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Compositions and Congruence: 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 Keep the intermediate image

Start at $P(6,-1)$ . First reflect across $y=x$ , then translate two units left and three units up. Give $P'$ and $P''$ , and explain which point receives the translation.
1. Hint 1
  A composition continues from the image produced by its first motion.
2. Hint 2
  Use the diagonal reflection rule first; then add the horizontal and vertical changes to the new coordinates.
Answer

$P'(-1,6)$ ; $P''(-3,9)$ ; the translation acts on $P'$ .

Full solution

The rising-diagonal reflection exchanges the coordinates, giving $P'(-1,6)$ .

Translate that intermediate point.
Its horizontal coordinate becomes
$-1-2=-3$

Its height becomes
$6+3=9$

The final image is $P''(-3,9)$ .
Applying the translation to P instead would restart rather than continue the sequence.

Answer

$P'(-1,6)$ ; $P''(-3,9)$ ; the translation acts on $P'$ .

Key idea

A second motion takes the first image as its starting point.
Problem 2 Work backward through the journey

Point $R$ is rotated $90^\circ$ counterclockwise about the origin and then translated four units left and two up. The final image is $R''(-7,1)$ . Recover $R'$ and $R$ , and check the forward journey.
1. Hint 1
  Undo the final motion before undoing the first one.
2. Hint 2
  Reverse the translation, then use a clockwise quarter-turn about the same origin.
Answer

$R'(-3,-1)$ and $R(-1,3)$ .

Full solution

Undo the slide with four right and two down.
The horizontal calculation is
$-7+4=-3$

The height calculation is
$1-2=-1$

Thus $R'(-3,-1)$ .
Undo the counterclockwise turn with the clockwise rule $(x,y)\mapsto(y,-x)$ , giving $R(-1,3)$ .

Check forward: the counterclockwise quarter-turn of $(-1,3)$ is $(-3,-1)$ .
Sliding four left and two up gives $(-7,1)$ , as required.

Answer

$R'(-3,-1)$ and $R(-1,3)$ .

Key idea

Reverse the order and invert each step to recover a preimage.
Problem 3 Choose the direction between parallel mirrors

First reflect across $y=1$ , then across $y=-3$ . Identify the single motion for the whole plane and find both images of $P(2,6)$ .

Restart at P and reverse the mirror order. Give the new final image and explain the change in direction.

Text description of this figure

Two horizontal mirrors have equations y=1 and y=-3. The upper mirror y=1 is first, the lower mirror y=-3 second. P is at (2,6), above both mirrors. A perpendicular guide crosses both mirrors. No images are drawn.
1. Hint 1
  Parallel mirrors give a translation perpendicular to them.
2. Hint 2
  Measure the gap, double it, and point from the first mirror toward the second. You can also apply the two horizontal reflection rules.
Answer

Original order: eight units down, $P'(2,-4)$ , $P''(2,-2)$ . Reversed order: eight units up, final image $(2,14)$ .

Full solution

The mirrors are four units apart.
From y=1 to y=-3 is downward, so the whole composition translates down by
$2(4)=8$

The first reflected height is
$2(1)-6=-4$

The second reflected height is
$2(-3)-(-4)=-2$

Horizontal position stays 2, so $P'(2,-4)$ and $P''(2,-2)$ .

In the reversed order, the directed gap runs upward.
The translation is eight up, so the final height is
$6+8=14$

The final image is $(2,14)$ .
Reversing the mirrors reverses the translation.

Answer

Original order: eight units down, $P'(2,-4)$ , $P''(2,-2)$ . Reversed order: eight units up, final image $(2,14)$ .

Key idea

The gap supplies the size; the mirror order supplies the direction.
Problem 4 Double the turn with its direction

Mirrors m and n meet at O. The directed angle from the chosen ray on m to the chosen ray on n is 24 degrees clockwise. Point P has OP=5.5 cm.

Case A. First reflect across m, then across n. Identify the single motion and give the final distance from O.

Case B. Restart at P and use n before m. Identify the single motion. Explain why the two mappings differ even though both preserve the distance from O.

