Dilations on the Coordinate Plane and Their Rules | Grade 8 Math | LevelKite

A dilation scales a figure from a fixed point called the centre of dilation. It uses a scale factor $k$ that is always positive ( $k > 0$ ). Each point $P$ other than the centre moves to a new point $P'$ , read P prime. $P'$ lies on the ray that starts at the centre and passes through $P$ , and it is $k$ times as far from the centre as $P$ is. The centre itself stays where it is, and the new figure is called the image. Recall that a scale factor greater than $1$ enlarges a figure and a scale factor between $0$ and $1$ shrinks it.

A dilation multiplies every length by $k$ and keeps every angle the same. So the image is similar to the original: corresponding angles are equal and corresponding side lengths are proportional. In this lesson, lengths are in grid units. You find each image point from exact coordinates, never by judging a picture by eye.

To find $P'$ with coordinates, think of the move from the centre to $P$ as a step with two parts. One part is an amount across, positive to the right and negative to the left. The other part is an amount up, positive upward and negative downward. The parts of a step can be negative, but a distance never is.

If you take the same step twice from the centre, you land twice as far away on the same ray. If you take half of the step, you land halfway to $P$ . In the same way, a dilation multiplies both parts of the step by $k$ . Multiplying both parts by the same positive number keeps the step pointing the same way, so $P'$ stays on the ray through $P$ .

When the centre is the origin $(0, 0)$ , the step to a point $P(x, y)$ is just $x$ across and $y$ up. So the image of $P$ is $P'(kx, ky)$ . In the figures below, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. Grid lines are $1$ unit apart.

Centre

(0, 0)

, scale factor

2

: Point A

(2, 2)

, Point B

(5, 2)

and Point C

(2, 4)

map to

A'(4, 4)

,

B'(10, 4)

and

C'(4, 8)

.

Worked example

Dilating from the origin

Triangle $ABC$ has vertices $A(2, 2)$ , $B(5, 2)$ and $C(2, 4)$ . Find its image $A'B'C'$ after a dilation with centre $(0, 0)$ and scale factor $2$ .

Name the centre and the scale factor: the centre is the origin and $k = 2$ . From the origin, the step to a point is just its pair of coordinates. So this dilation doubles both coordinates of every point.
$P(x, y) \to P'(2x,\ 2y)$
Apply this to $A$ . Point $A$ is $2$ across and $2$ up from the origin, so the step to $A'$ is twice that: $4$ across and $4$ up.
$A'(2 \cdot 2,\ 2 \cdot 2) = A'(4, 4)$
Multiply the coordinates of $B$ and $C$ by $2$ as well, since the same doubling applies to every point.
$B'(2 \cdot 5,\ 2 \cdot 2) = B'(10, 4),\quad C'(2 \cdot 2,\ 2 \cdot 4) = C'(4, 8)$
Check with side lengths in grid units. $AB$ and $A'B'$ are horizontal because each pair of endpoints shares a $y$ -coordinate, so subtract $x$ -coordinates. $AC$ and $A'C'$ are vertical, so subtract $y$ -coordinates. Each image side is $2$ times as long as its original side, as $k = 2$ requires.
$AB = 5 - 2 = 3,\ A'B' = 10 - 4 = 6;\quad AC = 4 - 2 = 2,\ A'C' = 8 - 4 = 4$

Answer

The image has vertices $A'(4, 4)$ , $B'(10, 4)$ and $C'(4, 8)$ , which are the points labelled with those coordinates in the figure.

Key concept

Scale the step from the centre

Take a dilation with centre $(a, b)$ and scale factor $k > 0$ . The step from the centre to any point $(x, y)$ is $x - a$ across and $y - b$ up. Multiply both parts by $k$ , then add the results to the centre's coordinates to reach the image.

When the centre is the origin, $a = 0$ and $b = 0$ , so the formula becomes $(x, y) \to (kx, ky)$ . Use that shortcut only when the centre is the origin.

(x, y) \to \big(a + k(x - a),\ b + k(y - b)\big)

Centre

(2, 1)

, scale factor

2

: Point D

(4, 3)

, Point E

(6, 3)

and Point F

(4, 5)

map to

D'(6, 5)

,

E'(10, 5)

and

F'(6, 9)

.

Worked example

Dilating from a centre that is not the origin

Triangle $DEF$ has vertices $D(4, 3)$ , $E(6, 3)$ and $F(4, 5)$ . Find its image $D'E'F'$ after a dilation with centre $(2, 1)$ and scale factor $2$ .

