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Lesson Explainer: Rotations about a Point Mathematics • 8th Grade

In this explainer, we will learn how to rotate points, line segments, and shapes about given points.

In geometry, we often use the congruence of shapes, angles, and line segments to prove results. This means that we want to find methods of checking the congruence of these objects. Often, this involves comparing the properties of two of these objects, but this is not the only way of studying congruence.

We can also ask, β€œWhat transformation can we apply to an object that does not affect its size or shape?” If a transformation does not affect these properties, then we can apply the transformation to any object to construct a congruent object. In particular, if we can show that two objects differ by one or more of these transformations, then we can conclude they are congruent.

There are many different transformations that do not affect the size or shape of an object; we can slide the object around, changing its position (translation), we can find a mirror image of the object (reflection), and we can also rotate the object about a point (rotation).

We will focus on rotating objects about a point. To start this, let’s consider exactly what rotation is. A rotation is defined by the center of the rotation and the amount of the rotation. In particular, we rotate each point around a circle centered at the center of the rotation. For example, let’s say we have the following rotation of a point 𝐴 centered at 𝑀.

We call the image of a point 𝐴 after rotation 𝐴′. We can also add more prime symbols to represent further rotations. We rotate point 𝐴 along the circle of radius 𝑀𝐴 centered at 𝑀.

This means that 𝑀𝐴 and 𝑀𝐴′ are radii of the same circle, so they have the same length. This is true for any point after any rotation.

We can describe the direction of a rotation in two ways; clockwise and counterclockwise. Clockwise means the rotation is in the same direction that the hands of a clock move in; counterclockwise is the opposite direction.

We can also note that the rotation is given in a counterclockwise direction.

To describe the size of the rotation, we need a measure of how far a point moves after its rotation. We might be tempted to do this by measuring the length of the arc between the point and its image or the distance between the point and its image. However, neither of these measures will work since different points move different lengths.

Instead, we can measure the amount of rotation by considering the measure of βˆ π΄π‘€π΄β€².

This will tell us the size of the rotation to rotate 𝐴𝑀 onto 𝐴′𝑀. We can also use the size of the rotation to tell us the direction. We can say that a counterclockwise rotation is a positive rotation and that a clockwise rotation is a negative rotation.

For example, if we find that π‘šβˆ π΄π‘€π΄β€²=120∘, then we can say that this is a 120∘ counterclockwise rotation centered at 𝑀. We write this 𝑅(𝑀,120)∘. We can represent the clockwise rotation of the same measure as 𝑅(𝑀,βˆ’120)∘.

We can define all of this formally as follows.

Definition: Rotation about a Point

To rotate a point 𝐴 about a point 𝑀, we move 𝐴 along the circle of radius 𝐴𝑀 centered at 𝑀. We call the image of 𝐴 after this rotation 𝐴′.

We can rotate 𝐴 about 𝑀 a specific distance and direction by specifying the measure of βˆ π΄π‘€π΄β€², where we use a positive value to represent a counterclockwise rotation and a negative value to represent a clockwise rotation.

We use the notation 𝑅(𝑀,π‘₯)∘ to represent a rotation centered at 𝑀 of measure π‘₯∘. The center of a rotation is the only point always fixed under rotation.

Let’s now see an example of using this notation to determine the image of a point after a given rotation.

Example 1: Rotating a Point about Another Point

Which point represents point 𝐴 after a rotation of 𝑅(𝑀,20)∘?

  1. Point 𝐹
  2. Point 𝐡
  3. Point 𝐢
  4. Point 𝐷
  5. Point 𝐸

Answer

We start by recalling that the notation 𝑅(𝑀,20)∘ tells us that the rotation is centered at 𝑀 and that the measure of the rotation is 20∘. In fact, since this value is positive, we know that this rotation is counterclockwise.

A rotation centered at 𝑀 that is 20∘ counterclockwise will move 𝐴 along the circle centered at 𝑀 of radius 𝑀𝐴 to a point 𝐴′ such that π‘šβˆ π΄π‘€π΄β€²=20∘.

We see that π‘šβˆ π΄π‘€π΅=20∘ and that this is in a counterclockwise direction.

Hence, the answer is point 𝐡.

Before we move on to our next example, we can consider some of the properties of rotations. For example, we stated at the start of the explainer that rotations do not affect the size or shape of the rotated object. We should justify this to make sure that it is true.

To do this, let’s first consider what happens when we rotate a line segment 𝐴𝐡 π‘₯∘ about 𝑀. We can note that we can rotate a line segment by just rotating its endpoints, so the rotation of 𝐴𝐡 will be 𝐴′𝐡′. This allows us to compare the lengths of these line segments.

We want to show that 𝐴𝐡=𝐴′𝐡′, and we can do this by proving the congruence of triangles 𝑀𝐴𝐡 and 𝑀𝐴′𝐡′. We note that 𝑀𝐴=𝑀𝐴′ and 𝑀𝐡=𝑀𝐡′ since they are radii of the same circles. We can show that π‘šβˆ π΄π‘€π΅=π‘šβˆ π΄β€²π‘€π΅β€² by noting that both sides of angle βˆ π΄π‘€π΅ are rotated the same amount. Therefore, △𝑀𝐴𝐡≅△𝑀𝐴′𝐡′ by the SAS criterion, and we can conclude that 𝐴𝐡=𝐴′𝐡′.

