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′(−4,1)
, 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)
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,−3)
, Y′(−2,−7)
, Z′(0,0)

Answer 1

Assuming the magnitude is 1/2, dilate each point about center C(4,7) by k = 1/2:

- X − C = (0−4, 4−7) = (−4, −3). Multiply by 1/2 → (−2, −1.5). Add to C → X′ = (4−2, 7−1.5) = (2, 5.5).
- Y − C = (2−4, 0−7) = (−2, −7). Multiply by 1/2 → (−1, −3.5). Add to C → Y′ = (3, 3.5).
- Z is the center so Z′ = (4, 7).

So X′(2, 5.5), Y′(3, 3.5), Z′(4, 7).