Lesson 6: Rotations of 180 Degrees

```NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 6
8β’2
Lesson 6: Rotations of 180 Degrees
Student Outcomes
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Students learn that a rotation of 180 degrees moves a point on the coordinate plane (π, π) to (βπ, βπ).
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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)
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Rotations of 180 degrees are special. Recall, a rotation of 180 degrees around π is a rigid motion so that if π
is any point in the plane, π, π, and πππ‘ππ‘πππ(π) are collinear (i.e., they lie on the same line).
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Rotations of 180 degrees occur in many situations. For example, the frequently cited fact that vertical angles
(vert. β βs) at the intersection of two lines are equal, follows immediately from the fact that 180-degree
rotations are angle-preserving. More precisely, let two lines πΏ1 and πΏ2 intersect at π, as shown:
Example 1
The picture below shows what happens when there is a rotation of πππ° around center πΆ.
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We want to show that the vertical angles (vert. β βs), β π and β π, are equal in measure (i.e., β π = β π). If we
let πππ‘ππ‘πππ0 be the 180-degree rotation around π, then πππ‘ππ‘πππ0 maps β π to β π. More precisely, if π
and π are points on πΏ1 and πΏ2 , respectively (as shown above), let πππ‘ππ‘πππ0 (π) = πβ² and πππ‘ππ‘πππ0 (π) = πβ² .
Then, πππ‘ππ‘πππ0 maps β πππ (β π) to β πβ² ππβ² (β π), and since πππ‘ππ‘πππ0 is angle-preserving, we have
β π = β π.
Lesson 6:
Rotations of 180 Degrees
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Lesson 6
8β’2
Example 2 (5 minutes)
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Letβs look at a 180-degree rotation, πππ‘ππ‘πππ0 around the origin π of a coordinate system. If a point π has
coordinates (π, π), it is generally said that πππ‘ππ‘πππ0 (π) is the point with coordinates (βπ, βπ).
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Suppose the point π has coordinates (β4, 3); we show that the coordinates of πππ‘ππ‘πππ0 (π) are (4, β3).
Example 2
The picture below shows what happens when there is a rotation of πππ° around center πΆ, the origin of the coordinate
plane.
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Let πβ² = πππ‘ππ‘πππ0 (π). Let the vertical line (i.e., the line parallel to the π¦-axis) through π meet the π₯-axis at a
point π΄. Because the coordinates of π are (β4, 3), the point π΄ has coordinates (β4, 0) by the way coordinates
are defined. In particular, π΄ is of distance 4 from π, and since πππ‘ππ‘πππ0 is length-preserving, the point
π΄β² = πππ‘ππ‘πππ0 (π΄) is also of distance 4 from π. However, πππ‘ππ‘πππ0 is a 180-degree rotation around π, so
π΄β² also lies on the π₯-axis but on the opposite side of the π₯-axis from π΄. Therefore, the coordinates of π΄β² are
(4,0). Now, β ππ΄π is a right angle andβsince πππ‘ππ‘πππ0 maps it to β πβ² π΄β² π and also preserves degreesβwe
see that β πβ²π΄β²π is also a right angle. This means that π΄β² is the point of intersection of the vertical line through
πβ² and the π₯-axis. Since we already know that π΄β² has coordinates of (4, 0), then the π₯-coordinate of πβ² is 4, by
definition.
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Similarly, the π¦-coordinate of π being 3 implies that the π¦-coordinate of πβ² is β 3. Altogether, we have proved
that the 180-degree rotation of a point of coordinates (β4, 3) is a point with coordinates (4, β3).
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 (βπ, βπ), as desired.
Exercises 1β9 (20 minutes)
Students complete Exercises 1β2 independently. Check solutions. Then, let students work in pairs on Exercises 3β4.
Students complete Exercises 5β9 independently in preparation for the example that follows.
Lesson 6:
Rotations of 180 Degrees
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Lesson 6
8β’2
Exercises 1β9
1.
Using your transparency, rotate the plane πππ degrees, about the origin. Let this rotation be πΉππππππππ . What are
the coordinates of πΉππππππππ (π, βπ)?
