# Points A and B are at (3 ,7 ) and (7 ,2 ), respectively. Point A is rotated counterclockwise about the origin by pi  and dilated about point C by a factor of 5 . If point A is now at point B, what are the coordinates of point C?

Mar 21, 2017

The point $C$ is $\left(- 1 , - \frac{24}{5}\right)$

#### Explanation:

The point $A '$ is symmetric about the origin $O$

The coordinates of $A '$ is $= \left(- 3 , - 7\right)$

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

Then,

$\vec{A ' B} = 5 \cdot \vec{A ' C}$

$\vec{A ' B} = < 7 - \left(- 3\right) , 2 - \left(- 7\right) \ge < 10 , 9 >$

$\vec{A ' C} = < x - \left(- 3\right) , y - \left(- 7\right) > = < x + 3 , y + 7 >$

Therefore,

$5 \cdot < x + 3 , y + 7 > = < 10 , 9 >$

So,

$5 \left(x + 3\right) = 10$

$5 x + 15 = 10$

$5 x = - 5$

$x = - 1$

and

$5 \left(y + 7\right) = 9$

$5 y + 35 = 9$

$5 y = 9 - 35 = - 24$

$y = - \frac{24}{5}$

Therefore,

The point $C$ is $\left(- 1 , - \frac{24}{5}\right)$

Mar 21, 2017

$C = \left(- \frac{11}{2} , - \frac{37}{4}\right)$

#### Explanation:

Under a counterclockwise rotation about the origin of $\pi$

• "a point " (x,y)to(-x,-y)

$\Rightarrow A \left(3 , 7\right) \to A ' \left(- 3 , - 7\right)$ where A' is the image of A.

$\text{ Under a dilatation about C of factor 5}$

Taking a $\textcolor{b l u e}{\text{vector approach}}$

$\Rightarrow \vec{C B} = 5 \vec{C A '}$

$\Rightarrow \underline{b} - \underline{c} = 5 \left(\underline{a} ' - \underline{c}\right)$

$\Rightarrow \underline{b} - \underline{c} = 5 \underline{a} ' - 5 \underline{c}$

$\Rightarrow 4 \underline{c} = 5 \underline{a} ' - \underline{b}$

$\textcolor{w h i t e}{\Rightarrow 4 c} = 5 \left(\begin{matrix}- 3 \\ - 7\end{matrix}\right) - \left(\begin{matrix}7 \\ 2\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow 4 c} = \left(\begin{matrix}- 15 \\ - 35\end{matrix}\right) - \left(\begin{matrix}7 \\ 2\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow 4 c} = \left(\begin{matrix}- 22 \\ - 37\end{matrix}\right)$

$\Rightarrow \underline{c} = \frac{1}{4} \left(\begin{matrix}- 22 \\ - 37\end{matrix}\right) = \left(\begin{matrix}- \frac{11}{2} \\ - \frac{37}{4}\end{matrix}\right)$

$\Rightarrow C = \left(- \frac{11}{2} , - \frac{37}{4}\right)$