Core practice ← Back to lesson

Translations: 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 crosses an axis

Translate $G(-7,2)$ by displacement $(4,-6)$ . Give the image coordinates and describe the travel using right, left, up, or down.
1. Hint 1
  The displacement pair gives changes to the two coordinates.
2. Hint 2
  Add the first entry to the horizontal coordinate and the second entry to the vertical coordinate.
Answer

$G'(-3,-4)$ ; four units right and six units down.

Full solution

The horizontal change is positive 4, so move right.
The new horizontal coordinate is
$-7+4=-3.$

The vertical change is negative 6, so move down.
The new vertical coordinate is
$2-6=-4.$

Thus the image is $G'(-3,-4)$ .

Answer

$G'(-3,-4)$ ; four units right and six units down.

Key idea

A displacement changes each coordinate separately, including when the point crosses an axis.
Problem 2 Recover a common instruction

A translation sends $H(6,-5)$ to $H'(2,3)$ . Write its coordinate rule and describe the common horizontal and vertical changes.
1. Hint 1
  Compare the destination with the start in each coordinate.
2. Hint 2
  Each change is the image coordinate minus the corresponding starting coordinate.
Answer

$(x,y)\mapsto(x-4,y+8)$ ; four units left and eight units up.

Full solution

For the horizontal change, subtract the start from the image:
$2-6=-4.$

For the vertical change, calculate
$3-(-5)=8.$

Every point receives these same changes.
The rule is $(x,y)\mapsto(x-4,y+8)$ , four left and eight up.

Answer

$(x,y)\mapsto(x-4,y+8)$ ; four units left and eight units up.

Key idea

One point-image pair fixes a translation because it fixes both coordinate changes.
Problem 3 Draw the whole image

Quadrilateral $JKLM$ is the four-sided boundary joining $J(-2,-3)$ , $K(2,-2)$ , $L(0,1)$ , and $M(-3,-1)$ in that order. Translate it two units right and three units up. Give every image vertex and draw the image on graph paper.

Text description of this figure

A unit grid shows the quadrilateral JKLM with vertices J at (negative 2, negative 3), K at (2, negative 2), L at (0, 1), and M at (negative 3, negative 1), connected in that order. No image is drawn.
1. Hint 1
  A translation applies to every vertex, then to the sides between them.
2. Hint 2
  Use displacement pair $(2,3)$ at each vertex, and connect the resulting points in the original order.
Answer

$J'(0,0)$ , $K'(4,1)$ , $L'(2,4)$ , $M'(-1,2)$ ; join $J'K'L'M'$ in that order.

Full solution

The horizontal rule is $x'=x+2$ .
The vertical rule is $y'=y+3$ .

For J, adding 2 to negative 2 and 3 to negative 3 gives $J'(0,0)$ .

For K, the calculations are $2+2=4$ and $-2+3=1$ , so $K'(4,1)$ .

For L, the calculations are $0+2=2$ and $1+3=4$ , so $L'(2,4)$ .

For M, the calculations are $-3+2=-1$ and $-1+3=2$ , so $M'(-1,2)$ .

Plot all four image vertices and join them in the order $J'K'L'M'$ .
The dashed boundary in the solution figure shows the resulting image.

Answer

$J'(0,0)$ , $K'(4,1)$ , $L'(2,4)$ , $M'(-1,2)$ ; join $J'K'L'M'$ in that order.

Key idea

Translating every vertex and keeping the original connections describes the image of the whole polygon boundary.
Problem 4 The starting points are missing

A translation has rule $(x,y)\mapsto(x-4,y+5)$ . Find the missing preimages in both cases.

Case A: the image is $P'(-6,2)$ .

Case B: the image is $Q'(-2,-1)$ .
1. Hint 1
  The missing starting coordinates must produce the given image coordinates under the rule.
2. Hint 2
  For a starting horizontal coordinate, solve $x-4=x'$ . For a starting vertical coordinate, solve $y+5=y'$ .
Answer

Case A: $P(-2,-3)$ . Case B: $Q(2,-6)$ .

Full solution

Case A: solve $x-4=-6$ to get $x=-2$ .
Solve $y+5=2$ to get $y=-3$ .
Thus $P=(-2,-3)$ .

Case A check: $-2-4=-6$ and $-3+5=2$ , giving the required image.

Case B: solve $x-4=-2$ to get $x=2$ .
Solve $y+5=-1$ to get $y=-6$ .
Thus $Q=(2,-6)$ .

Case B check: $2-4=-2$ and $-6+5=-1$ .

