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Dilations and Scale Factors: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A point on its ray

A dilation with center $O$ and scale factor $\frac 47$ sends a point $P$ to $P'$ . The distance $OP$ is $7$ cm.

Find $OP'$ and state whether $P'$ lies between $O$ and $P$ or beyond $P$ .
1. Hint 1
  Use the distance from the center to the point.
2. Hint 2
  Compare the positive scale factor with one to decide the order.
Answer

$OP'=4$ cm; $P'$ lies between $O$ and $P$ .

Full solution

Multiply the original distance by the scale factor.
$\begin{aligned}OP'&=\frac 47\cdot 7,\\&=4\text{ cm}.\end{aligned}$

The factor is positive and below one, so the image lies on ray $OP$ , closer to $O$ than $P$ is.
Thus $P'$ lies between $O$ and $P$ .

Answer

$OP'=4$ cm; $P'$ lies between $O$ and $P$ .

Key idea

A reduction changes distance from the chosen center while keeping the original ray.
Problem 2 Three vertices from the origin

Triangle $ABC$ has vertices $A(-4,-2)$ , $B(2,-2)$ and $C(0, 6)$ . It is dilated about the origin with scale factor $\frac 34$ .

(a) Find $A'$ and $B'$ .

(b) Find $C'$ and classify the dilation.
1. Hint 1
  Identify the center before choosing the coordinate rule.
2. Hint 2
  Multiply both coordinates of each vertex by the factor, including negative coordinates.
Answer

Part (a).
$\begin{gathered}A'=(-3,-1.5)\\B'=(1.5,-1.5)\end{gathered}$
Part (b).
$C'=(0, 4.5)$
This is a reduction.

Full solution

(a) Multiply each coordinate by three quarters.
$\begin{aligned}A'&=(-3,-1.5),\\B'&=(1.5,-1.5).\end{aligned}$

(b) Zero stays zero when multiplied, and six becomes four and a half.
$C'=(0, 4.5).$

The positive factor $\frac 34$ is less than one, so this is a reduction.

The image vertices use the same factor on both coordinates.

Answer

Part (a).
$\begin{gathered}A'=(-3,-1.5)\\B'=(1.5,-1.5)\end{gathered}$
Part (b).
$C'=(0, 4.5)$
This is a reduction.

Key idea

The origin rule multiplies both coordinates by the same number.
Problem 3 Lengths and angles carry different information

A dilation centered at $O$ sends $P$ to $P'$ , with
$\begin{aligned}OP&=10\text{ mm}\\OP'&=16\text{ mm}.\end{aligned}$
(a) Find the scale factor.

(b) A segment $YZ$ has length $15$ mm. Find $Y'Z'$ .

(c) An angle at $Y$ measures $84^\circ$ . Find its image measure.
1. Hint 1
  First identify which given distance belongs to the image.
2. Hint 2
  The length rule and the angle rule have different effects.
Answer

Part (a).
$k=\frac 85$
Part (b).
$Y'Z'=24\text{ mm}$
Part (c). The image angle measures $84^\circ$ .

Full solution

(a) Divide image distance by original distance.
$\begin{aligned}k&=\frac{16}{10},\\&=\frac 85.\end{aligned}$

(b) The same factor applies to any segment.
$\begin{aligned}Y'Z'&=\frac 85\cdot 15,\\&=24\text{ mm}.\end{aligned}$

(c) Angles are preserved, so the image angle measures $84^\circ$ .

Answer

Part (a).
$k=\frac 85$
Part (b).
$Y'Z'=24\text{ mm}$
Part (c). The image angle measures $84^\circ$ .

Key idea

A length factor applies to lengths, not to angle measures.
Problem 4 Two points around a shifted center

A dilation has center $O(-3, 2)$ and scale factor $\frac 12$ .

(a) Find the image of $P(1, 8)$ .

(b) Find the image of $Q(-5, 0)$ .
1. Hint 1
  Measure each horizontal and vertical displacement from O.
2. Hint 2
  Halve the displacements, then add the center coordinates back.
Answer

Part (a).
$P'=(-1, 5)$
Part (b).
$Q'=(-4, 1)$

Full solution

(a) The displacement to $P$ is four right and six up.
After halving it, start two right and three up from $O$ .
$P'=(-1, 5).$

(b) The displacement to $Q$ is two left and two down.
Half of that is one left and one down from $O$ .
$Q'=(-4, 1).$

Both images lie between O and their original points.

Answer

Part (a).
$P'=(-1, 5)$
Part (b).
$Q'=(-4, 1)$

Key idea

Scale displacements from the center rather than the raw coordinates.
Problem 5 A line remains while its points move

A dilation has center $O(1,-2)$ and scale factor $0.4$ .

