Question

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

Answer 1

The centroid will move by about `d=4/3sqrt233=20.35245" "`units

Explanation:

We have a triangle with vertices or corners at the points `A(-6, 3)`and `B(3, -2)` and `C(5, 4)`.

Let `F(x_f, y_f)=F(-2, 6)" "`the fixed point

Compute the centroid `O(x_g, y_g)` of this triangle, we have

`x_g=(x_a+x_b+x_c)/3=(-6+3+5)/3=2/3`
`y_g=(y_a+y_b+y_c)/3=(3+(-2)+4)/3=5/3`

Centroid `O(x_g, y_g)=O(2/3, 5/3)`

Compute the centroid of the bigger triangle (scale factor =5)

Let `O'(x_g', y_g')=`the centroid of the bigger triangle

the working equation:

`(FO')/(FO)=5`

solve for `x_g'`:

`(x_g'--2)/(2/3--2)=5`

`(x_g'+2)=5*8/3`
`x_g'=40/3-2`
`x_g'=34/3`

solve for `y_g'`

`(y_g'-6)/(5/3-6)=5`

`y_g'=6+5(-13/3)=(18-65)/3`

`y_g'=-47/3`

Compute now the distance from centroid O(2/3, 5/3) to new centroid O'(34/3, -47/3).

`d=sqrt((x_g-x_g')^2+(y_g-y_g')^2)`

`d=sqrt((2/3-34/3')^2+(5/3--47/3)^2)`

`d=sqrt((-32/3)^2+(52/3)^2)`

`d=sqrt(((-4*8)/3)^2+((4*13)/3)^2)`

`d=4/3*sqrt(64+169)`

`d=4/3*sqrt(233)=20.35245`

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