πΉππππππππ (π, βπ) = (βπ, π)
2.
Let πΉππππππππ be the rotation of the plane by πππ degrees, about the origin. Without using your transparency,
find πΉππππππππ (βπ, π).
πΉππππππππ (βπ, π) = (π, βπ)
Lesson 6:
Rotations of 180 Degrees
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Lesson 6
3.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (βπ, π) parallel
to the π-axis. Find πΉππππππππ (π³). Use your transparency if needed.
4.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, π) parallel to
the π-axis. Find πΉππππππππ (π³). Use your transparency if needed.
Lesson 6:
Rotations of 180 Degrees
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8β’2
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5.
Lesson 6
8β’2
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, π) parallel to
the π-axis. Is π³ parallel to πΉππππππππ (π³)?
Yes, π³ β₯ πΉππππππππ (π³).
6.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, π) parallel to
the π-axis. Is π³ parallel to πΉππππππππ (π³)?
Yes, π³ β₯ πΉππππππππ (π³).
Lesson 6:
Rotations of 180 Degrees
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7.
Lesson 6
8β’2
Let πΉππππππππ be the rotation of πππ degrees around the origin. Let π³ be the line passing through (π, βπ) parallel
to the π-axis. Is π³ parallel to πΉππππππππ (π³)?
Yes, π³ β₯ πΉππππππππ (π³).
8.
Let πΉππππππππ be the rotation of πππ degrees around the origin. Is π³ parallel to πΉππππππππ (π³)? Use your
transparency if needed.
Yes, π³ β₯ πΉππππππππ (π³).
Lesson 6:
Rotations of 180 Degrees
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9.
8β’2
Let πΉππππππππ be the rotation of πππ degrees around the center πΆ. Is π³ parallel to πΉππππππππ (π³)? Use your
transparency if needed.
Yes, π³ β₯ πΉππππππππ (π³)
Example 3 (5 minutes)
MP.2
Scaffolding:
THEOREM: Let π be a point not lying on a given line πΏ. Then, the 180-degree rotation
around π maps πΏ to a line parallel to πΏ.
PROOF: Let πππ‘ππ‘πππ0 be the 180-degree rotation around π, and let π be a point on πΏ. As
usual, denote πππ‘ππ‘πππ0 (π) by πβ². Since πππ‘ππ‘πππ0 is a 180-degree rotation, π, π, and πβ²
lie on the same line (denoted by β).
After completing Exercises 5β9,
students should be convinced
that the theorem is true. Make
it clear that their observations
can be proven (by
something different will
happen (e.g., the lines will
intersect).
We want to investigate whether πβ² lies on πΏ or not. Keep in mind that we want to show that the 180-degree rotation
maps πΏ to a line parallel to πΏ. If the point πβ² lies on πΏ, then at some point, the line πΏ and πππ‘ππ‘πππ0 (πΏ) intersect,
meaning they are not parallel. If we can eliminate the possibility that πβ² lies on πΏ, then we have to conclude that πβ² does
not lie on πΏ (rotations of 180 degrees make points that are collinear). If πβ² lies on πΏ, then β is a line that joins two points,
πβ² and π, on πΏ. However, πΏ is already a line that joins πβ² and π, so β and πΏ must be the same line (i.e., β = πΏ). This is
Lesson 6:
Rotations of 180 Degrees
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trouble because we know π lies on β, so β = πΏ implies that π lies on πΏ. Look at the hypothesis of the theorem: Let π be
a point not lying on a given line πΏ. We have a contradiction. So, the possibility that πβ² lies on πΏ is nonexistent. As we
said, this means that πβ² does not lie on πΏ.
What we have proved is that no matter which point π we take from πΏ, we know πππ‘ππ‘πππ0 (π) does not lie on πΏ. But
πππ‘ππ‘πππ0 (πΏ) consists of all the points of the form πππ‘ππ‘πππ0 (π) where π lies on πΏ, so what we have proved is that no
point of πππ‘ππ‘πππ0 (πΏ) lies on πΏ. In other words, πΏ and πππ‘ππ‘πππ0 (πΏ) have no point in common (i.e., πΏ β₯ πππ‘ππ‘πππ0 (πΏ)).
