Question

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?

Answer 1

The point `C=(-10,5)`

Explanation:

The points are the following

`A=(4,1)`

`B=(8,3)`

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

`A'=(-1,4)`

Let the point

`C=(x,y)`

Then,

`vec(CA')=1/2vec(CB)`

`<-1-x,4-y>=1/2<8-x,3-y>`

Therefore,

`2(-1-x)=8-x`

`-2-2x=8-x`

`x=-2-8=-10`

and

`2(4-y)=3-y`

`8-2y=3-y`

`y=8-3=5`

The point `C=(-10,5)`

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]}

Answer 2

`C=(17,2)`

Explanation:

Under a counterclockwise rotation about the origin of `pi/2`

`• " a point " (x,y)to(-y,x)`

`rArrA(4,1)toA'(-1,4)" where A' is the image of A"`

`"Under a dilatation about C of factor "1/2`

Taking a `color(blue)"vector approach"`

`rArrvec(CB)=1/2vec(CA')`

`rArrulb-ulc=1/2(ula'-ulc)`

`rArrulb-ulc=1/2ula'-1/2ulc`

`rArr1/2ulc=ulb-1/2ula`

`color(white)(rArr1/2c)=((8),(3))-1/2((-1),(4))`

`color(white)(raArr1/2c)=((17/2),(1))`

`rArrulc=2((17/2),(1))=((17),(2))`

`rArrC=(17,2)`