Text description of this figure

Mirrors m and n meet at O. The chosen ray on m points right; the chosen ray on n lies 24 degrees clockwise from it. P is off O with OP=5.5 cm. The diagram contains only the given mirrors, ray OP, and directed mirror angle.
1. Hint 1
  The center is the mirror intersection. Direction belongs to the ordered pair of mirrors.
2. Hint 2
  Double the mirror angle in each direction and compare the resulting turns.
Answer

Case A: a $48^\circ$ clockwise rotation about O; final radius 5.5 cm. Case B: a $48^\circ$ counterclockwise rotation about O. The two mappings differ.

Full solution

Case A.
Both mirrors fix O, and two folds double the clockwise angle.
$2(24^\circ)=48^\circ$

The equivalent motion is a 48-degree clockwise rotation about O.
Each reflection preserves distance from O, so the final radius is 5.5 cm.

Case B.
Reversing the mirror order reverses the directed angle, giving a 48-degree counterclockwise rotation about O.

These are different final directions for every point other than O: the directions differ by 96 degrees, rather than a full turn.
Keeping a radius alone does not choose the final ray.

Answer

Case A: a $48^\circ$ clockwise rotation about O; final radius 5.5 cm. Case B: a $48^\circ$ counterclockwise rotation about O. The two mappings differ.

Key idea

A rotation needs both its center and its directed turn, even when the radius is known.
Problem 5 A glide and its zero case

Use mirror $x=2$ and a translation five units down. Start from $H(-1,4)$ separately in each case.

Case A. Reflect first, then translate. Give both images.

Case B. Translate first, then reflect. Give both images and explain why the final mapping agrees for every point.

Case C. Replace the translation by zero displacement. Classify the resulting map. Compare its fixed points with those of the nonzero glide.
1. Hint 1
  Separate motion across the mirror from motion along it.
2. Hint 2
  The reflection changes horizontal position; this parallel slide changes height. With zero displacement, one step does nothing.
Answer

A: $H'(5,4)$ , $H''(5,-1)$ . B: $H'(-1,-1)$ , $H''(5,-1)$ . C: reflection across $x=2$ , fixing every point on that line; the nonzero glide fixes no points.

Full solution

Case A.
Reflection across x=2 changes the horizontal coordinate to
$2(2)-(-1)=5$

The intermediate image is $H'(5,4)$ .
Five down gives $H''(5,-1)$ .

Case B.
Five down first gives $H'(-1,-1)$ .
Reflecting gives $H''(5,-1)$ .

For any $(x,y)$ , reflection changes horizontal position to $4-x$ , while the slide changes height to $y-5$ .
Neither step alters the coordinate needed by the other, so both orders give $(4-x,y-5)$ .

Case C.
Zero translation leaves just reflection across x=2, whose fixed points are exactly the points on that mirror.
The nonzero glide changes every height by negative 5, so no point can be fixed.

Answer

A: $H'(5,4)$ , $H''(5,-1)$ . B: $H'(-1,-1)$ , $H''(5,-1)$ . C: reflection across $x=2$ , fixing every point on that line; the nonzero glide fixes no points.

Key idea

The parallel condition makes glide steps commute; the nonzero condition distinguishes a glide from a reflection.
Problem 6 Move a printed cutout

A drawing for a paper cutout uses millimeters. A specified reflection followed by a translation carries the whole triangle $ABC$ exactly onto $DEF$ , with $A\mapsto D$ , $B\mapsto E$ , and $C\mapsto F$ .

Given $AB=17$ mm, $BC=9$ mm, and $m\angle ABC=106^\circ$ , find the length of path $DEF$ and the angle at E. Write the congruence statement.

The order A, B, C is counterclockwise. State the corresponding order around D, E, F and explain whether this changes the angle measure.
1. Hint 1
  Follow the named correspondence before using preservation.
2. Hint 2
  The two-segment image path corresponds to AB followed by BC. One reflection reverses orientation; a translation keeps that new orientation.
Answer

Path $DEF$ is 26 mm; $m\angle DEF=106^\circ$ ; $\triangle ABC\cong\triangle DEF$ ; the order D,E,F is clockwise, with unchanged angle measure.

