passing through (6, −1). 2
1 1 and thus m& = . We substitute 2 2 this slope and the given point into point-slope form. Point Slope
Solution: Here the given line has slope m =
(6, −1)
m& =
y − y1 = m ( x − x1 ) y − ( −1) = y +1 = y=
1 ( x − 6) 2
1 x −3 2
1 x−4 2
ANSWER: y =
38.
1 2
1 x−4 2
PROBLEM:
3 Parallel to y = − x − 3 and passing through (−8, 2). 4 3 3 SOLUTION: Here the given line has slope m = − and thus m& = − . We 4 4 substitute this slope and the given point into point-slope form. Point
Slope
(−8, 2)
m& = −
y − y1 = m ( x − x1 ) 3 ( x − ( −8) ) 4 3 y−2 = − x−6 4 3 y = − x−4 4 y−2= −
3 4
3 ANSWER: y = − x − 4 4 39.
PROBLEM: Perpendicular to y = 3x − 1 and passing through (−3, 2). 1 Solution: Here the given line has slope m = 3 and thus m⊥ = − . We substitute 3 this slope and the given point into point-slope form. Point Slope 1 (−3, 2) m⊥ = − 3 y − y1 = m ( x − x1 )
1 ( x − ( −3) ) 3 1 y − 2 = − x −1 3 1 y = − x +1 3 1 ANSWER: y = − x + 1 3 y−2= −
40.
PROBLEM:
1 Perpendicular to y = − x + 2 and passing through (4, −3). 3 1 1 and thus m⊥ = − = 3 . We 1 3 − 3 substitute this slope and the given point into point-slope form.
SOLUTION: Here the given line has slope m = −
Point (4, −3)
Slope m⊥ = 3
y − y1 = m ( x − x1 )
y − ( −3) = 3 ( x − 4 ) y + 3 = 3x − 12 y = 3x − 15 ANSWER: y = 3x − 15
41.
PROBLEM:
Perpendicular to y = −2 and passing through (−1, 5). 1 = undefined . 0 We substitute this slope and the given point into point-slope form. Point Slope
Solution: Here the given line has slope m = 0 and thus m⊥ = −
(−1, 5)
m⊥ = undefined
y − y1 = m ( x − x1 )
y −5 = x − ( −1) undefined x = −1 ANSWER: x = −1
42.
PROBLEM: Perpendicular to x = 15 and passing through (5, −3). SOLUTION: Here the given line has slope m = undefined and thus m⊥ = −
1 = 0 . We substitute this slope and the given point into pointundefined
slope form. Point (5, −3)
Slope m⊥ = 0
y − y1 = m ( x − x1 )
y − ( −3) = 0 ( x − 5 ) y+3= 0 y = −3 ANSWER: y = −3
43.
PROBLEM: Parallel to y = 3 and passing through (2, 4). Solution: Here the given line has slope m = 0 and thus m& = 0 . We substitute this slope and the given point into point-slope form. Point Slope
( 2, 4 )
m& = 0
y − y1 = m ( x − x1 ) y − 4 = 0 ( x − 2) y=4
ANSWER: y = 4
44.
PROBLEM: Parallel to x = 2 and passing through (7, −3). SOLUTION: Here the given line has slope m = undefined and thus m& = undefined . We substitute this slope and the given point into point-slope
form. Point (7, −3)
Slope m& = undefined
y − y1 = m ( x − x1 )
y − ( −3) = undefined ( x − 7 ) y − ( −3 ) undefined x−7 = 0
= x−7
x=7
ANSWER: x = 7
45.
PROBLEM: Perpendicular to y = x and passing through (7, −13). Solution: Here the given line has slope m = 1 and thus m⊥ = −1 . We substitute this slope and the given point into point-slope form. Point Slope (7, −13) m⊥ = −1
y − y1 = m ( x − x1 )
y − ( −13) = −1( x − 7 ) y + 13 = − x + 7 y = −x − 6 ANSWER: y = − x − 6
46.
PROBLEM: Perpendicular to y = 2 x + 9 and passing through (3, −1). 1 SOLUTION: Here the given line has slope m = 2 and thus m⊥ = − . We 2 substitute this slope and the given point into point-slope form.
Point
Slope 1 (3, −1) m⊥ = − 2 y − y1 = m ( x − x1 ) 1 ( x − 3) 2 1 3 y +1 = − x + 2 2 1 1 y =− x+ 2 2 y − ( −1) = −
1 1 ANSWER: y = − x + 2 2
47.
PROBLEM:
Parallel to y =
1 x − 5 and passing through (−2, 1). 4
1 1 and thus m& = . We substitute 4 4 this slope and the given point into point-slope form. Point Slope
Solution: Here the given line has slope m =
(−2, 1)
m& =
y − y1 = m ( x − x1 )
1 4
1 ( x − ( −2 ) ) 4 1 y −1 = ( x + 2) 4 1 3 y = x+ 4 2 1 3 ANSWER: y = x + 4 2 y −1 =
48.
PROBLEM:
3 Parallel to y = − x + 1 and passing through (4, 1/4). 4 3 3 and thus m& = − . We 4 4 substitute this slope and the given point into point-slope form.
SOLUTION: Here the given line has slope m = −
Point
Slope
1⎞ 3 ⎛ m& = − ⎜ 4, ⎟ 4⎠ 4 ⎝ y − y1 = m ( x − x1 ) 1 3 = − ( x − 4) 4 4 3 13 y =− x+ 4 4 y−
3 13 ANSWER: y = − x + 4 4
49.
PROBLEM: Parallel to 2 x − 3 y = 6 and passing through (6,−2). SOLUTION: 2x − 3 y = 6 2x − 2x − 3y = 6 − 2x 3y = 2x − 6 3 2 6 y = x− 3 3 3 2 y = x−2 3 2 2 and thus m& = . We substitute this slope 3 3 and the given point into point-slope form.
Here the given line has slope m =
Point
Slope 2 (6, −2) m& = 3 y − y1 = m ( x − x1 ) y − ( −2 ) = y=
2 ( x − 6) 3
2 x−6 3
ANSWER: y =
50.
2 x−6 3
PROBLEM: Parallel to − x + y = 4 and passing through (9, 7). SOLUTION: −x + y = 4 −x + x + y = 4 + x
y = x+4 Here the given line has slope m = 1 and thus m& = 1 . We substitute this slope and the given point into point-slope form. Point
( 9, 7 )
Slope m& = 1
y − y1 = m ( x − x1 ) y − 7 = 1( x − 9 ) y = x−2 ANSWER: y = x − 2
51.
PROBLEM: Perpendicular to 5 x − 3 y = 18 and passing through (−9, 10). SOLUTION: 5 x − 3 y = 18 5 x − 5 x − 3 y = 18 − 5 x 3 y = 5 x − 18 3 5 18 y = x− 3 3 3 5 y = x−6 3 5 3 and thus m⊥ = − . We substitute this slope 3 5 and the given point into point-slope form.
Here the given line has slope m =
Point
Slope
(−9, 10)
m⊥ = −
y − y1 = m ( x − x1 )
3 5
3 ( x − ( −9 ) ) 5 3 23 y =− x+ 5 5 3 23 ANSWER: y = − x + 5 5 y − 10 = −
52.
PROBLEM: Perpendicular to x − y = 11 and passing through (6, −8). SOLUTION: x − y = 11
x − x − y = 11 − x y = x − 11 Here the given line has slope m = 1 and thus m⊥ = −1 . We substitute this slope and the given point into point-slope form. Point Slope (6, −8) m⊥ = −1 y − y1 = m ( x − x1 )
y − ( −8 ) = −1( x − 6 ) y = −x − 2 ANSWER: y = − x − 2
53.
PROBLEM: 1 1 Parallel to x − y = 2 and passing through (−15, 6). 5 3 SOLUTION: 1 1 x− y = 2 5 3 1 1 1 1 x− x− y = 2− x 5 5 3 5 1 1 y = x−2 3 5 1 1 3⋅ y = 3⋅ x − 3⋅ 2 3 5 3 y = x−6 5 3 3 and thus m& = . We substitute this slope and 5 5 the given point into point-slope form.
Here the given line has slope m =
Point
Slope 3 (−15, 6) m& = 5 y − y1 = m ( x − x1 ) y−6 = y=
3 ( x − ( −15) ) 5
3 x + 15 5
ANSWER: y =
54.
3 x + 15 5
PROBLEM:
Parallel to −10 x −
5 1 y = and passing through (−1, 1/2). 7 2
SOLUTION: 5 1 −10 x − y = 7 2 5 1 −10 x + 10 x − y = + 10 x 7 2 5 1 y = −10 x − 7 2 7 5 7 7 1 ⋅ y = ⋅ ( −10 ) x − ⋅ 5 7 5 5 2 7 y = −14 x − 10 Here the given line has slope m = −14 and thus m& = −14 . We substitute this slope and the given point into point-slope form.
Point
Slope
1⎞ ⎛ m& = −14 ⎜ −1, ⎟ 2⎠ ⎝ y − y1 = m ( x − x1 ) 1 = −14 ( x − ( −1) ) 2 27 y = −14 x − 2 y−
ANSWER: y = −14 x −
27 2
55.
PROBLEM:
Perpendicular to
1 1 x − y = 1 and passing through (−10, 3). 2 3
SOLUTION: 1 1 x − y =1 2 3 1 1 1 1 x − x − y = 1− x 2 2 3 2 1 1 y = x −1 3 2 1 1 3 ⋅ y = 3 ⋅ x − 3 ⋅1 3 2 3 y = x −3 2 3 1 2 and thus m⊥ = − = − . We substitute this 3 2 3 2 slope and the given point into point-slope form.
Here the given line has slope m =
Point
Slope
(−10, 3)
m⊥ = −
y − y1 = m ( x − x1 )
2 3
2 ( x − ( −10 ) ) 3 2 11 y =− x− 3 3 2 11 ANSWER: y = − x − 3 3 y −3 = −
56.
PROBLEM: Perpendicular to −5 x + y = −1 and passing through (−4, 0).
SOLUTION: −5 x + y = −1
−5 x + 5 x + y = −1 + 5 x y = 5x −1 1 Here the given line has slope m = 5 and thus m⊥ = − . We substitute this slope 5 and the given point into point-slope form.
Point
Slope
(−4, 0)
m⊥ = −
y − y1 = m ( x − x1 )
1 5
1 ( x − ( −4 ) ) 5 1 4 y =− x− 5 5 y−0 = −
1 4 ANSWER: y = − x − 5 5 57.
PROBLEM: Parallel to x + 4 y = 8 and passing through (−1, −2). SOLUTION: x + 4y = 8 x − x + 4y = 8 − x
4 y = −x + 8 4 1 8 y =− x+ 4 4 4 1 y =− x+2 4 1 1 and thus m& = − . We substitute this slope 4 4 and the given point into point-slope form.
Here the given line has slope m = −
Point (−1, −2)
Slope 1 m& = − 4
y − y1 = m ( x − x1 ) y − ( −2 ) = −
1 ( x − ( −1) ) 4
1 9 y =− x− 4 4 1 9 ANSWER: y = − x − 4 4 58.
PROBLEM: Parallel to 7 x − 5 y = 35 and passing through (2, −3). SOLUTION: 7 x − 5 y = 35 7 x − 7 x − 5 y = 35 − 7 x
5 y = 7 x − 35 5 7 35 y = x− 5 5 5 7 y = x −5 5 7 7 and thus m& = . We substitute this slope 5 5 and the given point into point-slope form.
Here the given line has slope m =
Point
Slope 7 (2, −3) m& = 5 y − y1 = m ( x − x1 ) 7 ( x − 2) 5 7 29 y = x− 5 5 y − ( −3) =
ANSWER: y =
7 29 x− 5 5
59.
PROBLEM: Perpendicular to 6 x + 3 y = 1 and passing through (8, −2). SOLUTION: 6x + 3 y = 1 6x − 6x + 3y = 1 − 6x
3 y = −6 x + 1 3 6 1 y =− x+ 3 3 3 1 y = −2 x + 3 Here the given line has slope m = −2 and thus m⊥ = and the given point into point-slope form. Slope 1 (8, −2) m⊥ = 2 y − y1 = m ( x − x1 ) Point
y − ( −2 ) = y=
1 ( x − 8) 2
1 x−6 2
ANSWER: y =
1 x−6 2
1 . We substitute this slope 2
60.
PROBLEM: Perpendicular to −4 x − 5 y = 1 and passing through (−1, −1). SOLUTION: −4 x − 5 y = 1 −4 x + 4 x − 5 y = 1 + 4 x
5 y = −4 x − 1 5 4 1 y =− x− 5 5 5 4 1 y =− x− 5 5 4 1 5 and thus m⊥ = − = . We substitute 4 4 5 − 5 this slope and the given point into point-slope form.
Here the given line has slope m = −
Point
Slope 5 (−1, −1) m⊥ = 4 y − y1 = m ( x − x1 ) 5 ( x − ( −1) ) 4 5 1 y = x+ 4 4 y − ( −1) =
ANSWER: y =
5 1 x+ 4 4
61.
PROBLEM: ⎛1 1⎞ Parallel to −5 x − 2 y = 4 and passing through ⎜ , − ⎟ . ⎝5 4⎠
SOLUTION: −5 x − 2 y = 4 −5 x + 5 x − 2 y = 4 + 5 x
2 y = −5 x − 4 2 5 4 y =− x− 2 2 2 5 y = − x−2 2 5 5 and thus m& = − . We substitute this slope 2 2 and the given point into point-slope form.
Here the given line has slope m = −
Point
Slope
5 ⎛1 1⎞ m& = − ⎜ ,− ⎟ 2 ⎝5 4⎠ y − y1 = m ( x − x1 ) 5⎛ 1⎞ ⎛ 1⎞ y −⎜− ⎟ = − ⎜ x − ⎟ 2⎝ 5⎠ ⎝ 4⎠ 5 1 y =− x+ 2 4 5 1 ANSWER: y = − x + 2 4
62.
