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 unit(s).

The distance in the y-coordinates from A(–2, 2) to the center of dilation F(1, 1) is unit(s).

The vertex A' of the image is .

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

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

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)

(-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.

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:

edge 2020-21

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: