Question

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

Answer 1

The displacement from old centroid to the new centroid is
`d=sqrt596`
`d=24.413" "`units

Explanation:

From the given, let `P_1(x_1, y_1)=(3, 1)` and `P_2(x_2, y_2)=(5, 2)` and `P_3(x_3, y_3)=(9, 4)` and `R(1, 9)`

Compute the centroid `C_1(x_c, y_c)` of the triangle first

`x_c=(x_1+x_2+x_3)/3=(3+5+9)/3=17/3`
`y_c=(y_1+y_2+y_3)/3=(1+2+4)/3=7/3`

The centroid `C_1(x_c, y_c)=(17/3, 7/3)`

Determine the new centroid `C_2(x_c', y_c')`

The ratio:
`(C_2 R)/(C_1 R)=4/1`
Find `x_c'`
`(x_c'-1)/(17/3-1)=4/1`

`x_c'=1+4(17/3-1)`

`x_c'=1+56/3`

`x_c'=59/3`

Find `y_c'`

`(y_c'-9)/(7/3-9)=4/1`

`y_c'=9+4(7/3-9)`

`y_c'=9+(4(7-27))/3`
`y_c'=9-80/3`
`y_c'=(27-80)/3`

`y_c'=-53/3`

The New Centroid is at `C_2(59/3, -53/3)=(19.67, -17.67)`

Let us determine the displacement `d`

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

`d=sqrt((59/3-17/3)^2+(-53/3-7/3)^2)`

`d=sqrt((59/3-17/3)^2+(-53/3-7/3)^2)`

`d=sqrt(14^2+(-20)^2)`
`d=sqrt596`
`d=24.413" "`units

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