PROBLEM:
Parallel to 6 x −
3 y = 9 and passing through 2
⎛1 2⎞ ⎜ , ⎟. ⎝3 3⎠
SOLUTION: 3 6x − y = 9 2 3 6x − 6x − y = 9 − 6x 2 3 y = 6x − 9 2 2 3 2 2 ⋅ y = ⋅ 6x − ⋅ 9 3 2 3 3 y = 4x − 6 Here the given line has slope m = 4 and thus m& = 4 . We substitute this slope and the given point into point-slope form.
Point
Slope
⎛1 2⎞ m& = 4 ⎜ , ⎟ ⎝3 3⎠ y − y1 = m ( x − x1 ) 2 1⎞ ⎛ = 4⎜ x − ⎟ 3 3⎠ ⎝ 2 y = 4x − 3 y−
ANSWER: y = 4 x −
2 3
63.
PROBLEM: Perpendicular to y − 3 = 0 and passing through (−6, 12). SOLUTION: y −3 = 0
Here the given line has slope m = 0 and thus m⊥ = −
1 = undefined . We substitute 0
this slope and the given point into point-slope form. Point (−6, 12)
Slope m⊥ = undefined
y − y1 = m ( x − x1 )
y − 12 = ( x − ( −6 ) ) undefined x+6 = 0 x = −6 ANSWER: x = −6
64.
PROBLEM: Perpendicular to x + 7 = 0 and passing through (5, −10). SOLUTION: x+7 =0
Here the given line has slope m = undefined and thus m⊥ = − substitute this slope and the given point into point-slope form. Point Slope (5, −10) m⊥ = 0 y − y1 = m ( x − x1 ) y − ( −10 ) = 0 ( x − 5 ) y = −10 ANSWER: y = −10
1 = 0 . We undefined
3.7 Introduction to Functions Part A: Functions Does the following correspondence represent a function?
1.
PROBLEM: Algebra students to their scores on the first exam. ANSWER: An algebra student corresponds exactly to an algebra score. Hence the above statement represents a function.
2.
PROBLEM: Family members to their ages. ANSWER: A family member corresponds exactly to an age. Hence the above statement represents a function.
3.
PROBLEM: Lab computers to their users. ANSWER: A single lab computer does not correspond exactly to one user. Hence the above statement does not represent a function.
4.
PROBLEM: Students to the schools they have attended. ANSWER: A single student does not correspond exactly to a single school attended. Hence the above statement does not represent a function.
5.
PROBLEM: People to their citizenships. ANSWER: A single person does not correspond exactly to a single citizenship. Hence the above statement does not represent a function.
6.
PROBLEM: Local businesses to their number of employees. ANSWER: A local business corresponds exactly to a number of employees. Hence the above statement represents a function.
Determine the domain and range and state whether the relation is a function or not.
7.
PROBLEM: {(3, 2), (5, 3), (7, 4)} SOLUTION:
3 5 7
2 3 4
ANSWER: Domain: {3, 5, 7} Range: {2, 3, 4} The relation is a function because each x-value corresponds to exactly one of yvalue.
8.
PROBLEM: {(−5, −3), (0, 0), (5, 0)} SOLUTION:
−5 0
−3 0
5 ANSWER: Domain:{–5, 0, 5} Range: {–3, 0} The relation is a function because each x-value corresponds to exactly one of yvalue.
9.
PROBLEM: {(−10, 2), (−8, 1), (−8, 0)} SOLUTION:
–10 −8
2 1 0
ANSWER: Domain: {−10, −8} Range: {2, 1, 0}. The relation is not a function because each x-value does not correspond to exactly one of y-value.
10.
PROBLEM: {(9, 12), (6, 6), (6, 3)} SOLUTION:
9 6
−3 6 3
ANSWER: Domain: {9, 6} Range: {–3, 6, 3} The relation is not a function because each x-value does not correspond to exactly one of y-value.
11.
PROBLEM:
SOLUTION: The coordinates marked are (–4, 1), (1, 2), and (2, 3). ANSWER: Domain: {–4, 1, 2} Range: {1, 2, 3} The relation is a function because each x-value corresponds to exactly one of yvalue.
12.
PROBLEM:
SOLUTION: The coordinates marked are (–3, –2), (1, –2), and (3, –2). ANSWER: Domain: {–3, 1, 3} Range: {–2} The relation is a function because each x-value corresponds to exactly one of yvalue.
13.
PROBLEM:
SOLUTION: The coordinates marked are (–2, 3), (2, 2), and (2, 5). ANSWER: Domain: {–2, 2} Range: {3, 2, 5} The relation is not a function because each x-value does not correspond to exactly one of y-value.
14.
PROBLEM:
SOLUTION: The coordinates marked are (4, –3), (4, 1), and (4, 3). ANSWER: Domain: {4} Range: {–3, 1, 3} The relation is not a function because each x-value does not correspond to exactly one of y-value.
15.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that 2 is the maximum y-value and any y-value lesser than this is not represented in the relation. Hence, the range is 2. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is {2}. The relation is a function because it passes the vertical line test.
16.
PROBLEM:
SOLUTION: The graph indicates that −3 is the minimum x-value and any xvalue greater than that is not represented in the relation. Hence, the domain is equal to −3. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will not intersect the graph only once; therefore it is not a function. ANSWER: The domain is {–3} and the range is all real numbers R = (–∞,∞). The relation is not a function because it does not pass the vertical line test.
17.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is all real numbers too R = (–∞,∞). The relation is a function because it passes the vertical line test.
18.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that −3 is the minimum y-value and any y-value greater than that is represented in the relation. Hence, the range consists of all yvalues greater than or equal to −3, or in interval notation[−3, ∞) . Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is [–3,∞). The relation is a function because it passes the vertical line test.
19.
PROBLEM:
SOLUTION: The graph indicates that −2 is the minimum x-value and any xvalue greater than that is represented in the relation. Hence, the domain consists of all values greater than or equal to −2 or the interval notation [ −2, ∞ ) . If we
choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph twice; therefore it is not a function. ANSWER: The domain is [ −2, ∞ ) and the range is all real numbers R = (–
∞,∞).The relation is not a function because it does not pass the vertical line test.
20.
PROBLEM:
SOLUTION: The graph indicates that 1 is the maximum x-value and any x-value lesser than that is represented in the relation. Hence, the domain consists of all xvalues lesser than or equal to 1, or in interval notation ( −∞,1] . If we choose any
y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph twice; therefore it is not a function. ANSWER: The domain is ( −∞,1] and the range is all real numbers R = (–
∞,∞).The relation is not a function because it does not pass the vertical line test. 21.
PROBLEM:
SOLUTION: The graph indicates that −4 is the minimum x-value and any xvalue greater than that is represented in the relation. Hence, the domain consists of all values greater than or equal to −4 or the interval notation [ −4, ∞ ) . If we
choose any positive y-value, then we will be able to find a corresponding point on the graph; therefore, the range is greater than or equal to zero or the interval notation [0, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is [ −4, ∞ ) and the range is all positive integers [0,
∞).The relation is a function because it passes the vertical line test.
22.
PROBLEM:
SOLUTION: The graph indicates that 3 is the maximum x-value and any x-value lesser than that is represented in the relation. Hence, the domain consists of all xvalues lesser than or equal to 3, or in interval notation ( −∞,3] . If we choose any
negative y-value, then we will be able to find a corresponding point on the graph; therefore, the range is lesser than or equal to zero or in the interval notation (–∞, 0]. Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞,3] and the range is (–∞, 0].The relation is a
function because it passes the vertical line test. 23.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any positive y-value, then we will be able to find a corresponding point on the graph; therefore, the range is greater than or equal to zero or the interval notation [0, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞, ∞ ) and the range is all positive integers [0,
∞).The relation is a function because it passes the vertical line test.
24.
PROBLEM:
SOLUTION: If we choose any x-value we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any positive y-value we will be able to find a corresponding point on the graph; therefore, the range is greater than or equal to zero or in the interval notation [ 0, ∞ ) . Any vertical line will intersect the graph once; therefore it is a
function. ANSWER: The domain is ( −∞, ∞ ) and the range is all positive
integers [ 0, ∞ ) .The relation is a function because it passes the vertical line test. 25.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The minimum value of y is 2 and any value greater than that is represented in the relation; therefore, the range is greater than or equal to 2 or the interval notation [2, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞, ∞ ) and the range is [2, ∞).The relation is a function because it passes the vertical line test.
26.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The minimum value of y is –3 and any value greater than that is represented in the relation; therefore, the range is greater than or equal to –3 or the interval notation [–3, ∞). Any vertical line will intersect the graph once; therefore it is a function. ANSWER: The domain is ( −∞, ∞ ) and the range is [–3, ∞).The relation is a
function because it passes the vertical line test. Part B: Function Notation Given the following functions find the function values.
27.
PROBLEM: f ( x) = 3x find f (−2) SOLUTION: f ( x) = 3 x
f (−2) = 3 ⋅ ( −2 ) = −6 ANSWER: −6
28.
PROBLEM: f ( x) = −5 x + 1 find f (−1) SOLUTION: f ( x) = −5 x + 1
f ( −1) = −5 ⋅ ( −1) + 1 = 5 + 1 = 6 ANSWER: 6
29.
PROBLEM: f ( x) = 53 x − 4 find f (15) SOLUTION: f ( x) = 53 x − 4
f (15) = 53 ⋅15 − 4 = 9 − 4 = 5 ANSWER: 5
30.
PROBLEM: f ( x) = 52 x − 15 find f (3) SOLUTION: f ( x) = 52 x − 15 f (3) =
2 1 5 ⋅ ( 3) − = = 1 5 5 5
ANSWER: 1
31.
PROBLEM: f ( x) = 52 x − 13 find f (− 13 ) SOLUTION: f ( x) = 52 x − 13
5 ⎛ 1⎞ 1 7 ⋅⎜ − ⎟ − = − 2 ⎝ 3⎠ 3 6 7 ANSWER: − 6 f ( x) =
32.
PROBLEM: f ( x) = −6 find f (7) SOLUTION: f ( x) = −6
f ( 7 ) = −6 ANSWER: −6
33.
PROBLEM: g ( x) = 5 find g (−4) SOLUTION: g ( x) = 5 g (−4) = 5 ANSWER: 5
34.
PROBLEM: g ( x) = −5 x find g (−3) SOLUTION: g ( x) = −5 x
g (−3) = −5 ( −3) = 15 ANSWER: 15
35.
PROBLEM: g ( x) = − 18 x + 85 find g ( 85 ) SOLUTION: 1 5 g ( x) = − x + 8 8 1 ⎛ 5 ⎞ 5 35 ⎛5⎞ g ⎜ ⎟ = − ⋅⎜ ⎟ + = 8 ⎝ 8 ⎠ 8 64 ⎝8⎠ 35 ANSWER: 64
36.
PROBLEM: g ( x) = 53 x − 5 find g ( 3) SOLUTION: 5 g ( x) = x − 5 3 5 g ( 3) = ⋅ 3 − 5 = 0 3 ANSWER: 0
37.
PROBLEM: f ( x) = 5 x − 9 find x when f ( x) = 1 SOLUTION: f ( x) = 5 x − 9 When f(x) = 1, f ( x) = 5 x − 9
1 = 5x − 9 10 5 = x 5 5 x=2 ANSWER: x = 2
38.
PROBLEM: f ( x) = −7 x + 2 find x when f ( x) = 0 SOLUTION: f ( x) = −7 x + 2 When f ( x) = 0, f ( x) = −7 x + 2 0 = −7 x + 2 2 = 7x
2 7 = x 7 7 2 x= 7 ANSWER: x =
2 7
39.
PROBLEM: f ( x) = − 75 x − 2 find x when f ( x) = −9 SOLUTION: f ( x) = − 75 x − 2 When f ( x) = −9 , 7 −9 = − x − 2 5 ⎛ 5 ⎞ ⎛ 7 ⎞⎛ 5 ⎞ −7 ⎜ − ⎟ = ⎜ − x ⎟ ⎜ − ⎟ ⎝ 7 ⎠ ⎝ 5 ⎠⎝ 7 ⎠ x=5 ANSWER: x = 5
40.
PROBLEM: f ( x) = − x − 4 find x when f ( x) = SOLUTION: f ( x) = − x − 4 When f ( x) = 12 1 −x − 4 = 2 1 −x − 4 + 4 = + 4 2 9 −x = 2 9 x=− 2 ANSWER: x = −
9 2
1 2
41.
PROBLEM: g ( x) = x find x when g ( x) = 12 SOLUTION: g ( x) = x When g ( x) = 12 , 12 = x x = 12 ANSWER: x = 12
42.
PROBLEM: g ( x) = − x + 1 find x when g ( x) =
2 3
SOLUTION: g ( x) = − x + 1 When g ( x) = 23 2 = −x +1 3 1 x= 3 ANSWER: x =
43.
1 3
PROBLEM: g ( x) = −5 x + 13 find x when g ( x) = − 12 SOLUTION: g ( x) = −5 x + 13 When g ( x) = − 12 , 1 1 − = −5 x + 2 3 5 5x = 6 5 5 x= 5 6⋅5 1 x= 6 1 ANSWER: x = 6
44.
PROBLEM: g ( x) = − 85 x + 3 find x when g ( x) = 3 SOLUTION: g ( x) = − 85 x + 3 When g ( x) = 3 5 3= − x+3 8 5 x=0 8 x=0 ANSWER: x = 0
Given f ( x) = 23 x − 1 and g ( x) = −3 x + 2 calculate the following (#46 ‐ #54):
45.
PROBLEM: f (6) SOLUTION: 2 f ( x) = x − 1 3 2 f ( 6) = ( 6) −1 = 3 3 ANSWER: 3
46.
PROBLEM: f (−12) SOLUTION: 2 f ( x) = x − 1 3 2 f ( −12 ) = ( −12 ) − 1 = −9 3 ANSWER: −9
47.
PROBLEM: f (0) SOLUTION: 2 f ( x) = x − 1 3 2 f ( 0 ) = ( 0 ) − 1 = −1 3
48.
49.