Full solution

The image sides DE and EF keep the lengths 17 mm and 9 mm.
Their path length is
$17+9=26$

The vertex B maps to E.
Both rigid steps preserve the angle measure directly, giving
$m\angle DEF=106^\circ.$

The specified whole-figure match establishes $\triangle ABC\cong\triangle DEF$ .
The letter order records A to D, B to E, C to F.

The reflection changes the counterclockwise order to clockwise.
The translation keeps it clockwise.
This orientation change does not alter the corresponding angle opening.

Answer

Path $DEF$ is 26 mm; $m\angle DEF=106^\circ$ ; $\triangle ABC\cong\triangle DEF$ ; the order D,E,F is clockwise, with unchanged angle measure.

Key idea

Correspondence tells which measurements agree, while the rigid sequence establishes the whole-figure match.
Problem 7 Find the first incorrect line

Start with $P(-5,2)$ . First rotate $90^\circ$ clockwise about the origin, then reflect across the $y$ -axis. A proposed solution is shown below.

Line 1. The rotation gives $P'(2,5)$ .

Line 2. The reflection gives $P''(-2,-5)$ .

Line 3. Therefore the two-step mapping is a half-turn about the origin.

Identify the first incorrect line, repair the final coordinates, and replace Line 3 with the correct single motion for the whole plane.
1. Hint 1
  Check each line against the motion that is supposed to produce it.
2. Hint 2
  For the single motion, apply the two rules to an arbitrary point (x,y) and compare with the known reflection rules.
Answer

Line 2 is the first incorrect line; $P''(-2,5)$ . Replace Line 3 with reflection across $y=-x$ .

Full solution

Line 1 is correct: a clockwise quarter-turn uses $(y,-x)$ and sends $(-5,2)$ to $(2,5)$ .

Line 2 is the first error.
Reflection across the y-axis negates horizontal position only, so $P''(-2,5)$ .

For an arbitrary point $(x,y)$ , the first image is $(y,-x)$ .
The second image is $(-y,-x)$ because the y-axis reflection negates the new first coordinate.

The rule $(-y,-x)$ is reflection across the falling diagonal y=-x.
A half-turn would instead use $(-x,-y)$ , so Line 3 must also be replaced.

Answer

Line 2 is the first incorrect line; $P''(-2,5)$ . Replace Line 3 with reflection across $y=-x$ .

Key idea

Repair the first failed step, then recheck every conclusion that depends on it.
Problem 8 Test a stretch on more than one side

Triangle $ABC$ has $A(1,1)$ , $B(1,4)$ , and $C(4,1)$ . Apply $(x,y)\mapsto(3x,y)$ . Give the three images and compare the lengths AB and AC with their image lengths.

Classify the mapping as rigid or nonrigid. Could a rigid sequence realize all three specified point matches? Explain why the unchanged side alone does not decide this.
1. Hint 1
  Check a vertical and a horizontal segment separately.
2. Hint 2
  Use coordinate differences for these axis-parallel lengths. A rigid sequence would preserve every corresponding length.
Answer

$A'(3,1)$ , $B'(3,4)$ , $C'(12,1)$ ; AB and A-prime B-prime are 3 units, while AC is 3 and A-prime C-prime is 9. The map is nonrigid; no rigid sequence realizes all three specified matches.

Full solution

Multiply each horizontal coordinate by 3 and keep each height.
This gives $A'(3,1)$ , $B'(3,4)$ , and $C'(12,1)$ .

The vertical length AB is $4-1=3$ .
Its image length is also $4-1=3$ .

The horizontal length AC is $4-1=3$ .
Its image length is
$12-3=9$

The stretch is nonrigid because it changes this distance.
A rigid sequence making the specified A and C matches would have to preserve AC, which these images do not.
Thus it cannot realize all three matches.

The unchanged vertical side verifies one preserved length.
It does not establish preservation for every pair of points in the plane.

