Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8β’2
Lesson 6: Rotations of 180 Degrees Student Outcomes ο§ ο§
Students learn that a rotation of 180 degrees moves a point on the coordinate plane (π, π), to (βπ, βπ).
Students learn that a rotation of 180 degrees around a point, not on the line, produces a line parallel to the given line.
Classwork Example 1 (5 minutes) ο§ ο§
Rotations of 180 degrees are special. Recall, a rotation of 180 degrees around π is the rigid motion so that if π is any point in the plane, π, π and π
ππ‘ππ‘πππ(π) are collinear (i.e., they lie on the same line).
Rotations of 180 degrees occur in many situations. For example, the frequently cited fact that vertical angles [vert. β π ] at the intersection of two lines are equal, follows immediately from the fact that 180-degree rotations are degree-preserving. More precisely, let two lines πΏ1 and πΏ2 intersect at π, as shown: Example 1 The picture below shows what happens when there is a rotation of πππΛ around center π.
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We want to show that the vertical angles [vert. β π ], β π and β π are equal (i.e., β π = β π). If we let π
ππ‘ππ‘πππ0 be the 180-degree rotation around π, then π
ππ‘ππ‘πππ0 maps β π to β π. More precisely, if π and π are points on πΏ1 and πΏ2 , respectively, (as shown above), let π
ππ‘ππ‘πππ0 (π) = πβ² and π
ππ‘ππ‘πππ0 (π) = πβ² . Then π
ππ‘ππ‘πππ0 maps β πππ (β π) to β πβ² ππβ² (β π), and since π
ππ‘ππ‘πππ0 is degree preserving, we have β π = β π.
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Lesson 6
8β’2
Example 2 (5 minutes) ο§ ο§
Letβs look at a 180-degree rotation, π
ππ‘ππ‘πππ0 around the origin π of a coordinate system. If a point π has coordinates (π, π), it is generally said that π
ππ‘ππ‘πππ0 (π) is the point with coordinates (βπ, βπ).
Suppose the point π has coordinates (β4, 3), we will show that the coordinates of π
ππ‘ππ‘πππ0 (π) are (4, β3). Example 2 The picture below shows what happens when there is a rotation of πππΛ around center π, the origin of the coordinate plane.
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Let πβ² = π
ππ‘ππ‘πππ0 (π). Let the vertical line (i.e., the line parallel to the y-axis) through π meet the π₯-axis at a point π΄. Because the coordinates of π are (β4, 3), the point π΄ has coordinates (β4, 0) by the way coordinates are defined. In particular, π΄ is of distance 4 from π, and since π
ππ‘ππ‘πππ0 is length-preserving, the point π΄β² = π
ππ‘ππ‘πππ0 (π΄) is also of distance 4 from π. But π
ππ‘ππ‘πππ0 is a 180 degree rotation around π, so π΄β also lies on the π₯-axis, but on the opposite side of the π₯-axis from π΄. Therefore, the coordinates of π΄β are (4,0). Now β ππ΄π is a right angle andβsince π
ππ‘ππ‘πππ0 maps it to β πβ² π΄β² π, and also preserves degreesβwe see that β πβ²π΄β²π is also a right angle. This means that π΄β is the point of intersection of the vertical line through πβ and the π₯-axis. But we already know that π΄β has coordinates of (4, 0), so the π₯-coordinate of πβ is 4, by definition. Similarly, the π¦-coordinate of π being 3 implies that the y-coordinate of πβ is β 3. Altogether, we have proved that the 180-degree rotation of a point of coordinates (β4, 3), is a point with coordinates (4, β3).
The reasoning is perfectly general: the same logic shows that the 180-degree rotation around the origin of a point of coordinates (π, π), is the point with coordinates (βπ, βπ), as desired.
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 6
8β’2
Exercises 1β9 (16 minutes) Students complete Exercises 1β2 independently. Check solutions. Then let students work in pairs on Exercises 3β4. Students complete Exercises 5β9 independently in preparation for the example that follows. Exercises 1.
Using your transparency, rotate the plane πππ degrees, about the origin. Let this rotation be πΉππππππππ. What are the coordinates of πΉππππππππ (π, βπ)?
πΉππππππππ (π, βπ) = (βπ, π)
2.