ANSWER: −1 PROBLEM: f (1) SOLUTION: 2 f ( x) = x − 1 3 2 1 f (1) = (1) − 1 = − 3 3 1 ANSWER: − 3 PROBLEM: g ( 23 ) SOLUTION: g ( x ) = −3 x + 2
⎛2⎞ ⎛2⎞ g ⎜ ⎟ = −3 ⎜ ⎟ + 2 = 0 ⎝3⎠ ⎝3⎠ ANSWER: 0 50.
PROBLEM: g (0) SOLUTION: g ( x ) = −3 x + 2
g ( 0 ) = −3 ( 0 ) + 2 = 2 ANSWER: 2
51.
PROBLEM: g ( −1) SOLUTION: g ( x) = −3 x + 2
g ( −1) = −3 ( −1) + 2 = 5 ANSWER: 5 52.
PROBLEM: g ( − 12 ) SOLUTION: g ( x) = −3x + 2
7 ⎛ 1⎞ ⎛ 1⎞ g ⎜ − ⎟ = −3 ⎜ − ⎟ + 2 = 2 ⎝ 2⎠ ⎝ 2⎠ 53.
ANSWER:
7 2
PROBLEM: Find x when f ( x) = 0 . SOLUTION: 2 f ( x) = x − 1 3 When f ( x) = 0 2 0 = x −1 3 3 3 2 ⋅1 = ⋅ x 2 2 3 3 x= 2 ANSWER: x =
3 2
54.
55.
PROBLEM: Find x when f ( x) = −3 . SOLUTION: 2 f ( x) = x − 1 3 When f ( x) = −3 2 −3 = x − 1 3 3 3 2 ⋅ ( −2 ) = ⋅ x 2 2 3 x = −3 ANSWER: x = −3 PROBLEM: Find x when g ( x) = −1 . SOLUTION: g ( x) = −3x + 2 When g ( x) = −1 −1 = −3 x + 2
− 3 = −3 x −3 −3 = x −3 −3 x =1 ANSWER: x = 1
56.
PROBLEM: Find x when g ( x) = 0 . SOLUTION: g ( x ) = −3 x + 2 When g ( x) = 0 0 = −3 x + 2 −2 = −3 x −2 −3 x = −3 −3 2 x= 3
ANSWER: x =
2 3
Given the graph find the function values.
57.
PROBLEM: Given the graph of f ( x) find f (−4) , f (−1) , f (0) and f (2) .
SOLUTION: From the graph when x = –4, y = – 3. Hence, f (−4) = −3. From the graph when x = –1, y = – 0. Hence, f (−1) = 0. From the graph when x = 0, y =1. Hence, f (0) = 1. From the graph when x = 2, y = 3. Hence, f (2) = 3. ANSWER: f (−4) = −3. , f (−1) = 0. , f (0) = 1. , f (2) = 3.
58.
PROBLEM: Given the graph of g ( x ) find g (−3) , g (−1) , g (0) and g (1) .
ANSWER: From the graph when x = –3, y = 4. Hence, From the graph when x = –1, y = 0. Hence, From the graph when x = 0, y = –2. Hence, From the graph when x = 1, y = –4. Hence,
g (−3) = 4. g (−1) = 0. g (0) = −2. g (1) = −4.
59.
PROBLEM: Given the graph of f ( x) find f (−4) , f (−1) , f (0) and f (2) .
SOLUTION: From the graph when x = –4, y = –4 . Hence, f (−4) = −4. From the graph when x = –1, y = –4 . Hence, f (−1) = −4. From the graph when x = 0, y = –4 . Hence, f (0) = −4. From the graph when x = 2, y = –4 . Hence, f (2) = −4. ANSWER: f (−4) = −4. , f (−1) = −4. , f (0) = −4. , f (2) = −4.
60.
PROBLEM: Given the graph of g ( x ) find g (−4) , g (−1) , g (0) and g (2) .
ANSWER: From the graph when x = –4, y = 0. Hence, g (−4) = 0. From the graph when x = –1, y = 0. Hence, g (−1) = 0. From the graph when x = 0, y = 0. Hence, g (0) = 0. From the graph when x = 2, y = 0. Hence, g (2) = 0.
61.
PROBLEM: Given the graph of f ( x ) find f (−1) , f (0) , f (1) ,and f (3) .
SOLUTION: From the graph when x = –1, y = 1 . Hence, f (−1) = 1. From the graph when x = 0, y = –2 . Hence, f (0) = −2. From the graph when x = 1, y = –3 . Hence, f (1) = −3. From the graph when x = 3, y = 1 . Hence, f (3) = 1. ANSWER: f (−1) = 1. , f (0) = −2. , f (1) = −3. , f (3) = 1.
62.
PROBLEM: Given the graph of g ( x ) find g (−2) , g (0) , g (2) and g (6) .
ANSWER: From the graph when x = –2, y = 0. Hence, g (−2) = 0. From the graph when x = 0, y = 12. Hence, g (0) = 12. From the graph when x = 2, y = 16. Hence, g (2) = 16. From the graph when x = 6, y = 0. Hence, g (6) = 0.
63.
PROBLEM: Given the graph of g ( x ) find g (−4) , g (−3) , g (0) and g (4) .
SOLUTION: From the graph when x = –4, y = 0. Hence, g (−4) = 0. From the graph when x = –3, y = 1. Hence, g (−3) = 1. From the graph when x = 0, y = 1. Hence, g (0) = 2. From the graph when x = 4, y = 3. Hence, g (4) = 3. ANSWER: g (−4) = 0. , g (−3) = 1. , g (0) = 2. , g (4) = 3.
64.
PROBLEM: Given the graph of f ( x ) find f (−4) , f (0) , f (1) ,and f (3) .
ANSWER: From the graph when x = –4, y = 1 . Hence, f (−4) = 1. From the graph when x = 0, y = –3 . Hence, f (0) = −3. From the graph when x = 1, y = –2 . Hence, f (1) = −2. From the graph when x = 3, y = 0 . Hence, f (3) = 0.
Given the graph find the x‐values.
65.
PROBLEM: Given the graph of f ( x) find x when f ( x) = 3 , f ( x) = 1 and f ( x) = −3 .
SOLUTION: When f ( x) = 3 , y = 3. From the graph when y = 3, x = –1. When f ( x) = 1 , y = 1. From the graph when y = 1, x = 0. When f ( x) = –3 , y = –3. From the graph when y = –3, x = 2. ANSWER: x = –1, x = 0, x = 2
66.
PROBLEM: Given the graph of g ( x ) find x when g ( x) = −1 , g ( x) = 0 and g ( x) = 1 .
ANSWER: When g ( x) = −1 , y = –1. From the graph when y = –1, x = 1. When g ( x) = 0 , y = 0. From the graph when y = 0, x = 0. When g ( x) = 1 , y = 1. From the graph when y = 1, x = –1.
67.
PROBLEM: Given the graph of f ( x) find x when f ( x) = 3 .
ANSWER: When f ( x) = 3 , y = 3. From the graph when y = 3, x = 0. (Answers may vary)
68.
PROBLEM: Given the graph of g ( x) find x when g ( x) = −4 , g ( x) = 0 and g ( x) = 4 .
ANSWER: When g ( x) = −4 , y = –4. From the graph when y = –4, x = –-6. When g ( x) = 0 , y = 0. From the graph when y = 0, x = –2. When g ( x) = 4 , y = 4. From the graph when y = 4, x = 2.
69.
PROBLEM: Given the graph of f ( x) find x when f ( x) = −16 , f ( x) = −12 and f ( x) = 0 .
SOLUTION: When f ( x) = −16 , y = –16 . From the graph when y = –16, x = –4. When f ( x) = −12 , y = –12. From the graph when y = –12, x = –2 and x = –6. When f ( x) = 0 , y = 0. From the graph when y = 0, x = 0 and x = –8. ANSWER: x = –4, x = –2 and x = –6, x = 0 and x = –8
70.
PROBLEM: Given the graph of g ( x) find x when g ( x) = −3 , g ( x ) = 0 and g ( x) = 1 .
ANSWER: When g ( x) = −3 , y = –3. From the graph when y = –3, x = –2 and x = 2. When g ( x) = 0 , y = 0. From the graph when y = 0, x = –1 and x = 1. When g ( x) = 1 , y = 1. From the graph when y = 1, x = 0.
71.
PROBLEM: Given the graph of f ( x) find x when f ( x) = −4 , f ( x) = 0 and f ( x) = −2 .
SOLUTION: When f ( x) = −4 , y = –4. From the graph when y = –4, x = –4 and x = 4. When f ( x) = 0 , y = 0. From the graph when y = 0, x = 0. When f ( x) = −2 , y = –2. From the graph when y = –2, x = –2 and x = 2. ANSWER: x = –4 and x = 4, x = 0, x = –2 and x = 2
72.
PROBLEM: Given the graph of g ( x) find x when g ( x) = 5 , g ( x) = 3 and g ( x) = 2 .
ANSWER: When g ( x) = 5 , y = 5. From the graph when y = 5, x = –3 and x = 3. When g ( x) = 3 , y = 3. From the graph when y = 3, x = –1 and x = 1. When g ( x) = 2 , y = 2. From the graph when y = 2, x = 0.
73.
PROBLEM: The cost, in dollars, of producing pens with a company logo is given by the function C ( x) = 1.65 x + 120 where x is the number of pens produced. Use the function to calculate the cost of producing 200 pens. SOLUTION: Cost of producing a pen is given by the equation C ( x) = 1.65 x + 120. ANSWER: Cost of producing 200 pens, C (200) = 1.65 ( 200 ) + 120 = $450.
74.
PROBLEM: The revenue, in dollars, from selling sweatshirts is given by the function R( x) = 29.95 x where x is the number of sweatshirts sold. Use the function to determine the revenue if 20 sweatshirts are sold. SOLUTION: Cost of producing a single sweatshirt is given by the function, R( x) = 29.95 x. ANSWER: Cost of producing 20 sweatshirts is R(20) = 29.95 ( 20 ) = $599.
75.
PROBLEM: The value of a new car, in dollars, is given by the function V (t ) = −2500t + 18000 where t represents the age of the car in years. Use the function to determine the value of the car when it is 5 years old. What was the value of the car new? SOLUTION: Value of the car is given by the function, V (t ) = −2500t + 18000.
Value of the car after 5 years, V (5) = −2500 ( 5) + 18000 = $5,500. ANSWER: Value of the new car, V (0) = −2500 ( 0 ) + 18000 = $18, 000.
76.
PROBLEM: The monthly income, in dollars, of a commissioned car salesman is given by the function I (n) = 550n + 1250 where n represents the number of cars sold in the month. Use the function to determine the salesman’s monthly income if he sells 3 cars this month. What is his income if he does not sell any cars in one a month? SOLUTION: Monthly income of the salesman based on the number of cars sold is given by the function, I (n) = 550n + 1250.
Monthly income if he sells 3 cars, I (3) = 550 ( 3) + 1250 = $2,900. ANSWER: Monthly income if he does not sell any car, I (0) = 550 ( 0 ) + 1250 = $1, 250.
77.
PROBLEM: The perimeter of an isosceles triangle with base measuring 10 cm is given by the function P ( x) = 2 x + 10 where x represents the length of each of the equal sides. Find the length of each side if the perimeter measure 40 cm. SOLUTION: P(x) = 40 cm P(x) = 2x + 10 40 = 2x + 10 2x = 30 2 30 x= 2 2 x = 15 cm ANSWER: The length of each side measures 15 cm.
78.
PROBLEM: The perimeter of a square depends on the length of each side s and is modeled by the function P( s ) = 4s . If the perimeter of a square measures 140 meters, then use the function to calculate the length of each side. SOLUTION: P(s) = 4s P(s) = 140 meters 4s = 140 meters 4 140 s= meters 4 4 s = 35 meters ANSWER: The length of each side is 35 meters.
79.
PROBLEM: A certain cellular phone plan charges $18 per month and $0.10 per minute of use which is modeled by the function C ( x) = 0.10 x + 18 where x represents the number of minutes of usage per month. Determine the minutes of usage if the cost for the month was $36. SOLUTION: C ( x) = 0.10 x + 18 C(x) = $36 36 = 0.10x + 18 0.10x = 18 0.10 18 x= 0.10 0.10 x = 180 minutes ANSWER: If the cost for the month is $36, the minutes of usage is 180.
80.
PROBLEM: The monthly revenue generated by selling subscriptions to a tutoring website is given by the function R( x) = 29 x where x represents the number of subscription sales per month. How many subscriptions were sold if the revenues for the month totaled $1,508? SOLUTION: R( x) = 29 x
R( x) = $1,508 29 x = 1508 29 1508 x= 29 29 x = 52 ANSWER: For the month totaled $1,508, 52 subscriptions were sold. Graph the linear function and state the domain and range.
81.
PROBLEM: f ( x) = − 52 x + 10 SOLUTION: f ( x) = − 52 x + 10
y = − 52 x + 10 Set y = 0 y = − 52 x + 10 0 = − 52 x + 10 5 x = 10 2 2 5 2 ⋅ x = ⋅10 5 2 5 x=4 x-intercept = (4, 0) Set x = 0 y = − 52 x + 10 5 ( 0 ) + 10 2 y = 10 y-intercept = (0, 10) y=−
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any yvalue, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 82.
PROBLEM: f ( x) = 53 x − 10 SOLUTION: f ( x) = 53 x − 10
y = 53 x − 10 Set y = 0 y = 53 x − 10 0 = 53 x − 10 3 x = 10 5 5 3 5 ⋅ x = ⋅10 3 5 3 2 x = 16 3 ⎛ 2 ⎞ x-intercept = ⎜16 , 0 ⎟ ⎝ 3 ⎠
Set x = 0 y = 53 x − 10 3 ( 0 ) + 10 5 y = 10 y-intercept = (0, 10) y=
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any yvalue, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 83.
PROBLEM: g ( x) = 6 x + 2 SOLUTION: g ( x) = 6 x + 2 Set g ( x) = 0 g ( x) = 6 x + 2
0 = 6( x) + 2 6 2 x=− 6 6 1 x=− 3 ⎛ 1 ⎞ x-intercept = ⎜ − , 0 ⎟ ⎝ 3 ⎠ Set x = 0 y = 6x + 2 y = 6 ( 0) + 2 y=2 g ( x) - intercept = (0, 2)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any g(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 84.