Answer

Case A: $P(-2,-3)$ . Case B: $Q(2,-6)$ .

Key idea

A missing preimage can be found by solving the coordinate equations and checking the result in the stated rule.
Problem 5 A stencil shifts on a drawing

Coordinates on a design drawing are measured in millimeters. A rigid stencil slides without turning so that its reference mark at $(8,-2)$ moves to $(3,-1)$ . A small hole in the stencil has center $(6,3)$ before the slide. Find the new center. State the new length of a 12 mm edge and the new measure of a $65^\circ$ corner.
1. Hint 1
  The reference mark supplies the displacement for every point of the stencil.
2. Hint 2
  Subtract the reference coordinates to find the two changes, apply them to the hole, and use rigid-motion preservation for the measurements.
Answer

Hole center $(1,4)$ ; edge length $12$ mm; corner measure $65^\circ$ .

Full solution

The horizontal change is
$3-8=-5.$

The vertical change is
$-1-(-2)=1.$

The hole receives those same changes.
Its new horizontal coordinate is $6-5=1$ , and its new vertical coordinate is $3+1=4$ , so its center is $(1,4)$ .

A translation preserves distances within the stencil, so the edge remains 12 mm long.
It also preserves angle measures, so the corner remains $65^\circ$ .

Answer

Hole center $(1,4)$ ; edge length $12$ mm; corner measure $65^\circ$ .

Key idea

A reference point locates the slide, while preserved measurements describe the unchanged shape.
Problem 6 Two preserved measurements

A translation maps $A,B,C$ to $A',B',C'$ respectively. Find the unknown and the preserved measurement in each case.

Case A: $AB=5p-7$ cm and $A'B'=3p+9$ cm.

Case B: $m\angle ABC=2q+11$ degrees and $m\angle A'B'C'=79^\circ$ .
1. Hint 1
  Corresponding lengths are equal, and corresponding angle measures are equal.
2. Hint 2
  Write one equation for each case. Solve it, then substitute to obtain the requested measurement.
Answer

Case A: $p=8$ , length $33$ cm. Case B: $q=34$ , angle measure $79^\circ$ .

Full solution

Case A: distance preservation gives
$5p-7=3p+9.$

Subtract $3p$ and add 7 to get
$2p=16.$

Thus $p=8$ .
Substitution gives
$5(8)-7=33,$
so both lengths are 33 cm.

Case B: angle preservation directly gives
$2q+11=79.$

Subtract 11 to get $2q=68$ , so $q=34$ .
The angle measure is the given $79^\circ$ .

Answer

Case A: $p=8$ , length $33$ cm. Case B: $q=34$ , angle measure $79^\circ$ .

Key idea

Set corresponding measurements equal, then distinguish the unknown parameter from the geometric measurement.
Problem 7 Parallel images and an old position

The lines $\ell$ and $m$ have equations $y=-2$ and $y=3$ . Apply the translation $(x,y)\mapsto(x-4,y+5)$ . Give the equations of $\ell'$ and $m'$ . Are the two image lines distinct and parallel? Explain why one image can occupy the old position of another line.
1. Hint 1
  On each horizontal line, all points share one vertical coordinate.
2. Hint 2
  Apply the vertical change to each fixed height. Keep track of which preimage line supplies each image.
Answer

$\ell':y=3$ and $m':y=8$ ; the images are distinct and parallel, and $\ell'$ occupies the old position of $m$ .

Full solution

Every point of $\ell$ has vertical coordinate negative 2.
After the slide, its height is
$-2+5=3.$

The horizontal change moves points along a horizontal line, so $\ell'$ is the entire line $y=3$ .
Every horizontal coordinate is reached.

For m, the image height is
$3+5=8.$

Thus $m'$ is the entire line $y=8$ .
The two image heights differ by 5, so the image lines are distinct and parallel.

The old line m was at height 3, which is where $\ell'$ now lies.
Preservation compares the two images with each other; it does not require every image to avoid every preimage position.

Answer

$\ell':y=3$ and $m':y=8$ ; the images are distinct and parallel, and $\ell'$ occupies the old position of $m$ .

Key idea

Track correspondence: distinct parallel image lines may still share positions with lines from the original figure.
Problem 8 Construct a slide on paper

Copy the broken line $ABC$ onto paper with its right angle at B. Construct its image under a translation that moves every point 6 cm in the direction of the reference arrow. The arrow indicates direction only; the picture is not a centimeter scale. Label all three image points and state the image angle at B-prime. Explain what must agree in the three displacement arrows.

The reference arrow gives direction. Construct the stated distance with your own ruler.

