Can a scale factor be negative or zero?

Negative, yes; zero, no. A negative scale factor places the image on the opposite ray from the center, which looks like a dilation by | k | combined with a 180-degree rotation about the center. Angles and shape are still preserved. A scale factor of zero is not allowed, because it would collapse every point onto the center and the transformation could not be undone.

Why does area scale by k^2 instead of k?

Area is a two-dimensional measurement, so both dimensions get multiplied. A rectangle with base b and height h becomes one with base kb and height kh, giving area (kb)(kh) = k^2bh. Since any polygon can be cut into triangles and rectangles, the same factor applies to all of them. If you know the area ratio and want the scale factor, take the square root.

How do I find the center of dilation if it is not given?

Draw a line through each point and its image: through A and A', then through B and B'. Since the center is always collinear with a point and its image, every such line passes through the center. The single point where those lines intersect is the center of dilation. On a grid, you can then confirm it by checking that the ratio of distances from that point to B and to B' matches your scale factor.

Is a dilation a rigid motion?

No. Rigid motions — translations, rotations, reflections — preserve distance, but a dilation with k ≠ ± 1 changes every length. That is why a dilation produces a similar figure rather than a congruent one. A dilation combined with rigid motions is called a similarity transformation, and that combination is the formal definition of two figures being similar, which is exactly where this unit heads next.

GEOM-6.1

Dilations & Scale Factor

Learn how to perform dilations from any center, find the scale factor between two figures, and know what dilations preserve versus scale (lengths by k, areas by k²).

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Dilations & Scale Factor, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Every time you zoom in on a photo, project a slide onto a wall, or read a scale drawing, you are using a dilation. A dilation is the one transformation in Geometry that changes size without changing shape — it stretches or shrinks a figure away from (or toward) a fixed point called the center, by a fixed multiplier called the scale factor.

In this lesson you will learn three things. First, how to actually build the image of a figure given a center and a scale factor, whether that center is the origin or some other point. Second, how to work backwards: given a figure and its image, find the scale factor. Third — and this is what makes dilations the foundation of the whole similarity unit — what a dilation keeps the same (angle measures, parallel lines, collinearity) versus what it multiplies (every length by

k

, every area by

k^2

Dilations, Centers, and Scale Factors

A dilation is determined by two things: a fixed point

C

called the center of dilation, and a nonzero number

k

called the scale factor. The dilation sends each point

P

to the point

P'

on ray

CP

such that

CP' = k \cdot CP

. In other words, you travel from the center toward the point, and then keep going (or stop short) so that your distance from the center is multiplied by

k

.

The value of

k

tells you exactly what happens:

Scale factor	Effect on the figure
$k > 1$	Enlargement — image is bigger, farther from center
$k = 1$	Image equals preimage (identity)
$0 < k < 1$	Reduction — image is smaller, closer to center
$k < 0$	Image is on the opposite ray through the center, and scaled by $\lvert k \rvert$

When the center is the origin, the coordinate rule is beautifully simple:

(x, y) \to (kx, ky)

. So a dilation with

k = 3

centered at the origin sends

(2, -5)

(6, -15)

.

Two facts students often miss. The center of dilation is the only point that does not move — it is a fixed point, and if the center happens to lie on the figure, that vertex stays put. Also, a dilation is not a rigid motion: unless

k = \pm 1

, distances change, so a dilation is not a congruence. It belongs to a broader family called similarity transformations.

Dilating from a Center That Is Not the Origin

Most homework problems eventually move the center off the origin, and the rule

(x,y) \to (kx, ky)

no longer works. The reliable method is to think in terms of movement relative to the center.

If the center is

C(a, b)

and the scale factor is

k

, then

P'(x', y') = \big(a + k(x - a),\; b + k(y - b)\big).

Read that as three steps: subtract the center to find how far the point sits from it, multiply that displacement by

k

, then add the center back on.

