Question

A triangle has corners at `(6 ,3 )`, `(4 ,-1 )`, and `(2 ,-9 )`. If the triangle is dilated by a factor of `5` about point #(2 ,-4 ), how far will its centroid move?

Answer 1

It will move by distance`=4/3sqrt61=10.4137" "`units

Explanation:

We need to compute the old centroid and the new centroid first.

We compute the old centroid `(x_c, y_x)` first

`x_c=(x_1+x_2+x_3)/3=(6+2+4)/3=4`

`y_c=(y_1+y_2+y_3)/3=(3+(-9)+(-1))/3=-7/3`

Old centroid `(x_c, y_x)=(4, -7/3)`

Now we compute the new centroid `(x_c', y_c')`

The given problem is about dilated by a factor of 5, therefore the ratio of the distance of the new centroid from the reference point `(2, -4)` to the distance of the old centroid from the reference point is `5:1`

For `x_c'`

`(x_c'-2)/(x_c-2)=5/1`

`(x_c'-2)/(4-2)=5/1`

`x_c'=12`

For `y_c'`

`(y_c'--4)/(y_c--4)=5/1`

`(y_c'--4)/(-7/3--4)=5/1`

`y_c'=13/3`

New centroid `(x_c', y_c')=(12, 13/3)`

Distance `d` from old centroid to the new centroid

`d=sqrt((x_c-x_c')^2+(y_c-y_c')^2)`

`d=sqrt((4-12)^2+(-7/3-13/3)^2)`

`d=sqrt(976/9)=4/3sqrt(61)=10.4137`

God bless...I hope the explanation is useful.