Q28. - Algebra 2

Question

For exercises 26-28, use quadrilateral $QRST$ shown at the right.

28. What type of transformation does the graph represent?

Step-by-Step Solution

Answer

Graph represents the rotation of $180°$ counter clockwise.

Step 1 - Find type of transformation

Types of transformation are

• Translation- When figure in moved from one location to another location without changing its size shape or orientation.
• Dilation- When figure is reduced or enlarged.
• Reflection- When every point of a figure is mapped to a corresponding image across a line of symmetry.
• Rotation- when a figure is moved around a centre point, usually origin.

As given image is moved around a origin, so given transformation is rotation.

Step 2 - Types of rotation

To determine the vertices of image multiply the rotation matrix with vertex matrix of given image

Common rotation matrix are

For a counter clockwise rotation about the origin of:	$90°$	$180°$	$270°$
Multiply the vertex matrix on the left by:	$\left[\begin{array}{cc}0& -1\\ 1& 0\\ & \end{array}\right]$	$\left[\begin{array}{cc}-1& 0\\ 0& -1\\ & \end{array}\right]$	$\left[\begin{array}{cc}0& 1\\ -1& 0\\ & \end{array}\right]$

Step 3 - Find type of rotation

Multiply vertex matrix of pre-image with rotation matrices and compare it with vertex matrix of image

 $\begin{array}{c}\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\times \left[\begin{array}{cccc}2& 4& 2& -3\\ 3& -3& -5& -2\end{array}\right]=\left[\begin{array}{cccc}-3& 3& 5& 2\\ 2& 4& 2& -3\end{array}\right]\left(1\right)\\ \left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\times \left[\begin{array}{cccc}2& 4& 2& -3\\ 3& -3& -5& -2\end{array}\right]=\left[\begin{array}{cccc}-2& -4& -2& 3\\ -3& 3& 5& 2\end{array}\right]\left(2\right)\\ \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\times \left[\begin{array}{cccc}2& 4& 2& -3\\ 3& -3& -5& -2\end{array}\right]=\left[\begin{array}{cccc}3& -3& -5& -2\\ -2& -4& -2& 3\end{array}\right]\left(3\right)\end{array}$

As vertex matrix in product (2) is same as the vertex matrix of image.

So, pre-image is rotated $180°$ counter clockwise.