Question

Triangle $XYZ$ is drawn on a coordinate plane with vertices $X(0, 4)$, $Y(2, 0)$, and $Z(4, 7)$. If you dilate the triangle to form triangle $X'Y'Z'$ using a magnitude of $\frac{1}{2}$ and a center of dilation at $(4, 7)$, what are the coordinate points of the new vertices? (1 point)Responses(x) $X'(2, 5.5)$, $Y'(3, 3.5)$, $Z'(4, 7)$( ) $X'(-4, 1)$, $Y'(0, -7)$, $Z'(4, 7)$( ) $X'(4, 11)$, $Y'(6, 7)$, $Z'(8, 14)$( ) $X'(-4, -3)$, $Y'(-2, -7)$, $Z'(0, 0)$

Answer 1

Use P' = center + (1/2)(P − center).

X': (4,7) + 1/2((0,4)−(4,7)) = (4,7) + 1/2(−4,−3) = (2,5.5)
Y': (4,7) + 1/2((2,0)−(4,7)) = (4,7) + 1/2(−2,−7) = (3,3.5)
Z' = center = (4,7)

So the correct choice is: X'(2,5.5), Y'(3,3.5), Z'(4,7).