Step	What you do	Example: $C(3,0)$ , $k=2$ , $P(5,2)$
1	Find $P - C$	$(5-3,\;2-0) = (2,2)$
2	Multiply by $k$	$(4,4)$
3	Add $C$ back	$(3+4,\;0+4) = (7,4)$

The single most common error is skipping step 3 — students multiply the displacement by

k

and report that as the image point, forgetting to translate back to the center. A quick sanity check catches it: the center, the original point, and the image must all be collinear. Plot them; if they do not line up on a straight ray out of the center, something went wrong.

A second check: measure

CP

and

CP'

. Their ratio must equal

k

exactly. In the example above,

CP = \sqrt{2^2+2^2} = 2\sqrt{2}

and

CP' = \sqrt{4^2+4^2} = 4\sqrt{2}

, so the ratio is

2

as required.

What Is Preserved and What Is Scaled

This is the conceptual heart of the lesson, and it is what the rest of Unit 6 leans on. A dilation is a similarity transformation: it preserves shape but not size.

Preserved (unchanged): angle measure, parallelism (parallel lines map to parallel lines), collinearity (points on a line stay on a line), betweenness and midpoints, and the ratio of any two lengths within the figure. A key consequence: a segment maps to a parallel segment (unless the segment lies on a line through the center, in which case it maps onto that same line).

Scaled: every length — sides, diagonals, altitudes, radii, perimeter — is multiplied by

k

. Every area is multiplied by

k^2

. If you ever extend this to solids, volume is multiplied by

k^3

Quantity	Under a dilation with factor $k$
Angle measure	Unchanged
Side length, perimeter	Multiplied by $k$
Area	Multiplied by $k^2$
Slope of a segment	Unchanged (image is parallel)

Why

k^2

? Area depends on two dimensions. A rectangle with base

b

and height

h

has area

bh

; after dilation its base is

kb

and its height is

kh

, so the new area is

(kb)(kh) = k^2 bh

. The same argument works for triangles and, by decomposition, any polygon.

The classic mistake is multiplying area by

k

instead of

k^2

. If a figure with area 20 square units is dilated by

k = 3

, the new area is

20 \cdot 9 = 180

square units, not 60. Going backwards is just as common a trap: if areas are in a ratio of

25:4

, the scale factor is

\sqrt{25}/\sqrt{4} = 5/2

, not

25/4

Finding the Scale Factor from Two Figures

When you are handed a figure and its image, the scale factor is always

k = \frac{\text{image length}}{\text{preimage length}}

using a pair of corresponding parts. The order matters: image on top. If you flip it you get the reciprocal, which describes the dilation that undoes the original one. That inverse relationship is worth knowing — the dilation with center

C

and factor

1/k

maps the image back to the preimage.

Be sure your two lengths really correspond. In a triangle labeled

\triangle ABC \to \triangle A'B'C'

, compare

AB

with

A'B'

, not

AB

with

B'C'

. If the figure is not labeled with primes, match parts by position: longest side to longest side, and the side opposite the 40-degree angle to the side opposite the 40-degree angle.

You can also recover

k

from distances to the center:

k = \dfrac{CP'}{CP}

. And if you only know areas, take the square root of the area ratio.

To locate an unknown center of dilation, draw lines through each pair of corresponding points (

A

with

A'

B

with

B'

). Because the center, a point, and its image are always collinear, all of those lines meet at exactly one point — that intersection is the center.

One more check that saves grief: if the image is larger,

|k|

must be greater than 1; if smaller,

|k|

must be between 0 and 1. Students frequently compute

k = 4/6

for an enlargement and do not notice the contradiction.

Reading Dilations on the Coordinate Plane

On a grid, dilations become a counting exercise, which makes them easy to check by eye.

Suppose

\triangle ABC

has

A(2,1)

, and the center of dilation is

C(0,0)

with

k = 3

. From the origin you move right 2 and up 1 to reach

A

; to reach

A'

you move right

3 \cdot 2 = 6

and up

3 \cdot 1 = 3

, landing at

(6,3)

. The horizontal and vertical runs from the center both get multiplied by

k

, which is exactly why the image segment stays parallel to the original: the slope

\frac{k \cdot \text{rise}}{k \cdot \text{run}}

is unchanged.

