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′(4,11)
, 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′(−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′(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

Scale factor = 1/2 (the "12" in the prompt is 1/2). Use P' = C + (1/2)(P − C) with center C = (4,7).

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: (4,7) + 1/2[(4,7) − (4,7)] = (4,7)

So X'(2,5.5), Y'(3,3.5), Z'(4,7).