Text description of this figure

A broken line goes horizontally from A to B, then vertically down from B to C. A right-angle mark is at B. A separate reference arrow above points right and slightly upward. The diagram gives direction only, not a scale for the required six-centimeter travel.
1. Hint 1
  The reference direction must be copied without turning at each starting vertex.
2. Hint 2
  From each vertex, draw a parallel arrow pointing the same way, using your ruler to set the required travel length.
Answer

A correctly translated broken line $A'B'C'$ ; $m\angle A'B'C'=90^\circ$ . All three displacement arrows have length 6 cm and the given direction.

Full solution

From A, draw an arrow parallel to the reference arrow, pointing the same way, and make it 6 cm long.
Its endpoint is A-prime.

Repeat the identical direction and length from B and C to obtain B-prime and C-prime.
The parallel construction from the preceding chapter can copy the direction.

Connect A-prime to B-prime and B-prime to C-prime.
Each point follows the same slide.

Angle preservation keeps the right-angle opening, so
$m\angle A'B'C'=90^\circ.$

The three travel arrows must agree in both length and direction.
Measuring only their lengths would not check the direction requirement.

This sketch shows the construction pattern. Each actual paper arrow must have length 6 cm.

Answer

A correctly translated broken line $A'B'C'$ ; $m\angle A'B'C'=90^\circ$ . All three displacement arrows have length 6 cm and the given direction.

Key idea

A paper translation copies one displacement from every point, regardless of the directions of the original sides.
Problem 9 Is equal travel enough?

A figure consists of two distinct points. Someone claims that any assignment moving both points equal distances must be a translation of that figure. Is the claim true? If not, give explicit coordinates for two starting points and their images, and explain why they refute the claim.
1. Hint 1
  A translation controls direction as well as distance.
2. Hint 2
  Consider two points that move equal distances toward each other along a number line.
Answer

False. For example, $A(0,0)\mapsto A'(1,0)$ and $B(3,0)\mapsto B'(2,0)$ give equal travel distances of 1 but opposite directions.

Full solution

Choose A at $(0,0)$ and its image at $(1,0)$ .
A travels one unit right.

Choose B at $(3,0)$ and its image at $(2,0)$ .
B travels one unit left.

Both travel distances are 1, satisfying the claim's assumption.
Their displacement pairs are $(1,0)$ and $(-1,0)$ , which differ.

No translation can make both specified matches because a translation requires the same direction as well as the same distance.

The journeys share a length, but not a direction. The rows are separated to show both arrows.

Answer

False. For example, $A(0,0)\mapsto A'(1,0)$ and $B(3,0)\mapsto B'(2,0)$ give equal travel distances of 1 but opposite directions.

Key idea

A counterexample must satisfy the stated assumption while breaking the claimed conclusion.
Problem 10 Reconstruct the missing map

The following data come from one translation. Reconstruct its coordinate rule, find u and v, and verify all three specified matches. Then give the image of $D(3,-1)$ .

Pair A: $A(-4,3)\mapsto A'(2,-2)$ .

Pair B: $B(u,0)\mapsto B'(5,-5)$ .

Pair C: $C(1,v)\mapsto C'(7,4)$ .
1. Hint 1
  One complete point-image pair determines the common displacement.
2. Hint 2
  Use Pair A to find the changes. Then each missing starting coordinate must produce its listed image under those changes.
3. Hint 3
  Check both coordinates in Pairs B and C before applying the reconstructed rule to D.
Answer

$(x,y)\mapsto(x+6,y-5)$ ; $u=-1$ , $v=9$ ; all three pairs agree; $D'(9,-6)$ .

Full solution

Pair A gives horizontal change
$2-(-4)=6.$

Its vertical change is
$-2-3=-5.$

Therefore every image follows $x'=x+6$ and $y'=y-5$ .

Pair B needs $u+6=5$ , so $u=-1$ .
Its vertical check is $0-5=-5$ , as required.

Pair C needs $v-5=4$ , so $v=9$ .
Its horizontal check is $1+6=7$ , as required.

Check Pair A with the same rule: $-4+6=2$ and $3-5=-2$ .
Each of the three specified matches agrees.

For D, the new horizontal coordinate is $3+6=9$ .
The new vertical coordinate is $-1-5=-6$ .
Thus $D'(9,-6)$ .

Answer

$(x,y)\mapsto(x+6,y-5)$ ; $u=-1$ , $v=9$ ; all three pairs agree; $D'(9,-6)$ .

Key idea

Reconstruct one common rule, use it to fill the missing data, and verify every supplied correspondence.

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