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 12
and a center of dilation at (4,7)
, what are the coordinate points of the new vertices?(1 point)
Responses

X′(−4,−3)
, Y′(−2,−7)
, Z′(0,0)
upper X prime left parenthesis negative 4 comma negative 3 right parenthesis , upper Y prime left parenthesis negative 2 comma negative 7 right parenthesis , upper Z prime left parenthesis 0 comma 0 right parenthesis

X′(4,11)
, Y′(6,7)
, Z′(8,14)
upper X prime left parenthesis negative 4 comma 11 right parenthesis , upper Y prime left parenthesis 6 comma 7 right parenthesis , upper Z prime left parenthesis 8 comma 14 right parenthesis

X′(2,5.5)
, Y′(3,3.5)
, Z′(4,7)
upper X prime left parenthesis 2 comma 5.5 right parenthesis , upper Y prime left parenthesis 3 comma 3.5 right parenthesis , upper Z prime left parenthesis 4 comma 7 right parenthesis

X′(−4,1)
, Y′(0,−7)
, Z′(4,7)

Answer 1

Use P' = C + k(P − C) with center C = (4,7) and k = 12.

X − C = (0−4, 4−7) = (−4,−3) → X' = (4,7) + 12(−4,−3) = (4−48, 7−36) = (−44, −29)
Y − C = (2−4, 0−7) = (−2,−7) → Y' = (4,7) + 12(−2,−7) = (4−24, 7−84) = (−20, −77)
Z is the center so Z' = (4,7).

Answer: X'(−44,−29), Y'(−20,−77), Z'(4,7).