Question

Points A and B are at `(7 ,9 )` and `(6 ,2 )`, respectively. Point A is rotated counterclockwise about the origin by `(3pi)/2` and dilated about point C by a factor of `2`. If point A is now at point B, what are the coordinates of point C?

Answer 1

The image of `A` rotated by `{3pi}/2` is `D(9,-7).` We get `B=r(D-C)+C` and we solve giving `C=(12, -16).`

Explanation:

I don't get why these come up as recently asked when they're two years old.

`{3pi}/2 = 270^circ`

The image of rotating `A(7,9)` by `270^circ` around the origin lands in the fourth quadrant, so must be `D=(9,-7)`.

Let's see where a point D ends up after dilation around C by a factor of `r`. One way to think about it is to translate the dilation point to the origin, scale the translate D using `r`, and translate back:

`D ' = r (D-C) + C = (1-r) C + rD`

That is always interesting to me. It's the parametric equation for a line between C (`r=0`) and D (`r = 1`). D' is on that line, `r` times the distance from C than D is. That makes sense given the meaning of dilation.

We have `D'=B` which gives us an equation to solve for C:

`B = (1-r) C + r D`

`B - rD = (1-r) C`

`C = {B-rD}/(1-r)`

`C = ( (6,2) - 2(9,-7) )/(1 - 2) = (12, -16)`

Check:

`(1 - r) C + r D = (-12,16) + (18,-14)=(6,2)=B quad sqrt`

Answer 2

`C=(12,-16)`

Explanation:

`"under a counterclockwise rotation about the origin of "(3pi)/2`

`• " a point "(x,y)to(y,-x)`

`rArrA(7,9)toA'(9,-7)" where A' is the image of A"`

`rArrvec(CB)=color(red)(2)vec(CA')`

`rArrulb-ulc=2(ula'-ulc)`

`rArrulb-ulc=2ula'-2ulc`

`rArrulc=2ula'-ulb`

`color(white)(rArrulc)=2((9),(-7))-((6),(2))`

`color(white)(rArrulc)=((18),(-14))-((6),(2))=((12),(-16))`

`rArrC=(12,-16)`