11.1 Straight Lines Problem Set

STRAIGHT LINES | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: If the y-intercept of a line is 3 and the x-intercept is 2, the equation of the line in slopeintercept form is most close to: %

A. 𝑦 = − 𝑥 + 2 & %

B. 𝑦 = − 𝑥 + 3 & &

C. 𝑦 = − 𝑥 + 3 %

%

D. 𝑦 = 𝑥 + 3 &



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PROBLEM 2: The y-intercept of the line, given in point-slope form below, is most close to: 𝑦 + 1 = 3(𝑥 − 2) A. −7 B. −2 C. −6 D. 7



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PROBLEM 3: The line that is parallel to the line 𝑦 = 2𝑥 − 1 and passes through the point (−5, 1) is represented most accurately by: A. 𝑦 = 2𝑥 − 7 1

B. 𝑦 = − 𝑥 + 11 &

C. 𝑦 = 2𝑥 + 11 D. 𝑦 = −2𝑥 − 9



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PROBLEM 4: 1

The line that runs parallel to a line with the equation 𝑦 = 𝑥 + 6 is represented most 3

accurately as: 1

A. 𝑦 = − 𝑥 − 6 3

B. 𝑦 = −4𝑥 + 6 C. 𝑦 = 2𝑥 − 3 D. 𝑥 = 4𝑦 − 3



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PROBLEM 5: An undefined line goes through the point (4, −6) and runs perpendicular to the line 𝑦 = 4𝑥 + 10. The equation that best represents this line is: A. 𝑦 = 𝑚𝑥 − 20 1

B. 𝑦 = − 𝑥 − 5 3

1

C. 𝑦 = 𝑥 + 5 7 1

D. 𝑦 = 𝑥 + 5 3



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PROBLEM 6: The equation that represents a line passing through points (−2, −3) and (2,1) is most close to: A. 𝑦 = 𝑥 − 1 7

1%

%

%

B. 𝑦 = − 𝑥 + C. 𝑦 = 2𝑥 − 3 D. 𝑥 = 4𝑦 − 3



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PROBLEM 7: The equation that best represents a line that passes through point (2,1) and runs perpendicular to 3𝑥 − 5𝑦 = 10 is most close to: A. 𝑦 = 𝑥 − 1 7

1%

%

%

B. 𝑦 = − 𝑥 + C. 𝑦 = 2𝑥 − 3 D. 𝑥 = 4𝑦 − 3



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PROBLEM 8: The general equation y = a1 + b& x is an algebraic expression for which of the following: A. A Taylor Series B. Circular Motion C. Conic Section D. A Straight Line



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PROBLEM 9: The slope of the line defined by 𝑦 − 𝑥 = 5 is most close to: A. m = 5 + x B. m = − C. m =

1 &

1 3

D. m = 1



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PROBLEM 10: The equation of a line having a slope of 2 and a y-intercept of -3 is most close to: A. 𝑦 = −3𝑥 + 2 B. 𝑦 = 2𝑥 − 3 &

C. 𝑦 = 𝑥 + 1 %

D. 𝑦 = 2𝑥 + 3



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PROBLEM 11: The equation of the line that passes through the points (0,0) and 2, −2 is best represented by: A. 𝑦 = 𝑥 B. 𝑦 = −2𝑥 + 2 C. 𝑦 = −2𝑥 D. 𝑦 = −𝑥



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STRAIGHT LINES | SOLUTIONS SOLUTION 1: The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Recall that to define a line we need to specify two distinct pieces of information concerning that line. Although not initially obvious, in this case, we have two points, which is sufficient in defining the line. The problem statement defines at which point the line “intercepts” both the x and y axis. This information is sufficient in allowing us to define two distinct points, which are (0, 3) and (2, 0). Using these points, we can define the slope, which is:

𝑚=

0−3 3 =− 2−0 2

At this point, we typically would plug in a point, along with the slope, and solve for the 𝑦-intercept. However, we are already given the 𝑦-intercept as 3.

