Question

The coordinates of trapezoid EFGH are E(4, 6), F(2, 3), G(4, 2), and H(8, 4). If the image is dilated to E’F’G’H’ and G’ is (8,4) what is the coordinate of H’?
I want to know how to do it, step by step please!!

Answer 1

G=(4,2), G'=(8,4)
G'/G=((8/4),(4/2))
G'/G=2
The dilation scale factor is 2, so all lengths will be doubled.
Original distance from G to H: ((8-4)^2+(4-2)^2)^(1/2)
=(4^2+2^2)^(1/2)
=(20)^(1/2)
New distance=2((20)^(1/2))
Original x distance=G+4
Original y distance=G+2
New x distance=G'+8
New y distance=G'+4
x=8+8
y=4+4
H'=(16,8)