Answer: Dilation of 1/2, translation 5 1/2 left and 3 down, rotation 90° clockwise about point Z
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:
![W' = (1)/(2)(2, 5)\quad=\bigg(1,2(1)/(2)\bigg)\\\\X' = (1)/(2)(6, 5)\quad=\bigg(3,2(1)/(2)\bigg)\\\\Y' = (1)/(2)(5, 2)\quad=\bigg(2(1)/(2),1\bigg)\\\\Z' = (1)/(2)(3, 2)\quad=\bigg(1(1)/(2),1 \bigg)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ed1uk5v4041dbx6n93o188o1p3ikn12g0a.png)
Step 2: Translation 5 1/2 units left and 3 units down
![W'' = \bigg(1-5(1)/(2),2(1)/(2)-3\bigg)=\bigg(-4(1)/(2),-(1)/(2)\bigg)\\\\X'' = \bigg(3-5(1)/(2),2(1)/(2)-3\bigg)=\bigg(-2(1)/(2),-(1)/(2)\bigg)\\\\Y'' = \bigg(2(1)/(2)-5(1)/(2),1-3\bigg)=\bigg(-3,-2\bigg)\\\\Z'' = \bigg(1(1)/(2)-5(1)/(2),1-3 \bigg)=\bigg(-4,-2\bigg)](https://img.qammunity.org/2021/formulas/mathematics/high-school/j1ohkum1aji3mw46krmzpuos04a3h0q8ht.png)
Step 3: Rotate 90° clockwise about point Z
![W''' =90^o_(Zcc)\bigg(-4(1)/(2),-(1)/(2)\bigg)=\bigg(-2(1)/(2),-1(1)/(2)\bigg)\\\\X''' = 90^o_(Zcc)\bigg(-2(1)/(2),-(1)/(2)\bigg)=\bigg(-2(1)/(2),-3(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)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rp7g6yfy2juu6rn17urxkx1i2tsjr227tu.png)
Using these three steps, you have transformed the coordinates of C into the coordinates of D.