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Therefore, we can move straight to defining the line in slope-intercept form, which is: 3 𝑦 =− 𝑥+3 2 𝟑

Therefore, the correct answer choice is B. 𝒚 = − 𝒙 + 𝟑 𝟐

SOLUTION 2: The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We are to determine the y-intercept of the line that is given to us in point-slope form, which is: 𝑦 + 1 = 3(𝑥 − 2) To determine the 𝑦-intercept, we need to work this formula in to a more familiar form, that being the general slope-intercept form, which is: 𝑦 = 𝑚𝑥 + 𝑏



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Where: • 𝑚 is the slope • 𝑏 is the y-intercept •

𝑥, 𝑦 are coordinates

We can do this. First, let’s distribute the 3 through on the right side of the equation, giving us: 𝑦 + 1 = 3𝑥 − 6 We are getting closer. Looking at the general slope-intercept form, we need to isolate y on the left side, so our next move will be to subtract 1 from each side, giving us: 𝑦 = 3𝑥 − 7 We now have our formula in slope-intercept form, in line with the structure: 𝑦 = 𝑚𝑥 + 𝑏 Where b is the y-intercept and the variable in which we are looking to define.



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Taking a quick look at our formula, we can see that 𝑏 = −7 which means that the yintercept of this line occurs at (0, −7) Alternatively, without going through the work that we did, we could have substituted the value of 0 for the “𝑥” variable to obtain a result for the formula, which would be the 𝑦-intercept. Therefore, the correct answer choice is A. −𝟕

SOLUTION 3: The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To start, recall that the standard or slope-intercept form for an equation is represented as: 𝑦 = 𝑚𝑥 + 𝑏 With 𝑥, 𝑦 being the coordinates, “𝑚” being the slope, and “𝑏” being the 𝑦-intercept. The FORMULA for the SLOPE OF A PERPENDICULAR LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.



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Two line are considered PARALLEL if they have the same slopes such that: 𝑚1 = 𝑚& Parallel lines have the same slope, so if the slope of the line that is defined within our problem statement is equal to 2, then the slope of the parallel line must be the same, which gives us: 𝑚=2 The problem statement, also establishes that the line passes through the point coordinates −5,1 . Taking the given coordinates, along with the determined slope, the slope-intercept form of our parallel line can be written as: (1) = (2)(−5) + 𝑏 The only unknown is our y-intercept, b, so let’s rearrange the formula to isolate this variable and solve, such that: 𝑏 = 1 − 2 −5 = 11



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So the equation of the line that runs parallel to the line 𝑦 = 2𝑥 − 1 and passes through the point (−5,1) is: 𝑦 = 2𝑥 + 11 Therefore, the correct answer choice is C. 𝒚 = 𝟐𝒙 + 𝟏𝟏

SOLUTION 4: The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For two lines to be parallel, they must have share the same slope, so let’s rearrange each option we are given so that we can visualize each slope. Rewriting each equation in the standard form, we tabulate the following: Equation

Standard Form

1 y=− x−6 4

Same

−

y = −4x + 6

Same

−4

y = 2x − 3

Same

2

x = 4y − 3



y=

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Slope 1 4

1 4

Expanding further, we can now identify and define the slope of each line, such that: Equation

Standard Form

1 y=− x−6 4

Same

−

y = −4x + 6

Same

−4

y = 2x − 3

Same

2

x = 4y − 3

y=

1 3 x+ 4 4

Slope 1 4

1 4 1

We are asked to determine which line runs parallel to a line defined by 𝑦 = 𝑥 + 6. 3

1

The slope of this line is which tells us that the line that will run parallel with it will 3

have this same slope. Reviewing our table of equations, we find that the there is one equation that meets this criteria, that being x = 4y − 3. Therefore, the correct answer choice is D. 𝒙 = 𝟒𝒚 − 𝟑



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SOLUTION 5: The FORMULA for the SLOPE OF A PERPENDICULAR LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Two lines are considered PERPENDICULAR when the slope of one line is inversely proportional to the slope of the second line, or in formulaic terms:

𝑚1 = −

1 𝑚&

The slope of the line given in our problem statement has a slope of 4, so: 𝑚1 = 4 We know that in order for a line to be perpendicular to this, then:

𝑚1 = −

1 𝑚&

Or plugging in our known variables:

𝑚& = −



1 1 =− 𝑚1 4

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NOTE: We could stop here, after scanning the results and seeing that only one of the 1

equations presented to us has a slope equal to − which tells us that this would be our 3

answer. However, for the purposes of repetition, let’s move forward with fully defining this equation. The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The equation of a line given in slope-intercept form (standard form) can be written as: 𝑦 = 𝑚𝑥 + 𝑏 Where in our current scenario, we know that:

𝑚=−

1 4

𝑥, 𝑦 = (4, −6) Therefore, plugging this data in to our general formula, we find that:

−6 = −



1 4

4 +𝑏

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The y-intercept, “𝑏”, is our only unknown. Let’s rearrange to isolate and solve for this variable. Doing so we get:

𝑏 = −6— −

1 4

4 = −5

Therefore, with all of our information now defined, the line that goes through the point (4, −6) and runs perpendicular to the line 𝑦 = 4𝑥 + 10 is best represented by: 1 𝑦 =− 𝑥−5 4 𝟏

Therefore, the correct answer choice is B. 𝒚 = − 𝒙 − 𝟓 𝟒

SOLUTION 6: The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Recall that to define a line we need to specify two distinct pieces of information concerning that particular line. In this case, we are given two points, which is sufficient and allow us to define the line that passes through them.



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The first step is to calculate the slope given the general formula:

𝑚=

𝑟𝑖𝑠𝑒 (𝑦& − 𝑦1 ) = 𝑟𝑢𝑛 (𝑥& − 𝑥1 )

Where: • (x1 , y1 ) are the x and y coordinates of point 1 •

x& , y& are the x and y coordinates of point 2

Knowing that our two points are (−2, −3) and (2,1), we can plug them in such that:

𝑚=

(1 − (−3)) 4 = =1 (2 − −2 ) 4

With the slope now defined, we will move next to defining the intercept. We will do this by taking one of the points and plugging it in to the general slopeintercept form (standard form) equation, which is given as: 𝑦 = 𝑚𝑥 + 𝑏 Where: • 𝑚 is the slope • 𝑏 is the y-intercept



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Rearranging this formula to isolate the y-intercept, we get: 𝑏 = 𝑦1 − 𝑚𝑥1

Taking either one of the points defined, it’s now just a simple plug and play, such that: 𝑏 = −3 − 1 −2 = −1 We have now defined “𝑚” and “𝑏”, the components that will allow us to define precisely the line given the scenario. Plugging these values in to the slope-intercept form for a line, we get: 𝑦 =𝑥−1

Therefore, the correct answer choice is A. 𝒚 = 𝒙 − 𝟏



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SOLUTION 7: The TOPIC of STRAIGHT LINES can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Two lines are considered PERPENDICULAR if their slopes are inversely proportional such that:

𝑚1 = −

1 𝑚&

The FORMULA for the SLOPE OF A PERPENDICULAR LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step will be to take the equation we are given and determine what the slope must be of the associated perpendicular line. We are given the equation: 3𝑥 − 5𝑦 = 10 In this form, it’s not immediately apparent what the slope of the line is, therefore, we must do some rearranging.

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The FORMULA for the SLOPE-INTERCEPT FORM of a STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This general form is given as: 𝑦 = 𝑚𝑥 + 𝑏 Gathering our variables and moving them around, we can rewrite the given equation in slope-intercept form as:

𝑦=

3 𝑥−2 5

This results in slope of:

𝑚=

3 5

Knowing that in order for a line to be PERPENDICULAR, then it’s slope must be inversely proportional, such that:

𝑚1 = −



1 𝑚&

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Therefor, the perpendicular line will have a slope of:

𝑚1 = −

5 3

We can now use the point-slope equation to fully define the equation of the line using the provided point and calculated slope, the general formula states that: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) The FORMULA for the POINT-SLOPE FORM of a STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. With all the proper variables defined, we can quickly plug them in and get: 5 𝑦 − 1 = − (𝑥 − 2) 3 This form isn’t what we are looking for, we need to simplify a bit in to the standard slope-intercept form. Doing so we get:

𝑦=

5 13 𝑥+ 3 3 𝟓

𝟏𝟑

𝟑

𝟑

Therefore, the correct answer choice is B. 𝒚 = − 𝒙 +



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SOLUTION 8: The GENERAL FORMULA for the EQUATION A STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The formula of a straight line is generally given as: 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 An equation representing a line can be expressed in a number of different ways. One of which is the slope-intercept form, or also known as the STANDARD FORM, stated as: 𝑦 = 𝑚𝑥 + 𝑏

This problem provides us a formula in the form: 𝑦 = 𝑎1 + 𝑏& 𝑥 This is equivalent to that of the slope-intercept form of a straight line. Therefore, the correct answer choice is 𝐃. 𝐀 𝐒𝐭𝐫𝐚𝐢𝐠𝐡𝐭 𝐋𝐢𝐧𝐞



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SOLUTION 9: The FORMULA highlighting the SLOPE-INTERCEPT FORM of a STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. There are a number of ways in which a line can be expressed in formulaic terms, of which, the slope-intercept form is just one. The slope-intercept form for a straight line is written is terms of: 𝑦 = 𝑚𝑥 + 𝑏 Where 𝑚 is the slope of the line and 𝑏 is the y-intercept. In our problem statement, we are given the equation: 𝑦−𝑥 =5 Staying in line with the slope-intercept form of a straight line, we know that our variable y must be isolated on the left side of the equation with all the other variables falling on the right.



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We can simply rearrange this formula to isolate y by adding x to each side of the equation, such that: 𝑦 =𝑥+5 This is now an equation in slope-intercept form. We know that the slope of this line will be the coefficient that sits in front of our x variable, which in our case, is 1. Therefore, the correct answer choice is D. 𝐦 = 𝟏

SOLUTION 10: The FORMULA for an EQUATION OF A STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

There are a number of ways in which a line can be expressed in formulaic terms. The route we take in defining a line depends heavily on the information that we are given in the problem statement.



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In this case we are given: • 𝑆𝑙𝑜𝑝𝑒: 2 • 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡: −3 With the slope and the y-intercept defined for us, the quickest route to defining this line will be to use the slope-intercept general form. The slope-intercept form of the equation of a straight line is given as: 𝑦 = 𝑚𝑥 + 𝑏 Where 𝑚 is the slope and 𝑏 is the y-intercept. The FORMULA for the SLOPE-INTERCEPT FORM of a STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. With both the slope and the y-intercept defined in the problem statement, this becomes a plug and play scenario, giving us the slope-intercept form as: 𝑦 = 2𝑥 − 3 Therefore, the correct answer choice is B. 𝐲 = 𝟐𝐱 − 𝟑



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SOLUTION 11: The VARIOUS FORMULAS driving an EQUATION of a STRAIGHT LINE can be referenced under the SUBJECT of MATHEMATICS on page 22 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. There are a number of ways in which a line can be expressed in formulaic terms. The route we take in defining this line depends heavily on the information that we are given in the problem statement. In this particular case, we are given two points that we are told a line passes through, they are: • Point 1: (0,0) • Point 2: 2, −2 With two points defined we can determine what the slope will be of a line that passes through them. The SLOPE OF A LINE is a measure of how “tilted” the line is and can be equated using the standard formula:

𝑚=



𝑟𝑖𝑠𝑒 (𝑦& − 𝑦1 ) = 𝑟𝑢𝑛 (𝑥& − 𝑥1 )

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Where: • (𝑥1 , 𝑦1 ) are the “x” and “y” coordinates of point 1 •

𝑥& , 𝑦& are the “x” and “y” coordinates of point 2

From the defined points, we have: • 𝑦1 = 0 • 𝑦& = −2 • 𝑥1 = 0 • 𝑥& = 2 Plugging these coordinates in for the two points, the slope will calculate as:

𝑚=

(−2 − 0) = −1 2−0

The equation that best represents a line that passes through both of our points will have slope of −1. We could go on to further define the line, but we have enough to determine that there is only one option given that satisfies the slope that we have defined. Therefore, the correct answer choice is D. 𝐲 = −𝐱



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