# 12.4: Rotations

**At Grade**Created by: CK-12

## Learning Objectives

- Find the image of a figure in a rotation in a coordinate plane.
- Recognize that a rotation is an isometry.

## Review Queue

- Reflect
△XYZ with verticesX(9,2),Y(2,4) andZ(7,8) over they− axis. What are the vertices of△X′Y′Z′ ? - Reflect
△X′Y′Z′ over thex− axis. What are the vertices of△X′′Y′′Z′′ ? - How do the coordinates of
△X′′Y′′Z′′ relate to△XYZ ?

**Know What?** The international symbol for recycling appears below. It is three arrows rotated around a point. Let’s assume that the arrow on the top is the preimage and the other two are its images. Find the center of rotation and the angle of rotation for each image.

## Defining Rotations

**Rotation:** A transformation by which a figure is turned around a fixed point to create an image.

**Center of Rotation:** The fixed point that a figure is rotated around.

Lines can be drawn from the preimage to the center of rotation, and from the center of rotation to the image. The angle formed by these lines is the angle of *rotation.*

In this section, our center of rotation will always be the ** origin.** Rotations can also be clockwise or counterclockwise. We will only do

**rotations, to go along with the way the quadrants are numbered.**

*counterclockwise*
**Investigation 12-1: Drawing a Rotation of 100∘**

Tools Needed: pencil, paper, protractor, ruler

- Draw
△ABC and a pointR outside the circle. - Draw the line segment
RB¯¯¯¯¯ . - Take your protractor, place the center on
R and the initial side onRB¯¯¯¯¯ . Mark a100∘ angle. - Find
B′ such thatRB=RB′ . - Repeat steps 2-4 with points
A andC . - Connect
A′,B′, andC′ to form△A′B′C′ .

This is the process you would follow to rotate any figure

To rotate a figure

**Example 1:** Rotate

**Solution:** You can either use Investigation 12-1 or the hint given above to find ** left** of the origin and 4 units

**The vertices are:**

*down.*

The image has vertices that are the negative of the preimage. This will happen every time a figure is rotated

**Rotation of 180∘:** If

From this example, we can also see that ** a rotation is an isometry.** This means that

Similar to the

**Example 2:** Rotate

**Solution:** When we rotate something ** perpendicular** to the line from the origin to its image. Therefore, if \begin{align*}S\end{align*} is 6 units to the

**of the origin and 1 unit**

*right***\begin{align*}S'\end{align*} will be 6 units**

*down,***and 1 to the**

*up*

*right.*Using this pattern, \begin{align*}T'\end{align*} is (8, 2).

If you were to write the slope of each point to the origin, \begin{align*}S\end{align*} would be \begin{align*}\frac{-1}{6} \rightarrow \frac{y}{x}\end{align*}, and \begin{align*}S'\end{align*} must be \begin{align*}\frac{6}{1} \rightarrow \frac{y'}{x'}\end{align*}. Again, they are perpendicular slopes, following along with the \begin{align*}90^\circ\end{align*} rotation. Therefore, the \begin{align*}x\end{align*} and the \begin{align*}y\end{align*} values switch and the new \begin{align*}x-\end{align*}value is the opposite sign of the original \begin{align*}y-\end{align*}value.

**Rotation of \begin{align*}90^\circ\end{align*}:** If \begin{align*}(x, y)\end{align*} is rotated \begin{align*}90^\circ\end{align*} around the origin, then the image will be \begin{align*}(-y, x)\end{align*}.

**Rotation of \begin{align*}270^\circ\end{align*}**

A rotation of \begin{align*}270^\circ\end{align*} counterclockwise would be the same as a clockwise rotation of \begin{align*}90^\circ\end{align*}. We also know that a \begin{align*}90^\circ\end{align*} rotation and a \begin{align*}270^\circ\end{align*} rotation are \begin{align*}180^\circ\end{align*} apart. We know that for every \begin{align*}180^\circ\end{align*} rotation, the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values are negated. So, if the values of a \begin{align*}90^\circ\end{align*} rotation are \begin{align*}(-y, x)\end{align*}, then a \begin{align*}270^\circ\end{align*} rotation would be the opposite sign of each, or \begin{align*}(y, -x)\end{align*}.

**Rotation of \begin{align*}270^\circ\end{align*}:** If \begin{align*}(x, y)\end{align*} is rotated \begin{align*}270^\circ\end{align*} around the origin, then the image will be \begin{align*}(y, -x)\end{align*}.

**Example 3:** Find the coordinates of \begin{align*}ABCD\end{align*} after a \begin{align*}270^\circ\end{align*} rotation.

**Solution:** Using the rule, we have:

\begin{align*}(x,y) & \rightarrow (y,-x)\\ A(-4,5) & \rightarrow A'(5,4)\\ B(1,2) & \rightarrow B'(2,-1)\\ C(-6,-2) & \rightarrow C'(-2,6)\\ D(-8,3) & \rightarrow D'(3,8)\end{align*}

While we can rotate any image any amount of degrees, only \begin{align*}90^\circ, 180^\circ\end{align*} and \begin{align*}270^\circ\end{align*} have special rules. To rotate a figure by an angle measure other than these three, you must use Investigation 12-1.

