Points A and B are at `(1 ,1 )` and `(4 ,6 )`, 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?

First, rotate `A` around the origin like the problem describes. The rotation angle in degrees is:

`(3cancel(pi))/2*(180º)/cancel(pi) =`

`3/2*180º = 270º`

A 270º rotation is three-fourths of a circle, so the point `A(1, 1)` becomes `A'(1,-1)`.

Now, we can figure out the equation on the line between `A'` and `B` using the slope and point-slope formulae:

`m = (y_2-y_1)/(x_2-x_1)=>`

`m = (-1-6)/(1-4)=(-7)/(-3)=7/3`

`y-y_1=m(x-x_1)=>`

`y-6=7/3(x-4)`

`y-6=7/3x-28/3`

`y=7/3x-10/3`

Since the dilation from point `C` was a factor of two, we know that the change in `x` from `A'` to `B` is equal to the change in `x` from `C` to `A'`.

The change in `x` from `A'` to `B` is `|1-4|=3`
We now know that the `x`-coordinate of point `C` is `x`-coordinate of point `A'` - 3, or `1-3=-2` .

We can plug in -2 to our line that we found to get the `y`-coordinate:

`y=7/3x-10/3=>`

`y=7/3(-2)-10/3`

`y=(-14)/3-10/3`

`y=-24/3`

`y=-8`

Therefore, the point `C` is `(-2, -8)`.

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

`• " a point "(x,y)to(y,-x)`

`rArra(1,1)toA'(1,-1)" 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((1),(-1))-((4),(6))`

`color(white)(rArrulc)=((2),(-2))-((4),(6))=((-2),(-8))`

`rArrC=(-2,-8)`