Let πΉππππππππ be the rotation of the plane by πππ degrees, about the origin. Without using your transparency, find πΉππππππππ (βπ, π). πΉππππππππ (βπ, π) = (π, βπ)
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 6
3.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (βπ, π) parallel to the π-axis. Find πΉππππππππ (π³). Use your transparency if needed.
4.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, π) parallel to the π-axis. Find πΉππππππππ (π³). Use your transparency if needed.
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8β’2
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NYS COMMON CORE MATHEMATICS CURRICULUM
5.
Lesson 6
8β’2
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, π) parallel to the π-axis. Is π³ parallel to πΉππππππππ (π³)? Yes, π³ β₯ πΉππππππππ (π³).
6.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, π) parallel to the π-axis. Is π³ parallel to πΉππππππππ (π³)?
Yes, π³ β₯ πΉππππππππ (π³).
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NYS COMMON CORE MATHEMATICS CURRICULUM
7.
Lesson 6
8β’2
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, βπ) parallel to the π-axis. Is π³ parallel to πΉππππππππ (π³)?
Yes, π³ β₯ πΉππππππππ (π³).
8.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Is π³ parallel to πΉππππππππ (π³)? Use your transparency if needed. Yes, π³ β₯ πΉππππππππ (π³).
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
9.
8β’2
Let πΉππππππππ be the rotation of πππ degrees around the origin. Is π³ parallel to πΉππππππππ (π³)? Use your transparency if needed. Yes, π³ β₯ πΉππππππππ (π³).
Example 3 (5 minutes) MP.2
Scaffolding:
Theorem. Let π be a point not lying on a given line πΏ. Then the 180-degree rotation around π maps πΏ to a line parallel to πΏ.
Proof: Let π
ππ‘ππ‘πππ0 be the 180-degree rotation around π, and let π be a point on πΏ. As usual, denote π
ππ‘ππ‘πππ0 (π) by πβ. Since π
ππ‘ππ‘πππ0 is a 180-degree rotation, π, π, πβ lie on the same line (denoted by β).
After completing Exercises 5β9, students should be convinced that the theorem is true. Make it clear that their observations can be proven (by contradiction) if we assume something different will happen (e.g., the lines will intersect).
We want to investigate whether πβ lies on πΏ or not. Keep in mind that we want to show that the 180-degree rotation maps πΏ to a line parallel to πΏ. If the point πβ lies on πΏ, then at some point, the line πΏ and π
ππ‘ππ‘πππ0 (πΏ) intersect, meaning they are not parallel. If we can eliminate the possibility that πβ lies on πΏ, then we have to conclude that πβ does Lesson 6: Date: Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8β’2
not lie on πΏ (rotations of 180 degrees make points that are collinear). If πβ lies on πΏ, then β is a line that joins two points, πβ and π, on πΏ. But already πΏ is a line that joins πβ and π, so β and πΏ must be the same line (i.e., β = πΏ). This is trouble because we know π lies on β, so β = πΏ implies that π lies on πΏ. Look at the hypothesis of the theorem: βLet π be a point not lying on a given line πΏ.β We have a contradiction. So the possibility that πβ lies on πΏ is nonexistent. As we said, this means that πβ does not lie on πΏ.
What we have proved is that no matter which point π we take from πΏ, we know π
ππ‘ππ‘πππ0 (π) does not lie on πΏ. But π
ππ‘ππ‘πππ0 (πΏ) consists of all the points of the form π
ππ‘ππ‘πππ0 (π) where π lies on πΏ, so what we have proved is that no point of π
ππ‘ππ‘πππ0 (πΏ)lies on πΏ. In other words, πΏ and π
ππ‘ππ‘πππ0 (πΏ) have no point in common. (i.e., πΏ β₯ π
ππ‘ππ‘πππ0 (πΏ)). The theorem is proved.
Closing (5 minutes) Summarize, or have students summarize, the lesson. ο§
Rotations of 180 degrees are special: β’
β’ ο§
A point, π, that is rotated 180 degrees around a center π, produces a point πβ so that π, π, πβ are collinear.
When we rotate around the origin of a coordinate system, we see that the point with coordinates (π, π) is moved to the point (βπ, βπ).
We now know that when a line is rotated 180 degrees around a point not on the line, it maps to a line parallel to the given line.
Lesson Summary ο§
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A rotation of 180 degrees around πΆ is the rigid motion so that if π· is any point in the plane, π·, πΆ and
πΉπππππππ(π·) are collinear (i.e., lie on the same line).
