On a coordinate plane, triangle A B C has points (negative 2, 2), (negative 1, negative 2), and (negative 3, negative 2). Point F is at (1, 1).
The dilation rule DF,3(x, y) is applied to △ABC, where the center of dilation is at F(1, 1).
The distance in the x-coordinates from A(–2, 2) to the center of dilation F(1, 1) is
3
unit(s).
The distance in the y-coordinates from A(–2, 2) to the center of dilation F(1, 1) is
2
unit(s).
The vertex A' of the image is
.
Answers
Given : On a coordinate plane, triangle A B C has points
(-2 , 2) , (-1 , - 2) , ( -3 , - 2)
Point F is at ( 1 , 1)
The dilation rule DF,3(x, y) is applied to △ABC, where the center of dilation is at F(1, 1).
To Find : The distance in the x-coordinates from A(–2, 2) to the center of dilation F(1, 1)
The distance in the y-coordinates from A(–2, 2) to the center of dilation
The vertex A' of the image is
Solution:
The distance in the x-coordinates from A(–2, 2) to the center of dilation F(1, 1)
= | - 2 - 1 |
= 3 units
The distance in the y-coordinates from A(–2, 2) to the center of dilation F(1, 1)
= | 2 - 1|
= 1 unit
A = (-2 , 2)
F = 1 ( 1 , 1)
Dilation factor = 3
A' = ( - 2 - 1) * 3 + 1 , (2 - 1) * 3 + 1
A' = -8 , 4
Hence The vertex A' of the image is (-8 , 4)
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