The theorem is proved.
Closing (5 minutes)
Summarize, or have students summarize, the lesson.
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ο§
Rotations of 180 degrees are special:
β
A point, π, that is rotated 180 degrees around a center π, produces a point πβ² so that π, π, and πβ² are
collinear.
β
When we rotate around the origin of a coordinate system, we see that the point with coordinates
(π, π) is moved to the point (βπ, βπ).
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
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A rotation of πππ degrees around πΆ is the rigid motion so that if π· is any point in the plane, π·, πΆ, and
πΉπππππππ(π·) are collinear (i.e., lie on the same line).
ο§
Given a πππ-degree rotation around the origin πΆ of a coordinate system, πΉπ , and a point π· with
coordinates (π, π), it is generally said that πΉπ (π·) is the point with coordinates (βπ, βπ).
THEOREM: Let πΆ be a point not lying on a given line π³. Then, the πππ-degree rotation around πΆ maps π³ to a line
parallel to π³.
Exit Ticket (5 minutes)
Lesson 6:
Rotations of 180 Degrees
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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 (β2, β4), and point π΅ has coordinates
(β3, 1), as shown below.
1.
What are the coordinates of πππ‘ππ‘πππ(π΄)? Mark that point on the graph so that πππ‘ππ‘πππ(π΄) = π΄β². What are the
coordinates of πππ‘ππ‘πππ(π΅)? Mark that point on the graph so that πππ‘ππ‘πππ(π΅) = π΅β².
2.
What can you say about the points π΄, π΄β² , and π? What can you say about the points π΅, π΅β² , and π?
3.
Connect point π΄ to point π΅ to make the line πΏπ΄π΅ . Connect point π΄β² to point π΅β² to make the line πΏπ΄β² π΅β² . What is the
relationship between πΏπ΄π΅ and πΏπ΄β² π΅β² ?
Lesson 6:
Rotations of 180 Degrees
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Exit Ticket Sample Solutions
Let there be a rotation of πππ degrees about the origin. Point π¨ has coordinates (βπ, βπ), and point π© has coordinates
(βπ, π), as shown below.
1.
What are the coordinates of πΉπππππππ(π¨)? Mark that point on the graph so that πΉπππππππ(π¨) = π¨β². What are the
π¨β² = (π, π), π©β² = (π, βπ)
2.
What can you say about the points π¨, π¨β² , and πΆ? What can you say about the points π©, π©β², and πΆ?
The points π¨, π¨β², and πΆ are collinear. The points π©, π©β², and πΆ are collinear.
3.
Connect point π¨ to point π© to make the line π³π¨π© . Connect point π¨β² to point π©β² to make the line π³π¨β²π©β² . What is the
Lesson 6:
Rotations of 180 Degrees
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Problem Set Sample Solutions
Use the following diagram for Problems 1β5. Use your transparency as needed.
1.
Looking only at segment π©πͺ, is it possible that a πππ° rotation would map segment π©πͺ onto segment π©β²πͺβ²? Why or
why not?
It is possible because the segments are parallel.
2.
Looking only at segment π¨π©, is it possible that a πππ° rotation would map segment π¨π© onto segment π¨β²π©β²? Why or
why not?
It is possible because the segments are parallel.
3.
Looking only at segment π¨πͺ, is it possible that a πππ° rotation would map segment π¨πͺ onto segment π¨β²πͺβ²? Why or
why not?
It is possible because the segments are parallel.
4.
Connect point π© to point π©β², point πͺ to point πͺβ², and point π¨ to point π¨β². 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©πͺ onto triangle π¨β²π©β²πͺβ²? If so, define the rotation (i.e., degree and center). If not,
explain why not.
Let there be a rotation πππ° around point (π, βπ). Then, πΉπππππππ(β³ π¨π©πͺ) =β³ π¨β²π©β²πͺβ².
Lesson 6:
Rotations of 180 Degrees
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6.
Lesson 6
8β’2
ΜΜΜΜ is not parallel to ΜΜΜΜΜΜ
π¨β²π©β², is there a πππ° rotation that would map