Given a 180-degree rotation, πΉπ around the origin πΆ of a coordinate system, and a point π· with
coordinates (π, π), it is generally said that πΉπ (π·) is the point with coordinates (βπ, βπ).
Theorem. Let πΆ be a point not lying on a given line π³. Then the 180-degree rotation around πΆ maps π³ to a line parallel to π³.
Exit Ticket (5 minutes)
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
Name ___________________________________________________
8β’2
Date____________________
Lesson 6: Rotations of 180 Degrees Exit Ticket Let there be a rotation of 180 degrees about the origin. Point π΄ has coordinates (β2, β4), and point π΅ has coordinates (β3, 1), as shown below.
1.
What are the coordinates of π
ππ‘ππ‘πππ(π΄)? Mark that point on the graph so that π
ππ‘ππ‘πππ(π΄) = π΄β². What are the coordinates of π
ππ‘ππ‘πππ(π΅)? Mark that point on the graph so that π
ππ‘ππ‘πππ(π΅) = π΅β².
2.
What can you say about the points π΄, π΄β² and π? What can you say about the points π΅, π΅β² and π?
3.
Connect point π΄ to point π΅ to make the line πΏπ΄π΅ . Connect the point π΄β² to point π΅β² to make the line πΏπ΄β² π΅β² . What is the relationship between πΏπ΄π΅ and πΏπ΄β² π΅β² ?
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8β’2
Exit Ticket Sample Solutions Let there be a rotation of πππ degrees about the origin. Point π¨ has coordinates (βπ, βπ) and point π© has coordinates (βπ, π), as shown below.
1.
What are the coordinates of πΉπππππππ(π¨)? Mark that point on the graph so that πΉπππππππ(π¨) = π¨β². What are the coordinates of πΉπππππππ(π©)? Mark that point on the graph so that πΉπππππππ(π©) = π©β. π¨β² = (π, π), π©β² = (π, βπ)
2.
What can you say about the points π¨, π¨β² and πΆ? What can you say about the points π©, π©β² and πΆ? The points π¨, π¨β² and πΆ are collinear. The points π©, π©β² and πΆ are collinear.
3.
Connect point π¨ to point π© to make the line π³π¨π© Connect the point π¨β² to point π©β² to make the line π³π¨β²π©β² . What is the relationship between π³π¨π© and π³π¨β²π©β² ? π³π¨π© β₯ π³π¨β²πβ²
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
8β’2
Problem Set Sample Solutions Use the following diagram for problems 1β5. Use your transparency, as needed.
1.
Looking only at segment π©πͺ, is it possible that a πππΛ rotation would map π©πͺ onto π©β²πͺβ²? Why or why not?
It is possible because the segments are parallel. 2.
Looking only at segment π¨π©, is it possible that a πππΛ rotation would map π¨π© onto π¨β²π©β²? Why or why not? It is possible because the segments are parallel.
3.
Looking only at segment π¨πͺ, is it possible that a πππΛ rotation would map π¨πͺ onto π¨β²πͺβ²? Why or why not? It is possible because the segments are parallel.
4.
Connect point π© to point π©β², point πͺ to point πͺβ², and point π¨ to point π¨β². What do you notice? What do you think that point is? All of the lines intersect at one point. The point is the center of rotation, I checked by using my transparency.
5.
Would a rotation map triangle π¨π©πͺ onto triangle π¨β²π©β²πͺβ² ? If so, define the rotation (i.e., degree and center). If not, explain why not. Let there be a rotation πππΛ around point (π, βπ). Then πΉπππππππ(βπ¨π©πͺ) = βπ¨β²π©β²πͺβ².
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NYS COMMON CORE MATHEMATICS CURRICULUM
6.
Lesson 6
8β’2
The picture below shows right triangles π¨π©πͺ and π¨βπ©βπͺβ, where the right angles are at π© and π©β. Given that π¨π© = π¨β² π©β² = π, and π©πͺ = π©β² πͺβ² = π, π¨π© is not parallel to π¨βπ©β, is there a πππΛ rotation that would map βπ¨π©πͺ onto βπ¨β²π©β²πͺβ² ? Explain.
No, because a πππΛ rotation of a segment will map to a segment that is parallel to the given one. It is given that π¨π© is not parallel to π¨βπ©β, therefore a rotation of πππΛ will not map βπ¨π©πͺ onto βπ¨β²π©β²πͺβ²
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