# Dilation

How to find the image under the given dilation: formula, examples, and their solutions.

## Formula

The image of a point (*x*, *y*)

under the dilation of *k* is

(*kx*, *ky*).

If |*k*| > 1,

then the image shows an enlargement:

the image gets larger

and goes away from the origin.

If |*k*| < 1,

then the image shows a reduction:

the image gets smaller

and goes toward the origin.

If *k* = -1,

(*x*, *y*) → (-*x*, -*y*).

So the dilation shows the reflection in the origin.

Reflection in the origin

## Example 1: Enlargement

The image is under the dilation of [2].

So the image of *A*(-2, 1) is,

multiply [2],*A*'([2]⋅(-2), [2]⋅1) = (-4, 2).

The image of *B*(3, 2) is,

multiply [2],*B*'([2]⋅3, [2]⋅2) = (6, 4).

The image of *C*(0, -2) is,

multiply [2],*C*'([2]⋅0, [2]⋅(-2)) = (0, -4).

△*ABC* has vertices*A*(-2, 1), *B*(3, 2), and *C*(0, -2).

△*A'B'C'* has vertices*A*'(-4, -2), *B*'(6, 4), and *C*'(0, -4).

Use these vertices

to draw △*ABC* and its image △*A'B'C'*

on the coordinate plane.

As you can see,

△*A'B'C'* is under the dilation of 2.

The shape of the triangle is reserved.

But its size is enlarged. (*k* = 2)

So, under a dilation,

if *k* ≠ 1,

the length and the area are not reserved.

## Example 2: Reduction

The image is under the dilation of [1/3].

So the image of *A*(-4, 6) is,

multiply [1/3],*A*'([1/3]⋅(-4), [1/3]⋅6) = (-4/3, 2).

The image of *B*(6, 0) is,

multiply [1/3],*B*'([1/3]⋅6, [1/3]⋅0) = (2, 0).

The image of *C*(-3, -3) is,

multiply [1/3],*C*'([1/3]⋅(-3), [1/3]⋅(-3)) = (-1, -1).

△*ABC* has vertices*A*(-4, 6), *B*(6, 0), and *C*(-3, -3).

△*A'B'C'* has vertices*A*'(-4/3, 2), *B*'(2, 0), and *C*'(-1, -1).

Use these vertices

to draw △*ABC* and its image △*A'B'C'*

on the coordinate plane.

As you can see,

△*A'B'C'* is under the dilation of 1/3.

The shape of the triangle is reserved.

But its size is reduced. (*k* = 1/3)