Points A and B are at (6 ,7 ) and (3 ,9 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/2  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?

Oct 21, 2016

we know , if in two dimension the rotation of a point (x,y) about origin by an angle $\theta$ in anticlockwise direction transforms its coordinates into (x',y') then

$x ' = x \cos \theta - y \sin \theta$

$y ' = x \sin \theta + y \cos \theta$

Here $\theta = \frac{3 \pi}{2}$
$\cos \theta = \cos \left(\frac{3 \pi}{2}\right) = 0 \mathmr{and} \sin \theta = \sin \left(\frac{3 \pi}{2}\right) = - 1$

So transformed coordinates of $A \to \left(6 , 7\right)$ will be

A'->((6*0-7*(-1)),(6*(-1))+7*0))

$= \left(7 , - 6\right)$

Similarly transformed coordinates of $B \to \left(3 , 9\right)$ will be

B'->((3*0-9*(-1)),(3*(-1))+9*0))

$= \left(9 , - 3\right)$
Let the coordinates of center of dilation C be (h,k).

So A' on 5 times dilation about C will be transformed into

$A {'}_{\text{5xdilated}} = \left(5 \left(7 - h\right) + h , 5 \left(- 6 - k\right) + k\right)$

By the given condition $A {'}_{\text{5xdilated}} = B$
So

$5 \left(7 - h\right) + h = 3 \implies 4 h = 32 \implies h = 8$

Again

$5 \left(- 6 - k\right) + k = 9 \implies 4 k = - 39 \implies k = - \frac{39}{4}$

Hence coordinates of $C \to \left(8 , - \frac{39}{4}\right)$