Question

A triangle has corners at `(8 ,3 )`, `(4 ,-5 )`, and `(2 ,1 )`. If the triangle is dilated by a factor of `5` about point #(1 ,-3 ), how far will its centroid move?

Answer 1

The centroid will move by `=18.1u`

Explanation:

Let the corners of the triangle be `(x_1,y_1)`, `(x_2,y_2)` and `(x_3,y_3)`

The coordinates of the centroid are

`C=((x_1+x_2+x_3)/2,(y_1+y_2+y_3)/3)`

Here, we have `(8,3)`, `(4,-5)`, and `(2,1)`

So,

The coordinates of the centroid are `C=((8+4+2)/3,(3-5+1)/3)=(14/3,-1/3)`

Let the coordinates of the centroid after dilatation be `C'=(x,y)`

The fixed point is `D=(1,-3)`

Therefore,

`vec(DC')=5vec(DC)`

`((x-1),(y+3))=5((14/3-1),(-1/3+3))=5((11/3),(8/3))=((55/3),(40/3))`

So,

`x-1=55/3`, `=>`, `x=55/3+1=58/3`

`y+3=40/3`, `=>`, `y=40/3-3=31/3`

So, the coordinates of `C'=(58/3,31/3)`

The distance between the centroids is

`C C'=sqrt((58/3-14/3)^2+(31/3+1/3)^2)`

`=sqrt((44/3)^2+(32/3)^2)`

`=sqrt(44^2+32^2)/3`

`=18.1u`