# Points A and B are at (4 ,1 ) and (8 ,3 ), respectively. Point A is rotated counterclockwise about the origin by pi/2  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?

Mar 27, 2017

The point $C = \left(- 10 , 5\right)$

#### Explanation:

The points are the following

$A = \left(4 , 1\right)$

$B = \left(8 , 3\right)$

After the rotation counterclockwise about the origin, we get the point

$A ' = \left(- 1 , 4\right)$

Let the point

$C = \left(x , y\right)$

Then,

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

$< - 1 - x , 4 - y \ge \frac{1}{2} < 8 - x , 3 - y >$

Therefore,

$2 \left(- 1 - x\right) = 8 - x$

$- 2 - 2 x = 8 - x$

$x = - 2 - 8 = - 10$

and

$2 \left(4 - y\right) = 3 - y$

$8 - 2 y = 3 - y$

$y = 8 - 3 = 5$

The point $C = \left(- 10 , 5\right)$

graph{((x-4)^2+(y-1)^2-0.01)((x-8)^2+(y-3)^2-0.01)((x+1)^2+(y-4)^2-0.01)((x+10)^2+(y-5)^2-0.01)=0 [-14, 18.06, -5.07, 10.93]}

Mar 27, 2017

$C = \left(17 , 2\right)$

#### Explanation:

Under a counterclockwise rotation about the origin of $\frac{\pi}{2}$

• " a point " (x,y)to(-y,x)

$\Rightarrow A \left(4 , 1\right) \to A ' \left(- 1 , 4\right) \text{ where A' is the image of A}$

$\text{Under a dilatation about C of factor } \frac{1}{2}$

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

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

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

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

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

$\textcolor{w h i t e}{\Rightarrow \frac{1}{2} c} = \left(\begin{matrix}8 \\ 3\end{matrix}\right) - \frac{1}{2} \left(\begin{matrix}- 1 \\ 4\end{matrix}\right)$

$\textcolor{w h i t e}{r a A r r \frac{1}{2} c} = \left(\begin{matrix}\frac{17}{2} \\ 1\end{matrix}\right)$

$\Rightarrow \underline{c} = 2 \left(\begin{matrix}\frac{17}{2} \\ 1\end{matrix}\right) = \left(\begin{matrix}17 \\ 2\end{matrix}\right)$

$\Rightarrow C = \left(17 , 2\right)$