Question

Triangle XYZ
𝑋
𝑌
𝑍
is drawn on a coordinate plane with vertices X(0,4)
𝑋
(
0
,
4
)
, Y(2,0)
𝑌
(
2
,
0
)
, and Z(4,7)
𝑍
(
4
,
7
)
. If you dilate the triangle to form triangle X′Y′Z′
𝑋
′
𝑌
′
𝑍
′
using a magnitude of 12
1
2
and a center of dilation at (4,7)
(
4
,
7
)
, what are the coordinate points of the new vertices?(1 point)
Responses

X′(4,11)
𝑋
′
(
4
,
11
)
, Y′(6,7)
𝑌
′
(
6
,
7
)
, Z′(8,14)
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)
𝑋
′
(
−
4
,
−
3
)
, Y′(−2,−7)
𝑌
′
(
−
2
,
−
7
)
, Z′(0,0)
𝑍
′
(
0
,
0
)

upper X prime left parenthesis negative 4 comma negative 3 right parenthesis , upper Y prime left parenthesis negative 2 comma

Answer 1

Use P' = C + k(P − C) with center C = (4,7) and k = 12.5.

X: vector (0−4, 4−7) = (−4, −3). Multiply by 12.5: (−50, −37.5). Add center: X' = (4−50, 7−37.5) = (−46, −30.5) = (−46, −61/2).

Y: vector (2−4, 0−7) = (−2, −7). ×12.5 = (−25, −87.5). Add center: Y' = (4−25, 7−87.5) = (−21, −80.5) = (−21, −161/2).

Z is the center, so Z' = (4,7).

Answer: X'(−46, −30.5), Y'(−21, −80.5), Z'(4,7).