PROBLEM: g ( x ) = −4 x + 6 SOLUTION: g ( x) = −4 x + 6 Set g ( x) = 0 g ( x ) = −4 x + 6
0 = −4 x + 6 4x = 6 4 6 x= 4 4 1 x =1 2 ⎛ 1 ⎞ x-intercept = ⎜1 , 0 ⎟ ⎝ 2 ⎠ Set x = 0 g ( x) = −4 x + 6 g ( x) = −4 ( 0 ) + 6 g ( x) = 6 g(x) intercept = (0, 6)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any g(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 85.
PROBLEM: h(t ) = 12 t − 3 SOLUTION: h(t ) = 12 t − 3 Set h(t) = 0 h(t ) = 12 t − 3
0 = 12 t − 3 1 2
t =3
t =6 t-intercept = ( 6, 0 ) Set t = 0 h(t ) = 12 t − 3 h(t ) =
1 2
( 0) − 3
h(t ) = −3 h(t)-intercept = (0, –3 )
ANSWER:
If we choose any t-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any h(t)value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 86.
PROBLEM: h(t ) = − 34 t + 3 SOLUTION: h(t ) = − 34 t + 3 Set h(t) = 0 h(t ) = − 34 t + 3
0 = − 34 t + 3 3 t =3 4 4 3 4 ⋅ t = ⋅3 3 4 3 t=4 t-intercept = ( 4, 0 ) Set t = 0 h(t ) = − 34 t + 3
h(t ) = − 34 ( 0 ) + 3
h(t ) = 3 h(t)-intercept = (0, 3)
ANSWER:
If we choose any t-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any h(t)value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 87.
PROBLEM: C ( x) = 100 + 50 x SOLUTION: C ( x) = 100 + 50 x Set C ( x) = 0 C ( x) = 100 + 50 x
0 = 100 + 50 x 50 100 x=− 50 50 x = −2 x-intercept = ( −2, 0 ) Set x = 0 C ( x) = 100 + 50 x C ( x) = 100 + 5 ( 0 ) C ( x) = 100 C ( x) -intercept = (0, 100)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any C(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ 88.
PROBLEM: C ( x) = 50 + 100 x SOLUTION: C ( x) = 50 + 100 x Set C ( x) = 0 C ( x) = 50 + 100 x
0 = 50 + 100 x 100 x = −50 100 50 x=− 100 100 1 x=− 2 ⎛ 1 ⎞ x-intercept = ⎜ − , 0 ⎟ ⎝ 2 ⎠ Set x = 0 C ( x) = 50 + 100 x C ( x) = 50 + 100 ( 0 ) C ( x) = 50 C ( x) -intercept = (0, 50)
ANSWER:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any C(x)-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Domain and Range \ Part C: Discussion Board Topics
89.
PROBLEM: Is a vertical line a function? What are the domain and range of a vertical line? ANSWER: The vertical line is not a function. E.g:
The graph indicates that −3 is the minimum x-value and any x-value greater than that is not represented in the relation. Hence, the domain is equal to −3. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will not intersect the graph only once; therefore it is not a function. The domain is {–3} and the range is all real numbers R = (–∞,∞). The relation is not a function because it does not pass the vertical line test.
90.
PROBLEM: Is a horizontal line a function? What are the domain and range of a horizontal line? ANSWER: The horizontal line is a function. E.g.:
If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that 2 is the minimum y-value and any y-value greater than that is not represented in the relation. Hence, the range is 2. Any vertical line will intersect the graph only once; therefore it is a function. The domain is all real numbers R = (–∞,∞) and the range is {2}. The relation is a function because it passes the vertical line test. 91.
PROBLEM: Come up with your own correspondence between real world sets. Explain why it represents a function or not? ANSWER: 30 NBA teams from 30 different cities: This represents a function because each NBA team corresponds exactly to one city.
Coffee produced is the highest in Brazil and Mexico: This does not represent a function because coffee does not correspond exactly to one country.
3.8 Linear Inequalities (two variables)
Part A: Solutions to Linear Inequalities (two variables) Is the ordered pair a solution to the given inequality?
1.
PROBLEM: y < 5 x + 1 ; (0, 0) SOLUTION: y < 5x + 1
0 < 5 ( 0) + 1 0 − x − 4 ; (0, −2) 2 SOLUTION: 1 y > − x−4 2 1 −2 > − ( 0 ) − 4 2 −2 > −4
3.
1 ANSWER: (0, −2) is a solution to the inequality y > − x − 4. 2 PROBLEM: 2 y ≤ x + 1 ; (6, 5) 3 SOLUTION: 2 y ≤ x +1 3 2 5 ≤ ( 6) + 1 3 5≤5 ANSWER: (6, 5) is a solution to the inequality y ≤
2 x + 1. 3
4.
PROBLEM: 1 y ≥ − x − 5 ; (−3, −8) 3 SOLUTION: 1 y ≥ − x −5 3 1 −8 ≥ − ( −3) − 5 3 −8 ≥ −4 ( Incorrect )
1 ANSWER: (−3, −8) is not a solution to the inequality y ≥ − x − 5. 3 5.
PROBLEM: 1 1 ⎛ 1 ⎞ y < x − ; ⎜ − , −1 ⎟ 5 3 ⎝ 3 ⎠ SOLUTION: 1 1 y < x− 5 3 1⎛ 1⎞ 1 −1 < ⎜ − ⎟ − 5⎝ 3⎠ 3 6 −1 < − 15 1 1 ⎛ 1 ⎞ ANSWER: ⎜ − , −1⎟ is a solution to the inequality y < x − . 5 3 ⎝ 3 ⎠
6.
PROBLEM: 4 x − 3 y ≤ 2 ; (−2, −1) SOLUTION: 4x − 3 y ≤ 2
4 ( −2 ) − 3 ( −1) ≤ 2 −5 ≤ 2 ANSWER: (−2, −1) is a solution to the inequality 4 x − 3 y ≤ 2.
7.
PROBLEM: − x + 4 y > 7 ; (0, 0) SOLUTION: −x + 4 y > 7
0+0 > 7 0 > 7 ( Incorrect ) ANSWER: (0, 0) is not a solution to the inequality − x + 4 y > 7.
8.
PROBLEM: 7 x − 3 y < 21 ; (5, −3) SOLUTION: 7 x − 3 y < 21
7 ( 5) − 3 ( −3) < 21 44 < 21 (Incorrect ) ANSWER: (5, −3) is not a solution to the inequality 7 x − 3 y < 21.
9.
PROBLEM: y > −5 ; (−3, −1) SOLUTION: y > −5 −1 > −5
ANSWER: (−3, −1) is a solution to the inequality y > −5.
10.
PROBLEM: x ≤ 0 ; (0, 7) SOLUTION: x≤0 0≤0
ANSWER: (0, 7) is a solution to the inequality x ≤ 0.
Part B: Graphing solutions to Linear Inequalities Graph the solution set.
11.
PROBLEM: y < −3x + 3 SOLUTION: y = −3x + 3 Set y = 0 y = −3x + 3
0 = −3x + 3 3x = 3 x =1 x − intercept: (1, 0 ) Set x = 0 y = −3x + 3 y = −3 ( 0 ) + 3 y =3 y − intercept: ( 0, 3) Test Point
y < −3x + 3
(0, 0)
0 < 0+3 0 −5 + 3 1 > −2
We shade the region that contains the test point. ANSWER:
17.
PROBLEM: y > −x + 4 SOLUTION: y = −x + 4 Set y = 0 y = −x + 4 0 = −x + 4 x=4
x − intercept: ( 4, 0 ) Set x = 0
y = −x + 4 y = 0+4 y=4 y − intercept: ( 0, 4 ) Test Point (3 , 3)
y > −x + 4 y > −x + 4
3 > −3 + 4 3 >1
We shade the region that contains the test point. ANSWER:
18.
PROBLEM: y > x−2 SOLUTION: y = x−2 Set y = 0 y = x−2 0 = x−2 x=2
x − intercept : ( 2, 0 ) Set x = 0
y = x−2 0 = x−2 x=2 y − intercept = (0, 2) Test Point
y > x−2
(1, 1)
y > x−2 1 > 1− 2 1 > −1
We shade the region that contains the test point. ANSWER:
19.
PROBLEM: y ≥ −1 SOLUTION: y = −1 Set y = 0 x − intercept : none
Set x = 0 y = −1 y − intercept: (0, − 1) Test Point (0, 0)
y ≥ −1
0 ≥ −1
We shade the region that contains the test point. ANSWER:
20.
PROBLEM: y < −3 SOLUTION: y = −3 Set y = 0 x − intercept : none
Set x = 0 y = −3 y − intercept: (0, − 3) Test Point
y < −3
(–4 , –4)
−4 < −3
We shade the region that contains the test point.
ANSWER:
21.
PROBLEM: x− x+ 2 2 3 5 2 > − ( 2) + 2 2 1 2>− 2
We shade the region that contains the test point. ANSWER:
25.
PROBLEM: −2 x + 3 y > 6 SOLUTION: −2 x + 3 y = 6 Set y = 0 −2 x + 3 y = 6
−2 x + 3 ( 0 ) = 6 −2 6 x= = −3 −2 −2 x − intercept : ( −3, 0 ) Set x = 0 −2 x + 3 y = 6 −2 ( 0 ) + 3 y = 6 3y = 6 3 6 y= 3 3 y=2 y − intercept: ( 0, 2 ) Test Point (–2 , 2)
−2 x + 3 y > 6
−2 x + 3 y > 6 −2 ( −2 ) + 3 ( 2 ) > 6
10 > 6 We shade the region that contains the test point.
ANSWER:
26.
PROBLEM: 7 x − 2 y > 14 SOLUTION: 7 x − 2 y = 14 Set y = 0 7 x − 2 y = 14 7 x − 0 = 14 7 x = 14 7 14 x= 7 7 x=2 x − intercept : ( 2, 0 )
Set x = 0 7 x − 2 y = 14 0 − 2 y = 14 −2 14 y= −2 −2 y = −7 y − intercept: ( 0, −7 )
Test Point (3 , 2)
7 x − 2 y > 14 7 x − 2 y > 14 7 ( 3) − 2 ( 2 ) > 14 21 − 4 > 14 17 > 14
We shade the region that contains the test point. ANSWER:
27.
PROBLEM: 5 x − y < 10 SOLUTION: 5 x − y = 10 Set y = 0 5 x − y = 10 5 x − 0 = 10 5 x = 10 5 10 x= 5 5 x=2 x − intercept : ( 2, 0 )
Set x = 0
5 x − y = 10 0 − y = 10 y = −10 y − intercept: ( 0, −10 ) Test Point (0 , 0)
5 x − y < 10
5 x − y < 10 0 − 0 < 10 0 < 10
We shade the region that contains the test point. ANSWER:
28.
PROBLEM: x− y 0
47.
PROBLEM: Write an inequality that describes all ordered pairs whose y-coordinate is at least 2. SOLUTION: y ≥ 2 describes all ordered pairs whose y-coordinate is at least 2. ANSWER: y ≥ 2
48.
PROBLEM: Write an inequality that describes all ordered pairs whose x-coordinate is at most 5. SOLUTION: x ≤ 5 describes all ordered pairs whose x-coordinate is at most 5. ANSWER: x ≤ 5
3.9 Review Exercises and Sample Exam Chapter 3 Review Exercises
3.1 Rectangular Coordinate System Graph the given set of ordered pairs.
1.
PROBLEM: {( −3, 4 ) , ( −4, 0 ) , ( 0,3) , ( 2, 4 )} SOLUTION: The coordinates x = − 3 and y = 4 , indicate that we should mark a point 3 units left of and 4 units above the origin. The coordinates x = − 4 and y = 0 , indicate that we should mark a point 4 units on the x-axis left from the origin. The coordinates x = 0 and y = 3 , indicate that we should mark a point 3 units on the y-axis above the origin. The coordinates x = 2 and y = 4 , indicate that we should mark a point 2 units right of and 4 units above the origin.
ANSWER:
2.
PROBLEM: {( −5, 5) , ( −3, − 1) , ( 0, 0 ) , ( 3, 2 )} SOLUTION: The coordinates x = − 5 and y = 5 , indicate that we should mark a point 5 units left of and 5 units above the origin. The coordinates x = − 3 and y = −1 , indicate that we should mark a point 3 units left of and 1 unit below the origin. The coordinates x = 0 and y = 0 , indicate that we should mark a point on the origin. The coordinates x = 3 and y = 2 , indicate that we should mark a point 3 units right of and 2 units above the origin. ANSWER:
3.
PROBLEM: Graph the points (−3, 5), (−3, −3) and (3, −3) on a rectangular coordinate plane. Connect the points and calculate the area of the shape. SOLUTION: The coordinates x = − 3 and y = 5 , indicate that we should mark a point 3 units left of and 5 units above the origin. The coordinates x = − 3 and y = −3 , indicate that we should mark a point 3 units left of and 3 units below the origin. The coordinates x = 3 and y = −3 , indicate that we should mark a point 3 units right of and 3 units below the origin. ANSWER:
Base of the triangle is formed by the coordinates, (−3, −3) and (3, −3). Base = b − a = 3 − ( − 3 ) = 6
Height of the triangle is formed by the coordinates, (−3, −3) and (−3, 5). Height = b − a = 5 − ( − 3 ) = 8
1 × Base × Height 2 1 = ×6×8 2 = 24 sq.units
Area =
4.