Hence, rotations do not affect the lengths of line segments or the measures of angles. In fact, since we can rotate a line segment by just rotating its endpoints, we can rotate any polygon by rotating its vertices and have a congruent polygon.

We can also rotate any point using a compass and protractor. For example, let’s say we want to rotate π΄Β βˆ’60∘ about a point 𝑀. We can draw ray 𝑀𝐴 and then measure an angle 60∘ clockwise from this ray. We can call this angle βˆ π΄π‘€π‘‹.

If we then trace a circle of radius 𝐴𝑀 centered at 𝑀, we can note that the point of intersection between 𝑀𝑋 and the circle will be the image of 𝐴 after the rotation.

In our next example, we will determine the correct image of a triangle after a given rotation using a compass and protractor.

Example 2: Rotating a Triangle about a Point

When the green triangle 𝐴𝐡𝐢 is rotated 90∘ about 𝐴, which triangle is its new position?

  1. △𝐴𝐷𝐸
  2. △𝐴𝐹𝐺
  3. △𝐴𝐻𝐼

Answer

We rotate a polygon about a point by first rotating all of its vertices. We recall that we rotate a point about 𝐴 by moving it along a circle centered at 𝐴 such that for any point in the shape 𝑋, we have that the circle is of radius 𝐴𝑋 and that the measure of βˆ π‘‹π΄π‘‹β€² is equal to the measure of the rotation.

In this case, we want to rotate a triangle 90∘, and since this value is positive, we note that we will rotate the shape counterclockwise.

We want to determine the images of the vertices of the triangle, so let’s start with 𝐴. Since 𝐴 is the center of rotation, it will not move under this rotation.

We can rotate vertices on triangle 𝐴𝐡𝐢 by considering circles centered at 𝐴. We can first draw a ray 𝐴𝐡 and then measure another ray that is 90∘ in the counterclockwise direction of 𝐴𝐡, we call this οƒͺ𝐴𝐽. If we then sketch a circle centered at 𝐴 of radius 𝐴𝐡, we note that 𝐷 is the point of intersection between the circle and ray οƒͺ𝐴𝐽 and that π‘šβˆ π΅π΄π·=90∘.

Thus, the image of 𝐡 after the rotation is 𝐷.

We can follow the same process for 𝐢. We sketch ray 𝐴𝐢 and another ray from 𝐴 that is 90∘ counterclockwise from 𝐴𝐢. We then sketch a circle of radius 𝐴𝐢 centered at 𝐴, and find that the point of intersection between the ray and the circle is point 𝐸.

So, the image of 𝐴 is 𝐴, the image of 𝐡 is 𝐷, and the image of 𝐢 is 𝐸.

Hence, the image of △𝐴𝐡𝐢 is △𝐴𝐷𝐸, which is answer A.

In our next example, we will rotate a given triangle that makes up a tenth of a decagon.

Example 3: Understanding the Effect of a Rotation of a Triangle given the Angle of Rotation

What is the image of △𝐢𝐷𝑀 after a βˆ’108∘ rotation?

Answer

We are asked to rotate β–³πΆπ·π‘€Β βˆ’108∘ about 𝑀. To do this, we note that the given polygon is a regular decagon, so the angles around 𝑀 are all congruent. We can also note that the sum of the angles around 𝑀 must be 360∘. Thus, each angle at the center has a measure of 36010=36∘∘. We then note that βˆ’108=3Γ—(βˆ’36)∘∘. So, we can find angles of measure βˆ’108∘ on the diagram by rotating through 3 triangles in a clockwise direction.

We could now rotate the triangle by rotating its vertices using a compass and protractor; however, this is not necessary. We can note that each rotation of βˆ’36∘ about 𝑀 will rotate the triangles clockwise such that they are coincident. One way of seeing this is to consider rotating a single line segment. For example, we can rotate π‘€πΆΒ βˆ’36∘ about 𝑀.

We then see that the image of this line segment is 𝑀𝐡.

We can do the same for 𝑀𝐷. If we rotate it βˆ’108∘ about 𝑀, we get 𝑀𝐴.

Therefore, when we rotate β–³πΆπ·π‘€Β βˆ’108∘ about 𝑀, the triangle will be rotated onto the third triangle clockwise from △𝐢𝐷𝑀.

We see that this is △𝑃𝐴𝑀.

In our next example, we will use the given image of a triangle after a rotation to determine the measure of the rotation.

Example 4: Finding the Measure of the Angle of Rotation

△𝐴𝐡′𝐢′ is the image of △𝐴𝐡𝐢 by a counterclockwise rotation of π‘₯∘ about 𝐴. Find π‘₯.

