A(-2,2) B(-1,-2), C(-3,-2), F(1,1)
DF,3 is a dilation factor of 3 centered at F(1,1)
The distance in the x-coordinates from A(–2, 2) to the center of dilation F(1, 1) is ____ unit(s).
Here we're asking about A, before dilation. The distance in the x coordinate is the absolute difference of the x coordinate, so from A to F is |-2 - 1| = 3
Answer: 3
The distance in the y-coordinates from A(–2, 2) to the center of dilation F(1, 1) is ____ unit(s).
Here that's the absolute value of the difference of the y coordinate,
|2 - 1| = 1
Answer: 1
The image A' of A(-2,2).
First we translate the plane so F is the origin,
A'' = A - F = (-2,2) - (1,1) = (-3, 1)
Now we dilate by 3,
A'''=(-9, 3)
Now we translate the origin back to F
A' = A''' + F = (-9, 3) + (1, 1) = (-8, 4)
Answer: (-8,4)
We can check that our x and y distances have tripled;
A'-F = (-8 - 1, 4 - 1) = (-9, 3) which corresponds to coordinate distances that are indeed three times (3,1). Good.
Answer:
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
1
unit(s).
The vertex A' of the image is
(–8, 4)
.
Step-by-step explanation:
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