# Points A and B are at (2 ,1 ) and (4 ,7 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/2  and dilated about point C by a factor of 3 . If point A is now at point B, what are the coordinates of point C?

Jun 21, 2017

The coordinates of point $C = \left(- \frac{1}{2} , - \frac{13}{2}\right)$

#### Explanation:

The matrix of a rotation counterclockwise by $\frac{3}{2} \pi$ about the origin is

$\left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right)$

Therefore, the transformation of point $A$ is

$A ' = \left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) \left(\begin{matrix}2 \\ 1\end{matrix}\right) = \left(\begin{matrix}1 \\ - 2\end{matrix}\right)$

Let point $C$ be $\left(x , y\right)$, then

$\vec{C B} = 2 \vec{C A '}$

$\left(\begin{matrix}4 - x \\ 7 - y\end{matrix}\right) = 3 \left(\begin{matrix}1 - x \\ - 2 - y\end{matrix}\right)$

So,

$4 - x = 3 \left(1 - x\right)$

$4 - x = 3 - 3 x$

$2 x = - 1$

$x = - \frac{1}{2}$

and

$7 - y = 3 \left(- 2 - y\right)$

$7 - y = - 6 - 3 y$

$2 y = - 6 - 7$

$y = - \frac{13}{2}$

Therefore,

point $C = \left(- \frac{1}{2} , - \frac{13}{2}\right)$

Jun 21, 2017

$C = \left(- \frac{1}{2} , - \frac{13}{2}\right)$

#### Explanation:

$\text{under a counterclockwise rotation about the origin of } \frac{3 \pi}{2}$

• " a point " (x,y)to(y,-x)

$\Rightarrow A \left(2 , 1\right) \to A ' \left(1 , - 2\right) \text{ where A' is the image of A}$

$\text{under a dilatation about C of factor 3}$

$\vec{C B} = \textcolor{red}{3} \vec{C A '}$

$\Rightarrow \underline{b} - \underline{c} = \textcolor{red}{3} \left(\underline{a} ' - \underline{c}\right)$

$\Rightarrow \underline{b} - \underline{c} = 3 \underline{a} ' - 3 \underline{c}$

$\Rightarrow 2 \underline{c} = 3 \underline{a} ' - \underline{b}$

$\textcolor{w h i t e}{\Rightarrow 2 \underline{c}} = 3 \left(\begin{matrix}1 \\ - 2\end{matrix}\right) - \left(\begin{matrix}4 \\ 7\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow 2 \underline{c}} = \left(\begin{matrix}3 \\ - 6\end{matrix}\right) - \left(\begin{matrix}4 \\ 7\end{matrix}\right) = \left(\begin{matrix}- 1 \\ - 13\end{matrix}\right)$

$\Rightarrow \underline{c} = \frac{1}{2} \left(\begin{matrix}- 1 \\ - 13\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{2} \\ - \frac{13}{2}\end{matrix}\right)$

$\text{the components of " ulc" are the coordinates of C}$

$\Rightarrow C = \left(- \frac{1}{2} , - \frac{13}{2}\right)$