PROBLEM: Graph the points (−4, 1), (0, 1), (0,−2) and (−4, −2) on a rectangular coordinate plane. Connect the points and calculate the area of the shape. SOLUTION: The coordinates x = − 4 and y = 1 , indicate that we should mark a point 4 units left of and 1 unit above the origin. The coordinates x = 0 and y = 1 , indicate that we should mark a point 1 unit above the origin on the y-axis. The coordinates x = 0 and y = −2 , indicate that we should mark a point two units below on the y-axis. The coordinates x = − 4 and y = −2 , indicate that we should mark a point 4 units left of and 2 units below the origin. ANSWER:
Length is shown by the coordinates (0, 1) and (–4, 1). Length = b − a = − 4 − 0 = 4
Breadth is formed by the coordinates (0, 1) and (0, –2). Breadth = b − a = −2 − 1 = 3
Area = length × breadth =4×3 = 12 sq.units
5.
PROBLEM: Graph the points (1, 0), (4, 0), (1,−5) and (4, −5) on a rectangular coordinate plane. Connect the points and calculate the perimeter of the shape. SOLUTION: The coordinates x = 1 and y = 0 , indicate that we should mark a point 1 unit right on the x-axis. The coordinates x = 4 and y = 0 , indicate that we should mark a point 4 units right on the x-axis. The coordinates x = 1 and y = −5 , indicate that we should mark a point 1 units right of and 5 units below the origin. The coordinates x = 4 and y = −5 , indicate that we should mark a point 4 units right of and 5 units below the origin. ANSWER:
Perimeter = 2(length + breadth) Length is formed by the cordinates (1, 0) and (1, –5). Length = b − a = −5 − 0 = 5
Breadth is formed by the coordinates (1, 0) and (4, 0). Breadth = b − a = 4 − 1 = 3 Perimeter = 2 ( length + breadth ) = 2 ( 5 + 3 ) = 2 ⋅8 = 16 units
6.
PROBLEM: Graph the points (−5, 2), (−5, −3), (1, 2) and (1, −3) on a rectangular coordinate plane. Connect the points and calculate the perimeter of the shape. SOLUTION: The coordinates x = − 5 and y = 2 , indicate that we should mark a point 5 units left of and 2 units above the origin. The coordinates x = − 5 and y = −3 , indicate that we should mark a point 5 units left of and 3 units below the origin. The coordinates x = 1 and y = 2 , indicate that we should mark a point 1 unit right of and 2 units above the origin. The coordinates x = 1 and y = −3 , indicate that we should mark a point 1 units right of and 3 unit below the origin. ANSWER:
Perimeter = 2(length + breadth) Length is formed by the cordinates (1, 2) and (–5, 2). Length = b − a = −5 − 1 = 6
Breadth is formed by the coordinates (1, 2) and (1, –3). Breadth = −3 − 2 = 5 Perimeter = 2 ( length + breadth ) = 2 ( 6 + 5 ) = 2 ⋅ 11 = 22 units
Calculate the distance between the given two points.
7.
PROBLEM: (−1, −2) and (5, 6) SOLUTION: Distance = =
( x2 − x1 )
2
+ ( y 2 − y1 )
2
( 5 − ( − 1) ) + ( 6 − ( − 2 ) ) 2
2
= 36 + 64 = 100 =10 units
ANSWER: 10 units
8.
PROBLEM: (2, −5) and (−2, −2) SOLUTION: Distance = =
( x2 − x1 ) ( −2 − 2 )
2
+ ( y 2 − y1 )
2
+ ( −2 − ( −5 ) )
2
2
= 16 + 9 = 25 =5 units
ANSWER: 5 units
9.
PROBLEM: (−9, −3) and (−8, 4) SOLUTION: Distance = =
( x2 − x1 )
2
+ ( y 2 − y1 )
2
( −8 − ( −9 ) ) + ( 4 − ( −3 ) )
= 1 + 49 = 50 = 25 ⋅ 2 =5 2 units
ANSWER: 5 2 units
2
2
10.
PROBLEM: (−1, 3) and (1, −3) SOLUTION: Distance = =
( x2 − x1 )
2
+ ( y 2 − y1 )
(1 − ( −1) ) + ( −3 − 3 ) 2
2
2
= 4 + 36 = 40 = 10 ⋅ 4 =2 10 units
ANSWER: 2 10 units Calculate the midpoint between the given points.
11.
PROBLEM: (−1, 3) and (5, −7) SOLUTION: ⎛ x + x2 y1 + y 2 Midpoint = ⎜ 1 , 2 ⎝ 2 = ( 2, − 2 )
⎞ ⎛ 5 − 1 −7 + 3 ⎞ , ⎟=⎜ ⎟ 2 ⎠ ⎠ ⎝ 2
ANSWER: ( 2, −2 )
12.
PROBLEM: (6, −3) and (−8, −11) SOLUTION: ⎛ x + x2 y1 + y 2 Midpoint = ⎜ 1 , 2 ⎝ 2 = ( − 1, − 7 )
ANSWER: ( −1, −7 )
⎞ ⎛ 6 − 8 − 3 − 11 ⎞ , ⎟=⎜ ⎟ 2 ⎠ ⎠ ⎝ 2
13.
PROBLEM: (7, −2) and (−6, −1) SOLUTION: ⎛ x + x2 y1 + y 2 Midpoint = ⎜ 1 , 2 ⎝ 2 ⎛1 3⎞ = ⎜ ,− ⎟ ⎝2 2⎠
⎞ ⎛ 7 − 6 −1 − 2 ⎞ , ⎟=⎜ ⎟ 2 ⎠ ⎠ ⎝ 2
⎛1 3⎞ ANSWER: ⎜ , − ⎟ ⎝2 2⎠ 14.
PROBLEM: (−6, 0) and (0, 0) SOLUTION: ⎛ x + x2 y1 + y 2 ⎞ ⎛ − 6 − 0 0 − 0 ⎞ Midpoint = ⎜ 1 , , ⎟=⎜ ⎟ 2 ⎠ ⎝ 2 2 ⎠ ⎝ 2 = ( − 3, 0 )
ANSWER: ( −3, 0 )
15.
PROBLEM: Show that the points (−1, −1), (1, −3), and (2, 0) form an isosceles triangle. SOLUTION: Points: (−1, −1) and (1, −3) a=
=
( x2 − x1 )
2
+ ( y 2 − y1 )
2
(1 − ( −1) ) + ( −3 − ( −1) ) 2
= 4+4 = 2⋅4 = 2 2units
Points: (1, −3) and (2, 0) b=
=
( x2 − x1 ) ( 2 − 1)
2
2
+ ( y 2 − y1 )
+ ( 0 − ( −3 ) )
2
= 1+ 9 = 10 units
Points: (2, 0) and (−1, −1)
2
2
c=
( x2 − x1 )
2
+ ( y 2 − y1 )
=
( −1 − 2 )
2
+ ( −1 − 0 )
2
2
= 9 +1 = 10 units
ANSWER: a = c. Hence, the points form an isosceles triangle.
16.
PROBLEM: Show that the points (2, −1), (6, 1), and (5, 3) form a right triangle. SOLUTION: Points: (2, −1) and (6, 1) a=
=
( x2 − x1 ) (6 − 2)
2
+ ( y 2 − y1 )
2
+ (1 − ( − 1) )
2
2
= 16 + 4 = 20 =
4 ⋅5
= 2 5units
Points: (6, 1) and (5, 3) b=
=
( x2 − x1 ) (5 − 6 )
2
2
+ ( y 2 − y1 )
+ ( 3 − 1)
2
2
= 1+ 4 = 5 units
Points: (5, 3) and (2, −1) c=
=
( x2 − x1 ) (2 − 5)
2
2
+ ( y 2 − y1 )
+ ( −1 − 3 )
2
2
= 9 + 16 = 25 = 5 units
Now to check, a2 + b2 = c2
(2 5 ) + ( 5 ) 2
2
= 52
20 + 5 = 25 25 = 25
ANSWER: Hence, the points form the vertices of a right triangle.
3.2 Graph by Plotting Points Determine whether or not the given point is a solution.
17.
PROBLEM: −5 x + 2 y = 7 ; (1, −1) SOLUTION: −5 x + 2 y = 7
−5 (1) + 2 ( −1) = 7 −5 − 2 = 7 −7 ≠ 7 ANSWER: (1, −1) is not a solution to −5 x + 2 y = 7 .
18.
PROBLEM: 6 x − 5 y = 4 ; ( −1, −2 ) SOLUTION: 6x − 5 y = 4
6 ( −1) − 5 ( −2 ) = 4 4=4 ANSWER: ( −1, −2 ) is a solution to 6 x − 5 y = 4 .
19.
PROBLEM: 3 y = x + 1 ; ( − 23 , 12 ) 4 SOLUTION: 3 y = x +1 4 1 3⎛ 2⎞ = ⎜ − ⎟ +1 2 4⎝ 3⎠ 1 1 = 2 2 ANSWER: ( − 23 , 12 ) is a solution to y =
3 x +1 . 4
20.
PROBLEM: 3 y = − x − 2 ; (10, −8 ) 5 SOLUTION: 3 y = − x−2 5 3 −8 = − (10 ) − 2 5 −8 = −8
3 ANSWER: (10, −8 ) is a solution to y = − x − 2 . 5 Find at least 5 ordered pair solutions and graph.
21.
PROBLEM: y = −x + 2 ANSWER:
x-value
y-value y = − x + 2 = − ( −2 ) + 2 = 4
Solution ( −2, 4)
x = −2 x = −1
y = − x + 2 = − ( −1) + 2 = 3
( −1,3)
y = − x + 2 = −0 + 2 = 2
( 0, 2)
x =1
y = − x + 2 = −1 + 2 = 1
(1,1)
x=2
y = − x + 2 = −2 + 2 = 0
( 2,0)
x=0
22.
PROBLEM: y = 2x − 3 ANSWER:
x-value
y-value y = 2 x − 3 = 2 ( −2 ) − 3 = −7
Solution ( −2, −7 )
x = −2 x = −1
y = 2 x − 3 = 2 ( −1) − 3 = −5
( −1, −5)
y = 2 x − 3 = 0 − 3 = −3
( 0, −3)
x =1
y = 2 x − 3 = 2 (1) − 3 = −1
(1, −1)
x=2
y = 2x − 3 = 2 ( 2) − 3 = 1
( 2,1)
x=0
23.
PROBLEM: 1 y = x−2 2 ANSWER:
x-value
x = −4
y-value 1 1 y = x − 2 = ( −4 ) − 2 = −4 2 2
Solution ( −4, −4)
x = −2
y=
1 1 x − 2 = ( −2 ) − 2 = −3 2 2
x=0
y=
1 1 x − 2 = ( 0 ) − 2 = −2 2 2
( 0, −2)
y=
1 1 x − 2 = ( 2 ) − 2 = −1 2 2
( 2, −1)
y=
1 1 x − 2 = ( 4) − 2 = 0 2 2
( 4,0)
x=2
x=4
( −2, −3)
24.
PROBLEM: 2 y=− x 3 ANSWER:
x-value
x = −3
y-value 2 2 y = − x = − ( −3) = 2 3 3
Solution ( −3, 2)
x=0
2 2 y = − x = − ( 0) = 0 3 3
x=3
2 2 y = − x = − ( 3 ) = −2 3 3
( 3, −2)
x=6
2 2 y = − x = − ( 6 ) = −4 3 3
( 6, −4)
x=9
2 2 y = − x = − ( 9 ) = −6 3 3
( 9, −6)
( 0,0)
25.
PROBLEM: y=3 ANSWER:
x-value
y-value
Solution
x = −2
y=3
( −2,3)
x = −1
y=3
( −1,3)
x=0
y=3
( 0,3)
x =1
y=3
(1,3)
x=2
y=3
( 2,3)
26.
PROBLEM: x = −3 ANSWER:
x-value
y-value
Solution
x = −3
y = −2
( −3, −2)
x = −3
y = −1
( −3, −1)
x = −3
y=0
( −3,0)
x = −3
y =1
( −3,1)
x = −3
y=2
( −3, 2)
27.
PROBLEM: x − 5 y = 15 SOLUTION: x − 5 y = 15 x − x − 5 y = 15 − x 5 y = x − 15 5 1 15 y = x− 5 5 5 1 y = x−3 5 ANSWER:
x-value
x=0
x=5
x = 10
x = 15
x = 20
y-value
Solution
y=
1 1 x − 3 = ( 0 ) − 3 = −3 5 5
( 0, −3)
y=
1 1 x − 3 = ( 5) − 3 = −2 5 5
( 5, −2)
y=
1 1 x − 3 = (10 ) − 3 = −1 5 5
(10, −1)
y=
1 1 x − 3 = (15) − 3 = 0 5 5
(15,0)
y=
1 1 x − 3 = ( 20 ) − 3 = 1 5 5
( 20,1)
28.
PROBLEM: 2 x − 3 y = 12 SOLUTION: 2 x − 3 y = 12 2 x − 2 x − 3 y = 12 − 2 x 3 y = 2 x − 12 3 2 12 y = x− 3 3 3 2 y = x−4 3 ANSWER:
x-value x = −3
x=0
x=3
x=6
x=9
y-value
Solution
y=
2 2 x − 4 = ( −3) − 4 = −6 3 3
( −3, −6)
y=
2 2 x − 4 = ( 0 ) − 4 = −4 3 3
( 0, −4)
y=
2 2 x − 4 = ( 3) − 4 = −2 3 3
( 3, −2)
y=
2 2 x − 4 = ( 6) − 4 = 0 3 3
( 6,0)
y=
2 2 x − 4 = (9) − 4 = 2 3 3
( 9, 2)
3.3 Graph using Intercepts Given the graph find the x‐ and y‐ intercepts
29.
PROBLEM:
ANSWER: The graph intersects the x-axis at –4. Hence the x-intercept is –4. The graph intersects the y-axis at –2. Hence the y-intercept is –2.
30.
PROBLEM:
SOLUTION: The graph intersects the x-axis at 1. Hence the x-intercept is 1. The graph intersects the y-axis at –3. Hence the y-intercept is –3. ANSWER: x-intercept is 1, y-intercept is –3
31.
PROBLEM:
ANSWER: The graph intersects the x-axis at 5. Hence the x-intercept is 5. The graph does not intersect the y-axis. Hence there is no y-intercept.
32.
PROBLEM:
SOLUTION: The graph intersects the x-axis at 0. Hence the x-intercept is 0. The graph intersects the y-axis at 0. Hence the y-intercept is 0. ANSWER: x-intercept is 0, y-intercept is 0. Find the intercepts and graph.