(a) Describe the image of the line $y=-2$ and find the image of $R(6,-2)$ on it.

(b) Find the image of the line $y=3$ and explain whether it is the same line or a distinct parallel line.
1. Hint 1
  Check whether each line contains the center.
2. Hint 2
  For a horizontal line, every point has the same y-coordinate. Apply the vertical coordinate rule to that number.
Answer

Part (a). The image line is $y=-2$ .
$R'=(3,-2)$
Part (b). The image is $y=0$ , a distinct parallel line.

Full solution

(a) The line $y=-2$ contains $O$ , so it maps onto itself as a set.
The point $R$ still moves.
$\begin{aligned}x'&=1+0.4(6-1),\\&=3,\\y'&=-2+0.4(-2+2),\\&=-2.\end{aligned}$

(b) All original points have vertical displacement five from the center.
$\begin{aligned}y'&=-2+0.4(3+2),\\&=0.\end{aligned}$

The horizontal coordinates still range over every real number, so the image is the entire line $y=0$ .
It is distinct from, and parallel to, $y=3$ .

Line locations and point locations obey the same dilation; only the set of points on the through-center line stays unchanged.

Answer

Part (a). The image line is $y=-2$ .
$R'=(3,-2)$
Part (b). The image is $y=0$ , a distinct parallel line.

Key idea

An unchanged line as a set need not have unchanged individual points.
Problem 6 Two labels to determine

A dilation has scale factor $1.5$ . All labeled lengths below are positive.

(a) A segment has length $5u-1$ and its image has length $7u+3$ . Find $u$ and both lengths.

(b) An original angle measures $(2v+9)^\circ$ and its image measures $(5v-39)^\circ$ . Find $v$ and the angle measure.
1. Hint 1
  Decide which quantities are multiplied and which stay equal.
2. Hint 2
  In part (a), multiply the whole original length expression by 1.5. In part (b), set the two angle expressions equal.
Answer

Part (a).
$u=9$
The original length is $44$ and the image length is $66$ . Part (b).
$v=16$
Both angles measure $41^\circ$ .

Full solution

(a) Apply the factor to the original length.
$\begin{aligned}7u+3&=1.5(5u-1),\\7u+3&=7.5u-1.5,\\4.5&=0.5u,\\u&=9.\end{aligned}$

Substitute to check the positive lengths.
$\begin{aligned}5(9)-1&=44,\\7(9)+3&=66.\end{aligned}$

(b) A dilation preserves the angle measure.
$\begin{aligned}2v+9&=5v-39,\\48&=3v,\\v&=16.\end{aligned}$

Substituting sixteen into either expression gives an angle of $41^\circ$ .

Answer

Part (a).
$u=9$
The original length is $44$ and the image length is $66$ . Part (b).
$v=16$
Both angles measure $41^\circ$ .

Key idea

The geometric quantity determines whether to scale an expression or equate it.
Problem 7 Test the fixed-point claim

Consider the claim “Every point on a line through a dilation's center is fixed.”

Refute it with one positive scale factor and one point away from the center. State the correct fixed-point rule, including the special factor.
1. Hint 1
  An unchanged line and an unchanged point are different statements.
2. Hint 2
  Choose a factor other than one and a point whose distance from the center is easy to multiply.
Answer

For example, center O at coordinate 0 on a number line, factor 3, and point P at 2 give image P′ at 6, so P is not fixed. For factor 1 every point is fixed; for any other positive factor only the center is fixed.

Full solution

Choose a number line with $O$ at $0$ and $P$ at $2$ .
Under scale factor $3$ , the image is at $6$ .
$\begin{aligned}OP'&=3\cdot 2,\\&=6.\end{aligned}$

The point at $2$ is not fixed, so the universal claim is false.
The line still maps onto itself as a set.

At factor $1$ , every distance from the center stays the same and every point is fixed.
For a positive factor different from $1$ , each positive distance changes, leaving only the center fixed.

The points move along the line even though the image of the entire line is the same line.

Answer

For example, center O at coordinate 0 on a number line, factor 3, and point P at 2 give image P′ at 6, so P is not fixed. For factor 1 every point is fixed; for any other positive factor only the center is fixed.

Key idea

A single moving point can refute a claim that every point is fixed.
Problem 8 Infer the missing center

A dilation with scale factor $\frac 32$ sends $P(4, 6)$ to $P'(7, 10)$ .

(a) Find the center $O(a,b)$ .

(b) Use that center to find the image of $Q(0, 2)$ .
1. Hint 1
  The displacement from the center is scaled before the center is added back.
2. Hint 2
  Write one equation using the horizontal coordinates and one using the vertical coordinates.
Answer

Part (a).
$O=(-2,-2)$
Part (b).
$Q'=(1, 4)$

Full solution

(a) Use the coordinate rule with the unknown center.
$\begin{aligned}7&=a+\tfrac 32(4-a),\\7&=6-\tfrac 12a,\\a&=-2.\end{aligned}$

Apply the same reasoning vertically.
$\begin{aligned}10&=b+\tfrac 32(6-b),\\10&=9-\tfrac 12b,\\b&=-2.\end{aligned}$

(b) Measure the displacement from $(-2,-2)$ to $Q(0, 2)$ , scale it, and add the center back.
$\begin{aligned}x'&=-2+\tfrac 32(0+2),\\&=1,\\y'&=-2+\tfrac 32(2+2),\\&=4.\end{aligned}$

Answer

Part (a).
$O=(-2,-2)$
Part (b).
$Q'=(1, 4)$

Key idea

One point-image pair and a factor different from one can determine a center.
Problem 9 Check three recorded images

The center is the origin. A record says that one dilation sends $A(2, 0)$ to $A'(5, 0)$ , $B(0, 2)$ to $B'(0, 5)$ , and $C(2, 2)$ to $C'(5, 4)$ .

Can all three records be correct? If not, keep the first two mappings and correct the image of $C$ . Explain your decision.
1. Hint 1
  The same factor must apply to every point.
2. Hint 2
  Find the factor from either of the first two point pairs, then apply it to both coordinates of C.
Answer

No. The first two mappings require $k=\frac 52$ , so the corrected image is $C'=(5, 5)$ .

Full solution

The first mapping takes distance $2$ from the origin to distance $5$ , giving factor $\frac 52$ .
The second mapping uses that same factor.

Both coordinates of $C$ must therefore be multiplied by $\frac 52$ .
$C'=(5, 5).$

The recorded point $(5, 4)$ is not even on ray $OC$ , whose points have equal positive coordinates.
It fails both the common-factor rule and the ray condition.

Answer

No. The first two mappings require $k=\frac 52$ , so the corrected image is $C'=(5, 5)$ .

Key idea

Check every mapping against the same center and factor.
Problem 10 Repair a stretched triangle

Triangle $ABC$ has vertices $A(0, 0)$ , $B(2, 0)$ and $C(0, 3)$ . A transformation records $A'(1,-3)$ , $B'(5,-3)$ and $C'(1, 0)$ .

(a) Explain why these three images cannot come from one dilation with any center.

(b) Keep $A'$ and $B'$ as recorded. Find a center and a positive scale factor that give these two mappings, and replace $C'$ so that the entire triangle image comes from that dilation.
1. Hint 1
  Compare an original horizontal length with its image, then do the same for a vertical length.
2. Hint 2
  For part (b), the horizontal segment determines the factor. Use A and its image to find the center.
3. Hint 3
  After finding the center, apply the same coordinate rule to C.
Answer

Part (a). The length factors are $2$ for $AB$ and $1$ for $AC$ , so no dilation works. Part (b). Use factor $2$ and center $(-1, 3)$ . The corrected third image is
$C'=(1, 3).$

Full solution

(a) The horizontal length $AB$ is $2$ , while $A'B'$ is $4$ , so that pair requires factor $2$ .
The vertical length $AC$ is $3$ , while $A'C'$ is also $3$ , so that pair requires factor $1$ .
One dilation cannot satisfy both.

(b) Keep the factor $2$ required by $AB$ and its image.
Use $A(0, 0)$ mapping to $(1,-3)$ to find the center.
$\begin{aligned}1&=a+2(0-a),\\a&=-1,\\-3&=b+2(0-b),\\b&=3.\end{aligned}$

This center sends $B(2, 0)$ to $(5,-3)$ as required.
Apply it to $C(0, 3)$ .
$\begin{aligned}x'&=-1+2(0+1),\\&=1,\\y'&=3+2(3-3),\\&=3.\end{aligned}$

Replace the recorded third image by $(1, 3)$ .
Now every vertex uses the same center and factor.

This picture illustrates part (a). The coordinate solution in part (b) repairs the third image.

Answer

Part (a). The length factors are $2$ for $AB$ and $1$ for $AC$ , so no dilation works. Part (b). Use factor $2$ and center $(-1, 3)$ . The corrected third image is
$C'=(1, 3).$

Key idea

Unequal length factors rule out a dilation; repairing it requires one consistent point rule.

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