Answer

We begin by recalling that a rotation will rotate a line segment onto a congruent line segment whose endpoints are the images of the endpoints of the line segment under a rotation. In particular, we can note that this rotation is about 𝐴, so 𝐴 will remain where it is after the rotation. Thus, 𝐴𝐡 is rotated onto 𝐴𝐡′.

We can determine the measure of the rotation by finding the measure of the angle between 𝐴𝐡 and 𝐴𝐡′. Since 37+69=106∘∘∘ and the rotation is counterclockwise, we can see that the rotation has a measure of 106∘.

In our final example, we will consider rotations of a regular octagon.

Example 5: Comparing Two Rotations of the Same Line Segment

Consider the octagon shape 𝐴𝐡𝐢𝐷𝐸𝐹𝐺𝐻 with 𝑀 as its center.

  1. What is the final position of 𝐴𝐡 when it is rotated 90∘?
    1. 𝐸𝐹
    2. 𝐺𝐻
    3. 𝐢𝐷
    4. None of the answers are correct.
  2. What is the final position of 𝐴𝐡 when it is rotated βˆ’270∘?
    1. 𝐸𝐹
    2. 𝐺𝐻
    3. 𝐢𝐷
    4. None of the answers are correct.
  3. Is there a difference between the two rotations?
    1. Yes
    2. No

Answer

Part 1

To rotate a line segment, we recall that we can just rotate the endpoints. In other words, the image of 𝐴𝐡 will be 𝐴′𝐡′. Let’s start by finding the image of 𝐴. Since the measure of the rotation is positive, the rotation is counterclockwise. So, we want to rotate 𝐴 90∘ counterclockwise about 𝑀.

We rotate a point 𝐴 90∘ about 𝑀 by finding point 𝐴′, which is counterclockwise on the circle centered at 𝑀 of radius 𝐴𝑀 such that π‘šβˆ π΄π‘€π΄β€²=90∘. We can find this point on the diagram by noting that the distance between 𝑀 and any vertex of the regular octagon is the same. So, all of the vertices lie on a circle of radius 𝐴𝑀 centered at 𝑀.

We can measure the angles at the center of the octagon by joining the center to all of the vertices and noting that the angles in the center are congruent and that their measures sum to 360∘. So, they are each of measure 45∘.

Thus, two of these angles combine to make 90∘, and so π‘šβˆ π΄π‘€πΆ=90∘.

Thus, 𝐴′=𝐢. We can follow the same reasoning to see that 𝐡′=𝐷.

Hence, the image of 𝐴𝐡 after a rotation of 90∘ is 𝐢𝐷, which is answer C.

Part 2

We can also rotate π΄π΅Β βˆ’270∘ using the same method. We need to note that since this value is negative, we rotate clockwise.

We see that the image of 𝐴 after a rotation of βˆ’270∘ is 𝐢 and that the image of 𝐡 after a rotation of βˆ’270∘ is 𝐷.

Hence, the image of 𝐴𝐡 After a rotation of βˆ’270∘ is 𝐢𝐷, which is answer C.

Part 3

To answer this question, we need to consider what happens to a point after each rotation. Let’s say we rotate a point 𝑋 by 90∘ about 𝑀. 𝑋 does not necessarily need to be on the octagon; it could be any point in space, but we rotate it in the same way. We trace a circle centered at 𝑀 and find the point on the circle such that π‘šβˆ π‘‹π‘€π‘‹β€²=90∘.

The reflex angle at 𝑀 will have a measure of 270∘ since it makes a full turn with the other angle. If we measure this clockwise, then its measure is βˆ’270∘.

We can see that 𝑋′ is also the image of 𝑋 after a rotation of βˆ’270∘.

Hence, the answer is that there is no difference between the transformations, which is answer B.

In the previous example, we say that a counterclockwise rotation of 90∘ is the same transformation as a clockwise rotation of 270∘. In general, if the absolute values of the two angles of rotation of opposite signs add to give 360∘, then these rotations will be equivalent. In particular, this means that a counterclockwise rotation of 180∘ is equivalent to a clockwise rotation of 180∘.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • To rotate a point 𝐴 about a point 𝑀, we move 𝐴 along the circle of radius 𝐴𝑀 centered at 𝑀. We call the image of 𝐴 after this rotation 𝐴′.
  • We can rotate 𝐴 about 𝑀 a specific distance and direction by specifying the measure of βˆ π΄π‘€π΄β€², where we use a positive value to represent a counterclockwise rotation and a negative value to represent a clockwise rotation.
  • We use the notation 𝑅(𝑀,π‘₯)∘ to represent a rotation centered at 𝑀 of measure π‘₯∘.
  • If 𝐴′ and 𝐡′ are the images of 𝐴 and 𝐡 after a rotation, then 𝐴′𝐡′ is the image of 𝐴𝐡 after the same rotation and 𝐴′𝐡′=𝐴𝐡.
  • We can rotate a polygon by rotating its vertices; the image of a polygon under a rotation is a congruent polygon.
  • A counterclockwise rotation of 180∘ is equivalent to a clockwise rotation of 180∘. In general, a counterclockwise rotation of π‘₯∘ is equivalent to a clockwise rotation of (360βˆ’π‘₯)∘.

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