33.
PROBLEM: 3x − 4 y = 12 SOLUTION: Set y = 0 3x − 4 y = 12
3x − 4 ( 0 ) = 12 3x = 12 3 12 x= 3 3 x=4 x − intercept: ( 4, 0 ) Set x = 0 3 x − 4 y = 12 3 ( 0 ) − 4 y = 12 −4 y = 12 −4 12 y= −4 −4 y = −3 y − intercept: ( 0, −3)
ANSWER: ( 4, 0 ) , ( 0, −3)
34.
PROBLEM: 2 x − y = −4 SOLUTION: Set y = 0 2 x − y = −4 2 x − 0 = −4
2 x = −4 2 4 x=− 2 2 x = −2 x − intercept: ( −2, 0 ) Set x = 0 2 x − y = −4 0 − y = −4 y=4 y − intercept: ( 0, 4 )
ANSWER: ( −2, 0 ) , ( 0, 4 )
35.
PROBLEM: 1 1 x − y =1 2 3 SOLUTION: Set y = 0 1 1 x − y =1 2 3 1 x −0 =1 2 1 x =1 2 x=2 x − intercept: ( 2, 0 )
Set x = 0 1 1 x − y =1 2 3 1 1 ( 0) − y = 1 2 3 1 − y =1 3 y = −3 y − intercept: ( 0, −3)
ANSWER: ( 2, 0 ) , ( 0, −3)
36.
PROBLEM: 1 2 − x+ y =2 2 3 SOLUTION: Set y = 0 1 2 − x+ y = 2 2 3 1 − x+0 = 2 2 1 − x=2 2 x = −4 x − intercept: ( −4, 0 )
Set x = 0 1 2 − x+ y =2 2 3 2 0+ y = 2 3 2 y=2 3 3 2 3 ⋅ y = ⋅2 2 3 2 y=3 y − intercept: ( 0,3)
ANSWER: ( −4, 0 ) , ( 0,3)
37.
PROBLEM: 5 y = − x+5 3 SOLUTION: Set y = 0 5 y = − x+5 3 5 0 = − x+5 3 5 x=5 3 3 5 3 ⋅ x = ⋅5 5 3 5 x=3 x − intercept: ( 3, 0 )
Set x = 0 5 y = − x+5 3 y = 0+5
y =5 y − intercept: ( 0,5 )
ANSWER: ( 3, 0 ) , ( 0,5 )
38.
PROBLEM: y = −3 x + 4 SOLUTION: Set y = 0 y = −3 x + 4 0 = −3 x + 4 3x = 4 3 4 x= 3 3 1 x =1 3
⎛ 1 ⎞ x − intercept: ⎜ 1 , 0 ⎟ ⎝ 3 ⎠ Set x = 0 y = −3 x + 4 y = 0+4 y=4 y − intercept: ( 0, 4 )
⎛ 1 ⎞ ANSWER: ⎜1 , 0 ⎟ , ( 0, 4 ) ⎝ 3 ⎠
3.4 Graph using the y‐intercept and Slope Given the graph, determine the slope and y‐intercept.
39.
PROBLEM:
SOLUTION: y = mx + b The points shown in the graph are, (0, 1) and (2, –3). y − y −3 − 1 2 Slope, m = 2 1 = = − = −1 x2 − x1 2 − 0 2 Substitute (0, 1) in the above equation. y = −x + b 1= 0+b 1= b
y − intercept: ( 0,1) ANSWER: −1 , y − intercept: ( 0,1)
40.
PROBLEM:
SOLUTION: y = mx + b The points shown in the graph are, (0, –4) and (2, 2).
y2 − y1 2 − ( −4 ) 6 = = =3 x2 − x1 2−0 2 Substitute (2, 2) in the above equation. y = 3x + b 2 = 6+b −4 = b Slope, m =
y − intercept: ( 0, −4 ) ANSWER: 3 , ( 0, −4 ) Determine the slope given two points.
41.
PROBLEM: (−3, 8) and (5, −6) SOLUTION: y −y −6 − 8 14 7 Slope, m = 2 1 = =− =− x2 − x1 5 − ( −3) 8 4 ANSWER: −
7 4
42.
PROBLEM: (0, −5) and (−6, 3) SOLUTION: y − y 3 − ( −5 ) 8 4 = =− Slope, m = 2 1 = x2 − x1 −6 − 0 −6 3 ANSWER: −
43.
4 3
PROBLEM: (1/2, −2/3) and (1/4, −1/3) SOLUTION:
1 ⎛ 2⎞ 1 − −⎜− ⎟ y2 − y1 4 3 ⎝ 3⎠ Slope, m = = = 3 =− 1 1 1 x2 − x1 3 − − 4 2 4 4 ANSWER: − 3 44.
PROBLEM: (5, −3/4) and (2, −3/4) SOLUTION:
3 ⎛ 3⎞ − −⎜− ⎟ y −y 4 ⎝ 4⎠ 0 Slope, m = 2 1 = = =0 x2 − x1 2−5 −3 ANSWER: 0
Express in slope‐intercept form and identify the slope and y‐intercept.
45.
PROBLEM: 12 x − 4 y = 8 SOLUTION: 12 x − 4 y = 8 12 x − 12 x − 4 y = 8 − 12 x 4 y = 12 x − 8 4 12 8 y = x− 4 4 4 y = 3x − 2 m = 3; b = –2
Slope, m = 3; y − intercept : ( 0, −2 ) ANSWER: Slope, m = 3; y − intercept : ( 0, −2 )
46.
PROBLEM: 3x − 6 y = 24 SOLUTION: 3x − 6 y = 24 3x − 3x − 6 y = 24 − 3x 6 y = 3x − 24 6 3 24 y = x− 6 6 6 1 y = x−4 2 1 m = ; b = –4 2 1 Slope, m = ; y − intercept : ( 0, −4 ) 2 ANSWER: y =
1 1 x − 4 , Slope, m = ; y − intercept : ( 0, −4 ) 2 2
47.
PROBLEM: 1 3 − x + y =1 3 4 SOLUTION: 1 3 − x + y =1 3 4 1 1 3 1 − x + x + y = 1+ x 3 3 4 3 3 1 y = 1+ x 4 3 4 3 4 4 1 ⋅ y = ⋅1 + ⋅ x 3 4 3 3 3 4 4 y = x+ 9 3 4 4 m = ;b = 9 3 4 ⎛ 4⎞ Slope, m = ; y − intercept : ⎜ 0, ⎟ 9 ⎝ 3⎠ 4 ⎛ 4⎞ ANSWER: Slope, m = ; y − intercept : ⎜ 0, ⎟ 9 ⎝ 3⎠
48.
PROBLEM: −5 x + 3 y = 0 SOLUTION: −5 x + 3 y = 0 −5 x + 5 x + 3 y = 0 + 5 x 3 y = 5x 3 5 y= x 3 3 5 y= x 3 5 m = ;b = 0 3 5 Slope, m = ; y − intercept : ( 0, 0 ) 3 ANSWER: y =
5 5 x , Slope, m = ; y − intercept : ( 0, 0 ) 3 3
Graph using the slope and y‐intercept.
49.
PROBLEM: y = −x + 3 SOLUTION: y = −x + 3 m = −1; b = 3 −1 rise Slope, m = ; y − intercept : ( 0,3) = 1 run Starting from the point (0, 3) we will use the slope to mark another point down 1 unit and to the right 1 unit. Join the two points with a straight line. ANSWER:
50.
PROBLEM: y = 4x −1 SOLUTION: y = 4x −1 m = 4; b = −1 4 rise Slope, m = = ; y − intercept : ( 0, −1) 1 run Starting from the point (0, –1) we will use the slope to mark another point up 4 units and to the right 1 unit. Join the two points with a straight line.
ANSWER:
51.
PROBLEM: y = −2 x SOLUTION: y = −2 x m = −2; b = 0 2 Slope, m = − ; y − intercept : ( 0, 0 ) 1 Starting from the point (0, 0) we will use the slope to mark another point down 2 units and to the right 1 unit. Join the two points with a straight line. ANSWER:
52.
PROBLEM: 5 y = − x+3 2 SOLUTION: 5 y = − x+3 2 5 m = − ;b = 3 2 5 rise Slope, m = − = ; y − intercept : ( 0,3) 2 run Starting from the point (0, 3) we will use the slope to mark another point down 5 units and to the right 2 units. Join the two points with a straight line. ANSWER:
53.
PROBLEM: 2x − 3 y = 9 SOLUTION: 2x − 3 y = 9 2x − 2x − 3y = 9 − 2x 3y = 2x − 9 3 2 9 y = x− 3 3 3 2 y = x −3 3 2 m = ; b = −3 3 2 Slope, m = ; y − intercept : ( 0, −3) 3
Starting from the point (0, –3) we will use the slope to mark another point up 2 units and to the right 3 units. Join the two points with a straight line. ANSWER:
54.
PROBLEM: 3 2x + y = 3 2 SOLUTION: 3 2x + y = 3 2 3 2x − 2x + y = 3 − 2x 2 3 y = −2 x + 3 2 2 3 2 2 ⋅ y = − ⋅ 2x + ⋅ 3 3 2 3 3 4 y =− x+2 3 4 m = − ;b = 2 3 4 rise Slope, m = − = ; y − intercept : ( 0, 2 ) 3 run Starting from the point (0, 2) we will use the slope to mark another point down 4 units and to the right 3 units. Join the two points with a straight line.
ANSWER:
55.
PROBLEM: y=0 SOLUTION: y=0 Slope, m = 0; y − intercept : ( 0, 0 )
Starting from the point (0, 0) we will draw an horizontal line with no slope. The xaxis will be the line. ANSWER:
56.
PROBLEM: x − 4y = 0 SOLUTION: x − 4y = 0 x − x − 4 y = −x 4y = x 4 1 y= x 4 4 1 y= x 4 1 m = ;b = 0 4 1 rise Slope, m = = ; y − intercept : ( 0, 0 ) 4 run Starting from the point (0, 0) we will use the slope to mark another point up 1 unit and to the right 4 units. Join the two points with a straight line. ANSWER:
3.5 Finding Linear Equations Given the graph, determine the equation of the line.
57.
PROBLEM:
SOLUTION: The points shown in the graph are (0, 1) and (2, –3). y = mx + b y − y −3 − 1 −4 = = −2 Slope, m = 2 1 = 2 x2 − x1 2 − 0 y = −2 x + b
Substitute (0, 1) in the equation above, y = −2 x + b 1= 0+b 1= b ANSWER: Completing the equation, y = −2 x + 1 .
58.
PROBLEM:
SOLUTION: The points shown in the graph are (2, 2) and (0, –4). y = mx + b y − y −4 − 2 −6 = =3 Slope, m = 2 1 = x2 − x1 0 − 2 −2 y = 3x + b
Substitute (2, 2) in the equation above, y = 3x + b 2 = 3( 2) + b 2 = 6+b −4 = b ANSWER: Completing the equation, y = 3 x − 4 .
59.
PROBLEM:
SOLUTION: The points shown in the graph are (0, –5 ) and (1, –5). y = mx + b y − y −5 − ( −5 ) 0 Slope, m = 2 1 = = =0 x2 − x1 1− 0 1 y = 0+b y=b
Substitute (0, –5 ) in the equation above, y=b −5 = b ANSWER: Completing the equation, y = −5 .
60.
PROBLEM:
SOLUTION: The points shown in the graph are (4, 0) and (4, 1). y = mx + b y − y 1− 0 = undefined Slope, m = 2 1 = x2 − x1 4 − 4 ANSWER: The slope is undefined. Hence, the equation of the line is a vertical line given by x = 4. Find the equation of a line given the slope and a point on the line.
61.
PROBLEM: m = 1/2; (−4, 8) SOLUTION: y = mx + b 1 y = x+b 2 Substitute (–4, 8 ) in the equation above, 1 y = x+b 2 1 8 = ( −4 ) + b 2 10 = b Completing the equation, y = mx + b 1 ANSWER: y = x + 10 2
62.
PROBLEM: m = −1/5; (−5, −9) SOLUTION: y = mx + b 1 y = − x+b 5 Substitute (−5, −9) in the above equation, 1 y = − x+b 5 1 −9 = − ( −5 ) + b 5 −9 − 1 = b −10 = b Completing the equation, y = mx + b 1 ANSWER: y = − x − 10 5
63.
PROBLEM: m = 2/3; (1, −2) SOLUTION: y = mx + b 2 y = x+b 3 Substitute (1, −2) in the equation above, 2 y = x+b 3 2 −2 = + b 3 8 − =b 3 Completing the equation, y = mx + b 2 8 ANSWER: y = x − 3 3
64.
PROBLEM: m = −3/4; (2, −3) SOLUTION: y = mx + b 3 y = − x+b 4 Substitute (2, −3) in the above equation, 3 y = − x+b 4 3 −3 = − ( 2 ) + b 4 3 − =b 2 Completing the equation, y = mx + b 3 3 ANSWER: y = − x − 4 2
Find the equation of the line given two points on the line.
65.
PROBLEM: (−5, −5) and (10, 7) SOLUTION: y = mx + b 7 − ( −5 ) 12 4 y −y = = Slope, m = 2 1 = x2 − x1 10 − ( −5 ) 15 5
4 x+b 5 Substitute (10, 7) in the equation above, 4 y = x+b 5 4 7 = (10 ) + b 5 −1 = b Completing the equation y = mx + b , 4 ANSWER: y = x − 1 5 y=
66.
PROBLEM: (−6, 12) and (3, −3) SOLUTION: y = mx + b y −y 15 5 −3 − 12 Slope, m = 2 1 = =− =− 9 3 x2 − x1 3 − ( −6 )
5 y = − x+b 3 Substitute (3, −3) in the equation above, 5 y = − x+b 3 5 −3 = − ( 3 ) + b 3 2=b Completing the equation y = mx + b , 5 ANSWER: y = − x + 2 3 67.
