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′(−4,1)
, Y′(0,−7)
, Z′(4,7)

upper X prime left parenthesis negative 4 comma 1 right parenthesis , upper Y prime left parenthesis 0 comma negative 7 right parenthesis , upper Z prime left parenthesis 4 comma 7 right parenthesis

X′(2,5.5)
, Y′(3,3.5)
, Z′(4,7)

Answer 1

Use P' = C + k(P − C) with center C=(4,7) and scale k=1/2.

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) (center stays fixed)

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