**Example 4:** ** Algebra Connection** The rotation of a quadrilateral is shown below. What is the measure of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}?

**Solution:** Because a rotation is an isometry, we can set up two equations to solve for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

\begin{align*}2y &= 80^\circ && \ 2x-3 =15\\ y &= 40^\circ && \qquad 2x = 18\\ &&& \qquad \ \ x = 9\end{align*}

**Know What? Revisited** The center of rotation is shown in the picture below. If we draw rays to the same point in each arrow, we see that the two images are a \begin{align*}120^\circ\end{align*} rotation in either direction.

## Review Questions

In the questions below, every rotation is *counterclockwise,* unless otherwise stated.

Using Investigation 12-1, rotate each figure around point \begin{align*}P\end{align*} the given angle measure.

- \begin{align*}50^\circ\end{align*}
- \begin{align*}120^\circ\end{align*}
- \begin{align*}200^\circ\end{align*}
- If you rotated the letter \begin{align*}p \ 180^\circ\end{align*} counterclockwise, what letter would you have?
- If you rotated the letter \begin{align*}p \ 180^\circ\end{align*}
*clockwise,*what letter would you have? Why do you think that is? - A \begin{align*}90^\circ\end{align*} clockwise rotation is the same as what counterclockwise rotation?
- A \begin{align*}270^\circ\end{align*} clockwise rotation is the same as what counterclockwise rotation?
- Rotating a figure \begin{align*}360^\circ\end{align*} is the same as what other rotation?

Rotate each figure in the coordinate plane the given angle measure. The center of rotation is the origin.

- \begin{align*}180^\circ\end{align*}
- \begin{align*}90^\circ\end{align*}
- \begin{align*}180^\circ\end{align*}
- \begin{align*}270^\circ\end{align*}
- \begin{align*}90^\circ\end{align*}
- \begin{align*}270^\circ\end{align*}
- \begin{align*}180^\circ\end{align*}
- \begin{align*}270^\circ\end{align*}
- \begin{align*}90^\circ\end{align*}

** Algebra Connection** Find the measure of \begin{align*}x\end{align*} in the rotations below. The blue figure is the preimage.

Find the angle of rotation for the graphs below. The center of rotation is the origin and the blue figure is the preimage.

** Two Reflections** The vertices of \begin{align*}\triangle GHI\end{align*} are \begin{align*}G(-2, 2), H(8, 2)\end{align*} and \begin{align*}I(6, 8)\end{align*}. Use this information to answer questions 24-27.

- Plot \begin{align*}\triangle GHI\end{align*} on the coordinate plane.
- Reflect \begin{align*}\triangle GHI\end{align*} over the \begin{align*}x-\end{align*}axis. Find the coordinates of \begin{align*}\triangle G'H'I'\end{align*}.
- Reflect \begin{align*}\triangle G'H'I'\end{align*} over the \begin{align*}y-\end{align*}axis. Find the coordinates of \begin{align*}\triangle G''H''I''\end{align*}.
- What
transformation would be the same as this double reflection?*one*

*Multistep Construction Problem*

- Draw two lines that intersect, \begin{align*}m\end{align*} and \begin{align*}n\end{align*}, and \begin{align*}\triangle ABC\end{align*}. Reflect \begin{align*}\triangle ABC\end{align*} over line \begin{align*}m\end{align*} to make \begin{align*}\triangle A'B'C'\end{align*}. Reflect \begin{align*}\triangle A'B'C'\end{align*} over line \begin{align*}n\end{align*} to get \begin{align*}\triangle A''B''C''\end{align*}. Make sure \begin{align*}\triangle ABC\end{align*} does not intersect either line.
- Draw segments from the intersection point of lines \begin{align*}m\end{align*} and \begin{align*}n\end{align*} to \begin{align*}A\end{align*} and \begin{align*}A''\end{align*}. Measure the angle between these segments. This is the angle of rotation between \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle A''B''C''\end{align*}.
- Measure the angle between lines \begin{align*}m\end{align*} and \begin{align*}n\end{align*}. Make sure it is the angle which contains \begin{align*}\triangle A'B'C'\end{align*} in the interior of the angle.
- What is the relationship between the angle of rotation and the angle between the two lines of reflection?

## Review Queue Answers

- \begin{align*}X'(-9, 2), Y'(-2, 4), Z'(-7, 8)\end{align*}
- \begin{align*}X''(-9, -2), Y''(-2, -4), Z''(-7, -8)\end{align*}
- \begin{align*}\triangle X''Y''Z''\end{align*} is the double negative of \begin{align*}\triangle XYZ; (x, y) \rightarrow (-x, -y)\end{align*}

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## Date Created:

Feb 23, 2012## Last Modified:

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