PROBLEM: (2, −1) and (−2, 2) SOLUTION: y = mx + b y − y 2 − ( −1) 3 =− Slope, m = 2 1 = −2 − 2 x2 − x1 4 3 y = − x+b 4 Substitute(2, −1) in the equation above, 3 y = − x+b 4 3 −1 = − ( 2 ) + b 4 1 =b 2 Completing the equation y = mx + b , 3 1 ANSWER: y = − x + 4 2
68.
PROBLEM: (5/2, −2) and (−5, 5/2) SOLUTION: y = mx + b 5 9 − −2 y2 − y1 2 ( ) 3 = =− 2 =− Slope, m = 5 15 x2 − x1 5 −5 − 2 2 3 y = − x+b 5 Substitute (−5, 5/2) in the equation above, 3 y = − x+b 5 5 3 = − ( −5 ) + b 2 5 1 − =b 2 Completing the equation y = mx + b , 3 1 ANSWER: y = − x − 5 2
69.
PROBLEM: (7, −6) and (3, −6) SOLUTION: y = mx + b y − y −6 − ( −6 ) =0 Slope, m = 2 1 = x2 − x1 3−7 y =b Substitute (3, −6) in the equation above, y=b −6 = b Completing the equation y = mx + b ,
ANSWER: y = −6
70.
PROBLEM: (10, 1) and (10, −3) SOLUTION: y = mx + b −3 − 1 y −y 4 = − = undefined Slope, m = 2 1 = x2 − x1 10 − 10 0 ANSWER: The slope is undefined. Hence the line is a vertical line x =10.
3.6 Parallel and Perpendicular Lines Determine if the lines are parallel, perpendicular or neither.
71.
PROBLEM: ⎧−3x + 7 y = 14 ⎨ ⎩ 6 x − 14 y = 42 SOLUTION: −3x + 7 y = 14 −3x + 3 x + 7 y = 14 + 3x 7 y = 3x + 14
7 3 14 y = x+ 7 7 7 3 y = x+2 7 3 m= 7
6 x − 14 y = 42 6 x − 6 x − 14 y = 42 − 6 x 14 y = 6 x − 42 14 6 42 y = x− 14 14 14 3 y = x −3 7 3 m= 7
ANSWER: Slopes are equal; hence the lines are parallel to each other.
72.
PROBLEM: ⎧2 x + 3 y = 18 ⎨ ⎩ 2 x − 3 y = 36 SOLUTION: 2 x + 3 y = 18 2 x − 2 x + 3 y = 18 − 2 x
3 y = −2 x + 18 3 2 18 y = − x+ 3 3 3 2 y = − x+6 3 2 m=− 3
2 x − 3 y = 36 2 x − 2 x − 3 y = 36 − 2 x 3 y = 2 x − 36 3 2 36 y = x− 3 3 3 2 y = x − 12 3 2 m= 3
ANSWER: Slopes are not equal; hence the lines are not parallel to each other. Slopes are not negative reciprocals; hence the lines are not perpendicular to each other.
73.
PROBLEM: ⎧ x + 4y = 2 ⎨ ⎩8 x − 2 y = −1 SOLUTION: x + 4y = 2 x − x + 4y = 2 − x
8 x − 2 y = −1 8 x − 8 x − 2 y = −1 − 8 x 2 y = 8x + 1 2 8 1 y = x+ 2 2 2 1 y = 4x + 2 m=4
4 y = −x + 2 4 1 2 y =− x+ 4 4 4 1 1 y =− x+ 4 2 1 m=− 4
ANSWER: Slopes are negative reciprocals; hence the lines are perpendicular to each other.
74.
PROBLEM: ⎧y = 2 ⎨ ⎩x = 2 SOLUTION: y=2
x=2
m=0
m = undefined
ANSWER: Slopes are negative reciprocals; hence the lines are perpendicular to each other. Find the equation of the line in slope‐intercept form:
75.
PROBLEM: Parallel to 5 x − y = 15 passing through (−10, −1). SOLUTION: 5 x − y = 15
5 x − 5 x − y = 15 − 5 x y = 5 x − 15 Here the given line has slope m = 5 and thus m& = 5 . We substitute this slope and the given point into point-slope form.
Point (−10, −1)
Slope m& = 5
y − y1 = m ( x − x1 )
y − ( −1) = 5 ( x − ( −10 ) ) y + 1 = 5 x + 50 y = 5 x + 49 ANSWER: y = 5 x + 49
76.
PROBLEM: Parallel to x − 3 y = 1 passing through (2, −2). SOLUTION: x − 3y = 1 x − x − 3y = 1− x
3y = x −1 3 1 1 y = x− 3 3 3 1 1 y = x− 3 3 1 1 and thus m& = . We substitute this slope and 3 3 the given point into point-slope form.
Here the given line has slope m =
Point
Slope 1 (2, −2) m& = 3 y − y1 = m ( x − x1 ) 1 ( x − 2) 3 1 2 y+2= x− 3 3 1 8 y = x− 3 3 y − ( −2 ) =
ANSWER: y =
1 8 x− 3 3
77.
PROBLEM: Perpendicular to 8 x − 6 y = 4 passing through (8, −1). SOLUTION: 8x − 6 y = 4
8x − 8x − 6 y = 4 − 8x 6 y = 8x − 4 6 8 4 y = x− 6 6 6 4 2 y = x− 3 3 4 1 3 and thus m⊥ = − = − . We substitute this 4 3 4 3 slope and the given point into point-slope form.
Here the given line has slope m =
Point
Slope 3 (8, −1) m⊥ = − 4 y − y1 = m ( x − x1 ) y − ( −1) = −
3 ( x − 8) 4
3 y +1 = − x + 6 4 3 y = − x+5 4
3 ANSWER: y = − x + 5 4 78.
PROBLEM: Perpendicular to 7 x + y = 14 passing through (5, 1). SOLUTION: 7 x + y = 14
7 x − 7 x + y = 14 − 7 x y = −7 x + 14 Here the given line has slope m = −7 a and thus m⊥ = and the given point into point-slope form.
1 . We substitute this slope 7
Point
Slope 1 m⊥ = ( 5, 1) 7 y − y1 = m ( x − x1 ) 1 ( x − 5) 7 1 2 y = x+ 7 7 y −1 =
ANSWER: y =
79.
1 2 x+ 7 7
PROBLEM: Parallel to y = 1 passing through (4, −1). SOLUTION: Here the given line has slope m = 0 and thus m& = 0 . We substitute this slope and
the given point into point-slope form. Point (4, −1)
Slope m& = 0
y − y1 = m ( x − x1 )
y − ( −1) = 0 ( x − 4 ) y = −1 ANSWER: y = −1
80.
PROBLEM: Perpendicular to y = 1 passing through (4, −1). SOLUTION: Here the given line has slope m = 0 a and thus m⊥ = undefined . We substitute
this slope and the given point into point-slope form. Point (4, −1)
Slope m⊥ = undefined
y − y1 = m ( x − x1 )
y − ( −1) = undefined ( x − 4 ) x−4 =0 x=4
ANSWER: x = 4 3.7 Introduction to Functions Determine the domain and range and state whether it is a function or not.
81.
PROBLEM: {(−10, −1), (−5, 2), (5, 2)} SOLUTION:
−10 −5
−1 2
5 ANSWER: Domain:{–10, –5, 5} Range: {–1, 2} The relation is a function because each x-value corresponds to exactly one of yvalue.
82.
PROBLEM: {(−12, 4), (−1, −3), (−1, −2)} SOLUTION:
−12 −1
4 –3 –2
ANSWER: Domain:{–12, –1} Range: {4, –3, –2} The relation is not a function because each x-value does not correspond to exactly one of y-value.
83.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is all real numbers R = (–∞,∞). The relation is a function because it passes the vertical line test.
84.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that 2 is the maximum y-value and any x-value greater than that is not represented in the relation. Hence, the domain is equal to 2. Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is R = (–∞,∞) and the range is {2}. The relation is a function because it passes the vertical line test.
85.
PROBLEM:
SOLUTION: The graph indicates that −3 is the minimum x-value and any x-value greater than that is represented in the relation. Hence, the domain is greater than or equal to −3 or the interval notation [−3, ∞) . If we choose any y-value, then we will be able to find a corresponding point on the graph; therefore, the range consists of all real numbers. Any vertical line will not intersect the graph only once; therefore it is not a function. ANSWER: The domain is [−3, ∞) and the range is all real numbers R = (–∞,∞). The relation is not a function because it does not pass the vertical line test.
86.
PROBLEM:
SOLUTION: If we choose any x-value, then we will be able to find a corresponding point on the graph; therefore, the domain consists of all real numbers. The graph indicates that −4 is the minimum y-value and any y-value greater than that is represented in the relation. Hence, the range is greater than or equal to −4 or the interval notation [−4, ∞) .Any vertical line will intersect the graph only once; therefore it is a function. ANSWER: The domain is all real numbers R = (–∞,∞) and the range is [−3, ∞). The relation is a function because it passes the vertical line test. Given the following:
87.
PROBLEM: f ( x) = 9 x − 4 find f (−1) . SOLUTION: f ( x) = 9 x − 4
f (−1) = 9 ( −1) − 4 = −9 − 4 = −13 ANSWER: −13
88.
PROBLEM: f ( x) = −5 x + 1 find f (−3) . SOLUTION: f ( x) = −5 x + 1
f (−3) = −5 ( −3) + 1 = 15 + 1 = 16 ANSWER: 16
89.
PROBLEM: 1 1 g ( x) = x − find g ( − 13 ) . 2 3 SOLUTION: 1 1 g ( x) = x − 2 3 1⎛ 1⎞ 1 1 g ( − 13 ) = ⎜ − ⎟ − = − 2⎝ 3⎠ 3 2 1 ANSWER: − 2
90.
PROBLEM: 3 1 ⎛2⎞ g ( x) = − x + find g ⎜ ⎟ . 4 3 ⎝3⎠ SOLUTION: 3 1 g ( x) = − x + 4 3 3⎛2⎞ 1 1 ⎛2⎞ g⎜ ⎟ = − ⎜ ⎟+ = − 4⎝3⎠ 3 6 ⎝3⎠ ANSWER: −
91.
1 6
PROBLEM: f ( x) = 9 x − 4 find x when f ( x) = 0 . SOLUTION: f ( x) = 9 x − 4 When f ( x) = 0 0 = 9x − 4 9x = 4 9 4 x= 9 9 4 x= 9 ANSWER: x =
4 9
92.
PROBLEM: f ( x) = −5 x + 1 find x when f ( x) = 2 . SOLUTION: f ( x) = −5 x + 1 When f ( x) = 2 f ( x) = −5 x + 1 2 = −5 x + 1 5x = 1 − 2 5 x = −1 5 1 x=− 5 5 1 x=− 5 ANSWER: x = −
93.
1 5
PROBLEM: 1 1 g ( x) = x − find x when g ( x) = 1 . 2 3 SOLUTION: 1 1 g ( x) = x − 2 3 When g ( x) = 1 1 1 1= x − 2 3 1 4 x= 2 3 8 x= 3 ANSWER: x =
8 3
94.
PROBLEM: 3 1 g ( x) = − x + find x when g ( x) = −1 . 4 3 SOLUTION: 3 1 g ( x) = − x + 4 3 When g ( x) = −1 3 1 −1 = − x + 4 3 3 4 x= 4 3 4 3 4 4 ⋅ x= ⋅ 3 4 3 3 16 x= 9 16 ANSWER: x = 9
Given the graph of a function f ( x) determine the following:
95.
PROBLEM: f (3) SOLUTION: From the graph when x = 3, y = – 2. Hence, f (3) is – 2. ANSWER: – 2
96.
PROBLEM: x when f ( x) = 4 ANSWER: If f ( x) = 4 , y = 4. From the graph when y = 4 , x = 0.
3.8 Linear Inequalities (two variables) Is the ordered pair a solution to the given inequality?
97.
PROBLEM: 6 x − 2 y ≤ 1 ; (−3, −7) SOLUTION: 6x − 2 y ≤ 1
6 ( −3) − 2 ( −7 ) ≤ 1 −18 + 14 ≤ 1 −4 ≤ 1 ANSWER: (−3, −7) is a solution to the inequality 6 x − 2 y ≤ 1 .
98.
PROBLEM: −3x + y > 2 ; (0, 2) SOLUTION: −3x + y > 2 0+2 > 2 2 > 2 Incorrect ANSWER: (0, 2) is not a solution to the inequality −3x + y > 2 .
99.
PROBLEM: 6 x − 10 y < −1 ; (5, −3) SOLUTION: 6 x − 10 y < −1
6 ( 5 ) − 10 ( −3) < −1 30 + 30 < −1 60 < −1
Incorrect
ANSWER: (5, −3) is not a solution to the inequality 6 x − 10 y < −1 .
100.
PROBLEM: 1 x − y > 0 ; (1, 4) 3 SOLUTION: 1 x− y >0 3 1 1 − ( 4) > 0 3 1 − > 0 Incorrect 3
1 ANSWER: (1, 4) is not a solution to the inequality x − y > 0 . 3 101.
PROBLEM: y > 0 ; (−3, −1) SOLUTION: y>0 −1 > 0 Incorrect ANSWER: (−3, −1) is not a solution to the inequality 6 x − 10 y < −1 .
102.
PROBLEM: x ≤ −5 ; (−6, 4) SOLUTION: x ≤ −5 −6 ≤ −5 ANSWER: (−6, 4) is as solution to the inequality x ≤ −5 .
Graph the solution set.
103.
PROBLEM: y ≥ −2 x + 1 SOLUTION: y = −2 x + 1 Set y = 0 y = −2 x + 1 0 = −2 x + 1
2x = 1 1 x= 2 ⎛1 ⎞ x − intercept : ⎜ , 0 ⎟ ⎝2 ⎠ Set x = 0 y = −2 x + 1 y = 0 +1 y =1 y − intercept = (0, 1)
Test Point
y ≥ −2 x + 1
(1, 1)
y ≥ −2 x + 1 1 ≥ −2 + 1 1 ≥ −1
We shade the region that contains the test point.
