Answer: Dilation of 1/2, translation 5 1/2 left and 3 down, rotation 90° clockwise about point Z
Step-by-step explanation:
Consider W = (2, 5)
X = (6, 5)
Y = (5, 2)
Z = (3, 2)
It was easier to name the points for the step-by-step illustration.
Step 1: Dilation of 1/2:
[tex]W' = \dfrac{1}{2}(2, 5)\quad=\bigg(1,2\dfrac{1}{2}\bigg)\\\\X' = \dfrac{1}{2}(6, 5)\quad=\bigg(3,2\dfrac{1}{2}\bigg)\\\\Y' = \dfrac{1}{2}(5, 2)\quad=\bigg(2\dfrac{1}{2},1\bigg)\\\\Z' = \dfrac{1}{2}(3, 2)\quad=\bigg(1\dfrac{1}{2},1 \bigg)[/tex]
Step 2: Translation 5 1/2 units left and 3 units down
[tex]W'' = \bigg(1-5\dfrac{1}{2},2\dfrac{1}{2}-3\bigg)=\bigg(-4\dfrac{1}{2},-\dfrac{1}{2}\bigg)\\\\X'' = \bigg(3-5\dfrac{1}{2},2\dfrac{1}{2}-3\bigg)=\bigg(-2\dfrac{1}{2},-\dfrac{1}{2}\bigg)\\\\Y'' = \bigg(2\dfrac{1}{2}-5\dfrac{1}{2},1-3\bigg)=\bigg(-3,-2\bigg)\\\\Z'' = \bigg(1\dfrac{1}{2}-5\dfrac{1}{2},1-3 \bigg)=\bigg(-4,-2\bigg)[/tex]
Step 3: Rotate 90° clockwise about point Z
[tex]W''' =90^o_{Zcc}\bigg(-4\dfrac{1}{2},-\dfrac{1}{2}\bigg)=\bigg(-2\dfrac{1}{2},-1\dfrac{1}{2}\bigg)\\\\X''' = 90^o_{Zcc}\bigg(-2\dfrac{1}{2},-\dfrac{1}{2}\bigg)=\bigg(-2\dfrac{1}{2},-3\dfrac{1}{2}\bigg)\\\\Y''' = 90^o_{Zcc}\bigg(-3,-2\bigg)=\bigg(-4,-2\bigg)\\\\Z''' =90^o_{Zcc}\bigg(-4,-2\bigg)=\bigg(-4,-3\bigg)[/tex]
Using these three steps, you have transformed the coordinates of C into the coordinates of D.