The centre is $(2, 1)$ , not the origin, and a dilation scales the step from the centre. So first find the step from the centre to $D$ by subtracting the centre's coordinates from those of $D$ .
$4 - 2 = 2 \text{ across},\quad 3 - 1 = 2 \text{ up}$
Multiply both parts of the step by $k = 2$ . Then add the results to the centre's coordinates, because you reach the image by starting at the centre.
$D' = (2 + 2 \cdot 2,\ 1 + 2 \cdot 2) = (2 + 4,\ 1 + 4) = (6, 5)$
Repeat for $E$ and $F$ . The step to $E$ is $6 - 2 = 4$ across and $3 - 1 = 2$ up. The step to $F$ is $4 - 2 = 2$ across and $5 - 1 = 4$ up.
$E' = (2 + 2 \cdot 4,\ 1 + 2 \cdot 2) = (10, 5),\quad F' = (2 + 2 \cdot 2,\ 1 + 2 \cdot 4) = (6, 9)$
Check the lengths as before. $DE$ and $D'E'$ are horizontal, and $DF$ and $D'F'$ are vertical. Each image side is twice its original side.
$DE = 6 - 4 = 2,\ D'E' = 10 - 6 = 4;\quad DF = 5 - 3 = 2,\ D'F' = 9 - 5 = 4$

Answer

The image has vertices $D'(6, 5)$ , $E'(10, 5)$ and $F'(6, 9)$ , which are the points labelled with those coordinates in the figure.

Now turn a dilation into a rule. A rule is written in arrow notation. The input point $(x, y)$ goes on the left of the arrow, and expressions for the coordinates of its image go on the right. Here $(x, y)$ stands for any point of the plane, so a rule works like a function: it gives exactly one image for each input point. To write a rule, state the centre and the scale factor, then substitute them into the formula in the key concept.

When the centre is the origin, each image coordinate is $k$ times the matching original coordinate, with the same $k$ throughout. In the first example, $A(2, 2) \to A'(4, 4)$ , $B(5, 2) \to B'(10, 4)$ and $C(2, 4) \to C'(4, 8)$ . Every coordinate is multiplied by $2$ and the centre is $(0, 0)$ , so the rule is $(x, y) \to (2x, 2y)$ . When the centre is another point, the rule needs the full formula, as the next example shows.

Worked example

Writing the rule from coordinates

Triangle $JKL$ has vertices $J(6, 8)$ , $K(10, 8)$ and $L(6, 2)$ . A dilation with centre $(2, 4)$ maps them to $J'(4, 6)$ , $K'(6, 6)$ and $L'(4, 3)$ . Find the scale factor and write the rule for the dilation.

Find $k$ from $J$ . The step from the centre to $J$ is $6 - 2 = 4$ across and $8 - 4 = 4$ up. The step to $J'$ is $4 - 2 = 2$ across and $6 - 4 = 2$ up. Each image part is $k$ times the original part, and the original parts are not zero, so you can divide.
$k = \frac{4 - 2}{6 - 2} = \frac{2}{4} = \frac{1}{2},\quad k = \frac{6 - 4}{8 - 4} = \frac{2}{4} = \frac{1}{2}$
Check $K$ and $L$ the same way, since a dilation uses one $k$ for every point. For $L$ , the up parts are $2 - 4 = -2$ and $3 - 4 = -1$ . A negative divided by a negative is positive, so $k$ is still $\tfrac{1}{2}$ .
$K:\ \frac{6 - 2}{10 - 2} = \frac{6 - 4}{8 - 4} = \frac{1}{2};\quad L:\ \frac{4 - 2}{6 - 2} = \frac{3 - 4}{2 - 4} = \frac{1}{2}$
Substitute $a = 2$ , $b = 4$ and $k = \tfrac{1}{2}$ into the formula in the key concept.
$(x, y) \to \left(2 + \tfrac{1}{2}(x - 2),\ 4 + \tfrac{1}{2}(y - 4)\right)$
Simplify each coordinate by distributing $\tfrac{1}{2}$ and combining the constant terms. The new expressions are equivalent to the old ones, so the rule still describes the same dilation. The constant terms $+1$ and $+2$ appear because the centre is not the origin.
$2 + \tfrac{1}{2}(x - 2) = 2 + \tfrac{1}{2}x - 1 = \tfrac{1}{2}x + 1,\quad 4 + \tfrac{1}{2}(y - 4) = 4 + \tfrac{1}{2}y - 2 = \tfrac{1}{2}y + 2$
Test the simplified rule on $L(6, 2)$ . It should give $L'(4, 3)$ , and it does.
$\left(\tfrac{1}{2} \cdot 6 + 1,\ \tfrac{1}{2} \cdot 2 + 2\right) = (3 + 1,\ 1 + 2) = (4, 3)$

Answer

The scale factor is $\tfrac{1}{2}$ , and the rule for the dilation is $(x, y) \to \left(\tfrac{1}{2}x + 1, \tfrac{1}{2}y + 2\right)$ .

Takeaway: a dilation multiplies the step from its centre to every point by the same positive scale factor $k$ . For centre $(a, b)$ , the rule is $(x, y) \to \big(a + k(x - a), b + k(y - b)\big)$ . The shortcut $(x, y) \to (kx, ky)$ is only for a centre at the origin. Next, you will describe a sequence of transformations that shows two figures are similar. You will also justify whether two figures are similar.