# Points A and B are at (2 ,6 ) and (1 ,9 ), respectively. Point A is rotated counterclockwise about the origin by pi  and dilated about point C by a factor of 1/2 . If point A is now at point B, what are the coordinates of point C?

Apr 21, 2018

$C = \left(1 , 6\right)$

#### Explanation:

$\text{under a counterclockwise rotation about the origin of } \pi$

• " a point "(x,y)to(-x,-y)

$\Rightarrow A \left(2 , 6\right) \to A ' \left(- 2 , - 6\right) \text{ where A' is the image of A}$

$\Rightarrow \vec{C B} = \textcolor{red}{\frac{1}{2}} \vec{C A '}$

$\Rightarrow \underline{b} - \underline{c} = \frac{1}{2} \left(\underline{a} ' - \underline{c}\right)$

$\Rightarrow \underline{b} - \underline{c} = \frac{1}{2} \underline{a} ' - \frac{1}{2} \underline{c}$

$\Rightarrow \frac{1}{2} \underline{c} = \underline{b} - \frac{1}{2} \underline{a} '$

$\textcolor{w h i t e}{\Rightarrow \frac{1}{2} \underline{c}} = \left(\begin{matrix}1 \\ 9\end{matrix}\right) - \frac{1}{2} \left(\begin{matrix}- 2 \\ - 6\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{1}{2} \underline{c}} = \left(\begin{matrix}1 \\ 9\end{matrix}\right) - \left(\begin{matrix}- 1 \\ - 3\end{matrix}\right) = \left(\begin{matrix}2 \\ 12\end{matrix}\right)$

$\Rightarrow \underline{c} = \frac{1}{2} \left(\begin{matrix}2 \\ 12\end{matrix}\right) = \left(\begin{matrix}1 \\ 6\end{matrix}\right)$

$\Rightarrow C = \left(1 , 6\right)$