Points A and B are at (4 ,5 ) and (7 ,2 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/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?

Apr 3, 2016

$C \left(9 , 8\right)$

Explanation:

Given: $A \left(4 , 5\right) \mathmr{and} B \left(7 , 2\right)$ A is rotated by $3 \frac{\pi}{2}$ and dilated by a factor of 1/2 about a point C. After rotation and dilation A's new location is on B, that is ${A}^{R D} \left(7 , 2\right)$.
Required: the coordinates of C such that ${A}^{R D} \left(7 , 2\right) = B \left(7 , 2\right)$
Solution Strategy: a) Rotate $A$, by $R \left(\frac{3}{2} \pi\right)$
b) Using Dilation about a $C$ construct and knowing ${A}^{R D}$ solve C
a) Rotation of $A$ by $\frac{3}{2} \pi$
${A}^{R} = R \left(\frac{3}{2} \pi\right) A$
$R \left(\frac{3}{2} \pi\right) = \left[\begin{matrix}\cos \left(\frac{3}{2} \pi\right) & - \sin \left(\frac{3}{2} \pi\right) \\ \sin \left(\frac{3}{2} \pi\right) & \cos \left(\frac{3}{2} \pi\right)\end{matrix}\right] = \left[\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right]$
${A}^{R} = \left[\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right] \left[\begin{matrix}4 \\ 5\end{matrix}\right] = \left[\begin{matrix}5 \\ - 4\end{matrix}\right]$
b) In order to dilate about C(x,y) we need to do do the following:
i) Translation ${A}^{R}$ by $C \left(x , y\right)$
ii) Dilate by 1/2,
iii) Undo the translation:
Putting i), ii) and iii) we can wrtite:
${A}^{R D} = \left[\begin{matrix}7 \\ 2\end{matrix}\right] = \left[\begin{matrix}\frac{1}{2} \left(5 - x\right) + x \\ \frac{1}{2} \left(- 4 - y\right) + y\end{matrix}\right]$ solve for $x \mathmr{and} y$
7=5/2-x/2+x; x=9
2=-2-y/2+y; y=8
$C \left(9 , 8\right)$