ANSWER:
104.
PROBLEM: y < 3x − 4 SOLUTION: y = 3x − 4 Set y = 0 y = 3x − 4 0 = 3x − 4
3x = 4 4 x= 3 ⎛ 1 ⎞ x − intercept : ⎜1 , 0 ⎟ ⎝ 3 ⎠ Set x = 0 y = 3x − 4 y = 0−4 y = −4 y − intercept : (0, − 4)
Test Point
y < 3x − 4
(2, –1)
y < 3x − 4 −1 < 6 − 4 −1 < 2
We shade the region that contains the test point. ANSWER:
105.
PROBLEM: −x + y ≤ 3 SOLUTION: −x + y = 3 Set y = 0 −x + y = 3
−x + 0 = 3 x = −3 x − intercept : ( −3, 0 ) Set x = 0 −x + y = 3 0+ y = 3 y =3 y − intercept = (0, 3)
Test Point
−x + y ≤ 3
(1, 1)
−x + y ≤ 3 −1 + 1 ≤ 3 0≤3
We shade the region that contains the test point.
ANSWER:
106.
PROBLEM: 5 1 x+ y ≤2 2 2 SOLUTION: 5 1 x+ y =2 2 2 Set y = 0 5 1 x+ y =2 2 2 5 x+0 = 2 2 4 x= 5
⎛4 ⎞ x − intercept : ⎜ , 0 ⎟ ⎝5 ⎠ Set x = 0 5 1 x+ y =2 2 2 1 0+ y = 2 2 y=4 y − intercept: (0, 4)
Test Point
5 1 x+ y ≤2 2 2
(–2, –2)
5 1 x+ y ≤2 2 2 5 1 ( −2 ) + ( −2 ) ≤ 2 2 2 −6 ≤ 2
We shade the region that contains the test point. ANSWER:
107.
PROBLEM: 3x − 5 y > 0 SOLUTION: 3x − 5 y = 0 Set y = 0 3x − 5 y = 0
3x − 0 = 0 x=0 x − intercept : ( 0, 0 ) Set x = 0 3x − 5 y = 0 0 − 5y = 0 y=0 y − intercept = (0, 0)
Test Point
3x − 5 y > 0
(1, –1 )
3x − 5 y > 0 3+5 > 0 8>0
We shade the region that contains the test point. ANSWER:
108.
PROBLEM: y>0 SOLUTION: y=0 x − intercept : none
y − intercept : (0, 0) Test Point
y>0
(1, –1)
y>0 1> 0
We shade the region that contains the test point.
ANSWER:
Chapter 3 Sample Exam
1.
PROBLEM: Graph the points (−4, −2), (−4, 1) and (0, −2) on a rectangular coordinate plane. Connect the points and calculate the area of the shape. SOLUTION: The coordinates x = − 4 and y = 2 , indicate that we should mark a point 4 units left of and 2 units above the origin. The coordinates x = − 4 and y = 1 , indicate that we should mark a point 4 units left of and 1 unit below the origin. The coordinates x = 0 and y = −2 , indicate that we should mark a point 2 units below the origin on the y-axis. ANSWER:
The shape formed is a , Triangle
1 × base × height 2 Base is formed by the coordinates (0, − 2) and (4, − 2). Area =
Base = 4 − 0 = 4 Height is formed by the coordinates ( − 4,1) and ( − 4, − 2) Height = 1 − ( −2 ) = 3 Area =
2.
1 × 4 × 3 = 6 sq.units 2
PROBLEM: Is (−2, 4) a solution to 3x − 4 y = −10 ? Justify your answer. SOLUTION: 3x − 4 y = −10
3 ( −2 ) − 4 ( 4 ) = −10 −6 − 16 = −10 −22 ≠ −10 ANSWER: (−2, 4) is not a solution to 3x − 4 y = −10 . Given the set of x‐values {−2, −1, 0, 1, 2} find the corresponding y‐values and graph the following:
3.
PROBLEM: y = x −1 ANSWER:
x-value
y-value y = x − 1 = −2 − 1 = −3
Solution ( −2, −3)
x = −2 x = −1
y = x − 1 = −1 − 1 = −2
( −1, −2)
y = x − 1 = 0 − 1 = −1
( 0, −1)
x =1
y = x −1 = 1 −1 = 0
(1,0)
x=2
y = x −1 = 2 −1 = 1
( 2,1)
x=0
4.
PROBLEM: y = −x +1 ANSWER:
x-value
y-value y = − x + 1 = − ( −2 ) + 1 = 3
Solution ( −2,3)
x = −2 x = −1
y = − x + 1 = − ( −1) + 1 = 2
( −1, 2)
y = −x +1 = 0 +1 = 1
( 0,1)
x =1
y = − x + 1 = −1 + 1 = 0
(1,0)
x=2
y = − x + 1 = −2 + 1 = −1
( 2, −1)
x=0
5.
PROBLEM: On the same set of axes graph y = 4 and x = −3 . Give the point where they intersect. SOLUTION: x-value
y-value y=4
Solution ( −2, 4)
x = −2 x = −1
y=4
( −1, 4)
y=4
( 0, 4)
x =1
y=4
(1, 4)
x=2
y=4
( 2, 4)
x=0
x-value
y-value y = −2
Solution ( −2, −3)
x = −3 x = −3
y = −1
( −1, −3)
y=0
( 0, −3)
x = −3
y =1
(1, −3)
x = −3
y=2
( 2, −3)
x = −3
ANSWER: The lines intersect at ( −3, 4 ) . Find the x‐ and y‐intercepts and use those points to graph the following:
6.
PROBLEM: 2x − y = 8 SOLUTION: Set y = 0 2x − y = 8
2x − 0 = 8 2x = 8 x=4 x − intercept : ( 4, 0 ) Set x = 0 2x − y = 8 2 ( 0) − y = 8 y = −8 y − intercept : ( 0, −8)
ANSWER: ( 4, 0 ) , ( 0, −8 )
7.
PROBLEM: 12 x + 5 y = 15 SOLUTION: Set y = 0 12 x + 5 y = 15 12 x = 15 15 5 1 x = = =1 12 4 4 ⎛ 1 ⎞ x − intercept : ⎜1 , 0 ⎟ ⎝ 4 ⎠ Set x = 0 12 x + 5 y = 15 0 + 5 y = 15 5 y = 15 y =3
y − intercept : ( 0,3)
⎛ 1 ⎞ ANSWER: ⎜1 , 0 ⎟ , ( 0,3) ⎝ 4 ⎠
8.
PROBLEM: Calculate the slope of the line passing through (−4, −5) and (−3, 1). SOLUTION: 1 − ( −5 ) y −y 6 = =6 Slope,m = 2 1 = x2 − x1 −3 − ( −4 ) 1 ANSWER: 6
Determine the slope and y‐intercept. Use them to graph the following:
9.
PROBLEM: 3 y = − x+6 2 SOLUTION: 3 y = − x+6 2 3 m = − ;b = 6 2 3 rise ; y − intercept : ( 0, 6 ) m=− = 2 run Starting from the point (0, 6) we will use the slope to mark another point down 3 units and to the right 2 units. Join the two points with a straight line. ANSWER: m = −
3 rise = ; y − intercept : ( 0, 6 ) 2 run
10.
PROBLEM: 5x − 2 y = 6 SOLUTION: 5x − 2 y = 6
5x − 5x − 2 y = 6 − 5x 2 y = 5x − 6 2 5 6 y = x− 2 2 2 5 y = x −3 2 5 m = ; b = −3 2 5 rise ; y − intercept : ( 0, −3) m= = 2 run Starting from the point (0, –3) we will use the slope to mark another point up 5 units and to the right 2 units. Join the two points with a straight line.
ANSWER: m =
11.
5 rise = ; y − intercept : ( 0, −3) 2 run
PROBLEM: Given m = −3 , determine m⊥ . SOLUTION: m = −3 1 1 m⊥ = = − m 3 ANSWER: −
12.
1 3
PROBLEM:
Are the given lines parallel, perpendicular, or neither?
SOLUTION: −2 x + 3 y = −12 −2 x + 2 x + 3 y = −12 + 2 x
3 y = 2 x − 12 2 y = x−4 3 2 m= 3
⎧−2 x + 3 y = −12 ⎨ ⎩ 4 x − 6 y = 30
4 x − 6 y = 30 4 x − 4 x − 6 y = 30 − 4 x 6 y = 4 x − 30 2 y = x −5 3 2 m= 3
ANSWER: Slopes are equal; hence the lines are parallel to each other.
13.
PROBLEM: Determine the slope of the given lines: a. y = −2 b. x = 13 c. Are these lines parallel, perpendicular, or neither? ANSWER: a. y = −2 m=0
b. x = 13 m = undefined c. The slopes are negative reciprocals. Hence, the lines are perpendicular to each other. 14.
PROBLEM: Given the graph, determine slope-intercept form of the line:
SOLUTION: y = mx + b The points shown in the graph are, (–4, 0) and (0, 6). y −y 6−0 6 3 = = Slope,m = 2 1 = x2 − x1 0 − ( −4 ) 4 2
3 x+b 2 Substitute (0, 6) in the above equation, 3 6 = (0) + b 2 6=b Completing the equation y = mx + b , y=
ANSWER: y =
3 x+6 2
15.
PROBLEM: Determine the equation of the line with slope m = − 34 passing through (8, 1). SOLUTION: y = mx + b m = − 34 ; Point : ( 8,1)
3 y = − x+b 4 Substitute ( 8,1) in the above equation, 1= −
3 (8) + b 4
7=b Completing the equation, y = mx + b 3 ANSWER: y = − x + 7 4 16.
PROBLEM: Find the equation to the line passing through (−2, 3) and (4, 1). SOLUTION: y = mx + b Points are (−2, 3) and (4, 1) y −y −2 1− 3 1 = =− Slope,m = 2 1 = x2 − x1 4 − ( −2 ) 6 3
1 y = − x+b 3 Substitute (4, 1) in the above equation, 1 y = − x+b 3 1 1 = − ( 4) + b 3 7 =b 3 Completing the equation y = mx + b , 1 7 ANSWER: y = − x + 3 3
17.
PROBLEM: Find the equation of the line parallel to 5 x − y = 6 passing through (−1, −2). SOLUTION: 5x − y = 6
5x − 5x − y = 6 − 5x y = 5x − 6 Here the given line has slope m = 5 and thus m& = 5 . We substitute this slope and the given point into point-slope form. Point (−1, −2)
Slope m& = 5
y − y1 = m ( x − x1 )
y − ( −2 ) = 5 ( x − ( −1) ) y + 2 = 5x + 5 y = 5x + 3 ANSWER: y = 5 x + 3
18.
PROBLEM: Find the equation of the line perpendicular to − x + 2 y = 4 passing through (1/2, 5). SOLUTION: −x + 2 y = 4 −x + x + 2 y = 4 + x
2y = x + 4 1 y = x+2 2 1 1 and thus m⊥ = − = −2 . We substitute this 1 2 2 slope and the given point into point-slope form.
Here the given line has slope m =
Point ⎛1 ⎞ ⎜ ,5 ⎟ ⎝2 ⎠
Slope m⊥ = −2
y − y1 = m ( x − x1 ) 1⎞ ⎛ y − 5 = −2 ⎜ x − ⎟ 2⎠ ⎝ y − 5 = −2 x + 1 y = −2 x + 6 ANSWER: y = −2 x + 6
4 Given a linear function f ( x) = − x + 2 determine the following: 5 19.
20.
PROBLEM: f (10) SOLUTION: 4 f ( x) = − x + 2 5 4 f (10 ) = − (10 ) + 2 = −6 5 ANSWER: −6 PROBLEM: x where f(x) = 0 SOLUTION: 4 f ( x) = − x + 2 5 Where f(x) = 0 4 0=− x+2 5 4 x=2 5 5 x= 2 ANSWER: x =
5 2
21.
PROBLEM: Graph the solution set: 3x − 4 y > 4 SOLUTION: 3x − 4 y = 4 Set y = 0 3x − 0 = 4
3x = 4 4 x= 3 ⎛ 1 ⎞ x − intercept : ⎜1 , 0 ⎟ ⎝ 3 ⎠ Set x = 0 3x − 4 y = 4 0 − 4y = 4 y = −1 y − intercept : ( 0, −1) Test Point (1, –1)
3x − 4 y > 4 3+ 4 > 4 7>4
We shade the region that contains the test point. ANSWER:
22.
PROBLEM: Graph the solution set: y − 2 x ≥ 0 SOLUTION: y − 2x = 0 Set y = 0 y − 2x = 0
0 − 2x = 0 x=0 x − intercept : ( 0, 0 ) Set x = 0 y − 2x = 0 y−0 = 0 y=0 y − intercept : ( 0, 0 ) Test Point
y − 2x ≥ 0
(1, –1)
1+ 2 ≥ 0 3≥0
We shade the region that contains the test point. ANSWER:
23.
PROBLEM: A rental car company charges $32.00 to rent a car plus $0.52 per mile driven. Write an equation that gives the cost of renting the car in terms of the number of miles driven. Use the formula to determine the cost of renting the car and driving it 46 miles. SOLUTION: Let the number of miles driven by the car be represented by x. Total cost = $32 + $0.52x f(x) = $32 + $0.52x f(46) = 32 + 0.52(46) = $55.92 ANSWER: Cost of renting the car and driving it 46 miles is $55.92.
24.
PROBLEM: A car was purchased new for $12,000 and was sold 5 years later for $7,000. Write a linear equation that gives the value of the car in terms of its age in years. SOLUTION:
Given: (0,12000) and (5,7000) 7000 − 12000 5000 Depreciation= =− 5−0 5 = −1000 Total value = Original value – Depreciating amount × Number of years ANSWER: Total value = $12,000 – $1,000x
25.
PROBLEM: The area of a rectangle is 72 square meters. If the width measures 4 meters, then determine the length of the rectangle. SOLUTION: Length of the rectangle is represented by x. Area = Length × Width 72 sq. meters = x × 4 meters x = 18 meters ANSWER: Length of the rectangle is 18 meters.
