Question

A triangle has corners at `(-3 ,5 )`, `(7 ,6 )`, and `(1 ,-4 )`. If the triangle is dilated by a factor of `2/3` about point #(-3 ,4 ), how far will its centroid move?

Answer 1

Thus the new corners of the dilated triangle `A'B'C"` are
`A'(-3, 14/3), B'(11/3, 16/3), C'(-1/3, -4/3)`

Explanation:

For each vertex we need to find the dilation of that point by 2/3 around (-3,4):
`A(-3,5), B(7, 6), C(1, -4)`
Dilate `B'=D_((-3,4), 2/3) *B' (7,6)=>[7-(-3), 6-4] *2/3 + [-3, 4]= (11/3, 16/3)`
`A'=D_((-3,4), 2/3)* A'(-3,5) =>[-3-(-3), 5-4] *2/3 + [-3, 4]= (-3, 14/3)`
`C'=D_((-3,4), 2/3) *C'(1,-4) =>[1-(-3), -4-4] *2/3 + [-3, 4]= (-1/3, -4/3)`

Thus the new corners of the dilated triangle `A'B'C"` are
`A'(-3, 14/3), B'(11/3, 16/3), C'(-1/3, -4/3)`