# 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)