A negative scale factor sends the point through the center to the other side. Dilating

A(2,1)

about the origin with

k = -2

gives

(-4,-2)

. The image is the same size as it would be with

k = 2

, but rotated

180

degrees about the center. Shape and angle measures are still preserved.

Useful habits when working on a grid: always plot the center first, and always sketch the ray from the center through each vertex before you place the image point. If a vertex is dilated correctly, its image sits on that ray. Finally, verify one length with the distance formula and one angle by comparing slopes. Two quick checks will catch nearly every arithmetic slip, and they build the intuition you will need when the unit moves on to proving triangles similar.

Key terms

Dilation.: A transformation that maps each point $P$ to a point $P'$ on ray $CP$ so that $CP' = k \cdot CP$ , where $C$ is the center and $k$ is the scale factor. It changes size but not shape.
Center of dilation.: The fixed point from which the figure is stretched or shrunk. It is the only point that maps to itself, and it is always collinear with each point and its image.
Scale factor ( $k$ ).: The nonzero number that every distance from the center is multiplied by. Computed as image length divided by corresponding preimage length.
Enlargement.: A dilation with $\lvert k \rvert > 1$ ; the image is larger than the preimage and farther from the center.
Reduction.: A dilation with $0 < \lvert k \rvert < 1$ ; the image is smaller than the preimage and closer to the center.
Preimage and image.: The original figure is the preimage; the result after the transformation is the image, usually labeled with prime marks such as $A'$ .
Similarity transformation.: A transformation, or sequence of them, made of rigid motions and dilations. It preserves angle measure and ratios of lengths, producing a figure similar to the original.
Area ratio.: For two similar figures related by scale factor $k$ , the ratio of their areas is $k^2$ ; conversely, $k$ equals the square root of the area ratio.

Worked example

Triangle

ABC

has vertices

A(1,2)

B(5,2)

, and

C(1,4)

. Perform a dilation with center

D(3,0)

and scale factor

k = 2

. Find the coordinates of

A'

B'

, and

C'

, then verify that lengths scale by 2 and area scales by 4.

Use

P' = \big(a + k(x-a),\; b + k(y-b)\big)

with center

D(3,0)

and

k = 2

.

For

A(1,2)

: the displacement from

D

(1-3,\; 2-0) = (-2, 2)

. Multiply by 2:

(-4, 4)

. Add the center back:

A' = (3-4,\; 0+4) = (-1, 4)

.

For

B(5,2)

: displacement is

(2, 2)

, doubled is

(4,4)

, so

B' = (7, 4)

.

For

C(1,4)

: displacement is

(-2, 4)

, doubled is

(-4, 8)

, so

C' = (-1, 8)

.

Now check lengths.

AB

runs from

(1,2)

(5,2)

, so

AB = 4

A'B'

runs from

(-1,4)

(7,4)

, so

A'B' = 8

, and

8/4 = 2

. Likewise

AC = 2

(vertical, from

y=2

y=4

) and

A'C' = 4

(from

y=4

y=8

), again a ratio of 2. Notice

\overline{AB}

and

\overline{A'B'}

are both horizontal — parallelism preserved — and both triangles have a right angle at

A

and

A'

— angle measure preserved.

Now check area. The preimage is a right triangle with legs 4 and 2, so its area is

\frac{1}{2}(4)(2) = 4

square units. The image has legs 8 and 4, so its area is

\frac{1}{2}(8)(4) = 16

square units. The ratio is

16/4 = 4 = k^2

, exactly as predicted.

Final check:

D

A

, and

A'

should be collinear. From

D(3,0)

A(1,2)

the run is

-2

and the rise is

+2

; from

D(3,0)

A'(-1,4)

the run is

-4

and the rise is

+4

. Same slope of

-1

, so they lie on one ray. The dilation is correct.

Practice questions

A pentagon has an area of 30 square centimeters. It is dilated about a point with scale factor

k = \frac{1}{2}

. What is the area of the image?

15 square centimeters
7.5 square centimeters
60 square centimeters
120 square centimeters

Answer: 7.5 square centimeters

Areas scale by

k^2

, not

k

. Here

k^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}

, so the image area is

30 \cdot \frac{1}{4} = 7.5

square centimeters. Choosing 15 is the classic slip of multiplying the area by

k

instead of

k^2

; that value would be correct only for a length, such as a side or the perimeter.

Point

P(6, -9)

is dilated about the origin to

P'(-4, 6)

. Find the scale factor, and state whether the image is on the same side of the center as the preimage.

Answer: $k = -\frac{2}{3}$ ; the image is on the opposite side of the center.

With center at the origin the rule is

(x,y) \to (kx, ky)

. From the

x

-coordinates,

6k = -4

, so

k = -\frac{2}{3}

. Check with the

y

-coordinates:

-9 \cdot \left(-\frac{2}{3}\right) = 6

, which matches. Because

k

is negative,

P'

lies on the ray opposite

\overrightarrow{OP}

, so it is on the other side of the center. Because

\lvert k \rvert = \frac{2}{3} < 1

, the image is also closer to the center than the preimage was.

Triangle

RST

is dilated with center

R

and scale factor 3 to form triangle

R'S'T'

. Explain why

R' = R

, why

m\angle S = m\angle S'

, and why

\overline{ST}

is parallel to

\overline{S'T'}

Answer: The center is fixed, so $R' = R$ ; dilations preserve angle measure, so $m\angle S = m\angle S'$ ; and a segment not passing through the center maps to a parallel segment, so $\overline{ST} \parallel \overline{S'T'}$ .

The center of a dilation is the only point whose distance from itself is multiplied by

k

and still equals zero, so it never moves — that is why choosing a vertex as the center pins that vertex in place. Angle measure is preserved because a dilation multiplies all distances from the center by the same factor, producing a figure with identical shape; formally, the image triangle is similar to the original by SSS with ratio 3, so corresponding angles are congruent. For the parallel segments:

R

S

S'

are collinear and

R

T

T'

are collinear with

RS' = 3RS

and

RT' = 3RT

, so by the converse of the side-splitter idea the segment

\overline{S'T'}

must be parallel to

\overline{ST}

FAQ

Can a scale factor be negative or zero?: Negative, yes; zero, no. A negative scale factor places the image on the opposite ray from the center, which looks like a dilation by $\lvert k \rvert$ combined with a $180$ -degree rotation about the center. Angles and shape are still preserved. A scale factor of zero is not allowed, because it would collapse every point onto the center and the transformation could not be undone.
Why does area scale by $k^2$ instead of $k$ ?: Area is a two-dimensional measurement, so both dimensions get multiplied. A rectangle with base $b$ and height $h$ becomes one with base $kb$ and height $kh$ , giving area $(kb)(kh) = k^2bh$ . Since any polygon can be cut into triangles and rectangles, the same factor applies to all of them. If you know the area ratio and want the scale factor, take the square root.
How do I find the center of dilation if it is not given?: Draw a line through each point and its image: through $A$ and $A'$ , then through $B$ and $B'$ . Since the center is always collinear with a point and its image, every such line passes through the center. The single point where those lines intersect is the center of dilation. On a grid, you can then confirm it by checking that the ratio of distances from that point to $B$ and to $B'$ matches your scale factor.
Is a dilation a rigid motion?: No. Rigid motions — translations, rotations, reflections — preserve distance, but a dilation with $k \neq \pm 1$ changes every length. That is why a dilation produces a similar figure rather than a congruent one. A dilation combined with rigid motions is called a similarity transformation, and that combination is the formal definition of two figures being similar, which is exactly where this unit heads next.

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