Answer

$A'(3,1)$ , $B'(3,4)$ , $C'(12,1)$ ; AB and A-prime B-prime are 3 units, while AC is 3 and A-prime C-prime is 9. The map is nonrigid; no rigid sequence realizes all three specified matches.

Key idea

A changed distance refutes rigidity even when some other distances stay unchanged.
Problem 9 Build two useful mirror sequences

Choose mirrors from $y=0$ , $y=4$ , $y=x$ , and $x=0$ . Mirrors may be reused between the separate cases. Specify an ordered pair for each requested whole-plane mapping, and check it on $U(2,-3)$ .

Case A. A translation eight units up.

Case B. A 90-degree clockwise rotation about the origin.
1. Hint 1
  Choose parallel mirrors for a slide and intersecting mirrors for a turn.
2. Hint 2
  Use half the required translation distance as the gap, or half the required directed turn as the mirror angle.
Answer

One choice in each case. Case A: first $y=0$ , then $y=4$ ; final $U''(2,5)$ . Case B: first $y=x$ , then $y=0$ ; final $U''(-3,-2)$ .

Full solution

Case A.
The mirrors y=0 and y=4 have a four-unit upward gap.
In that order the doubled gap is eight up.

Check U: reflection across y=0 gives $U'(2,3)$ .
Reflection across y=4 gives height
$2(4)-3=5$

Thus $U''(2,5)$ , eight units above the start.

Case B.
From the rising ray of y=x to the rightward ray of y=0 is 45 degrees clockwise.
These mirrors therefore give a 90-degree clockwise rotation in that order.

Check U: exchange coordinates to get $U'(-3,2)$ .
Then negate the height to get $U''(-3,-2)$ , the clockwise quarter-turn image.

Answer

One choice in each case. Case A: first $y=0$ , then $y=4$ ; final $U''(2,5)$ . Case B: first $y=x$ , then $y=0$ ; final $U''(-3,-2)$ .

Key idea

Design the mirror gap or directed angle before computing sample images.
Problem 10 Make an ordered mapping storyboard

Triangle $ABC$ has $A(-3,1)$ , $B(-1,1)$ , and $C(-2,4)$ . Triangle $DEF$ has $D(1,-3)$ , $E(1,-1)$ , and $F(-2,-2)$ .

Use one quarter-turn about the origin and one translation, in an order you specify, to map $A$ to $D$ , $B$ to $E$ , and $C$ to $F$ . Draw a three-panel storyboard on paper with the start, intermediate image, and final image. Label every vertex and state all intermediate coordinates.

Verify all three final matches. Write the congruence statement and explain why the sequence maps the whole triangle, rather than only the listed vertices.

Text description of this figure

The starting triangle has A (-3,1), B (-1,1), C (-2,4). The target has D (1,-3), E (1,-1), F (-2,-2). Separate panels use the same length scale. No intermediate image or motion is given.
1. Hint 1
  A useful first step may make corresponding sides point the same way.
2. Hint 2
  Try an origin quarter-turn, then compare the resulting points with their respective targets. All three must need the same translation.
3. Hint 3
  Retain the intermediate labels with one prime; label the final images with two primes and their target names.
Answer

One valid sequence: $90^\circ$ counterclockwise about origin, then two right; $A'(-1,-3)$ , $B'(-1,-1)$ , $C'(-4,-2)$ ; $A''=D$ , $B''=E$ , $C''=F$ ; $\triangle ABC\cong\triangle DEF$ .

Full solution

First rotate counterclockwise through 90 degrees about the origin, using $(x,y)\mapsto(-y,x)$ .

The intermediate vertices are $A'(-1,-3)$ , $B'(-1,-1)$ , and $C'(-4,-2)$ .
Draw and join these in the middle panel.

From A-prime to D the horizontal change is
$1-(-1)=2$

Their heights agree.
Check the other matches with the same two-right translation.
B-prime gives $(1,-1)=E$ , and C-prime gives $(-2,-2)=F$ .

The final images are therefore $A''=D$ , $B''=E$ , and $C''=F$ .
Draw that triangle in the third panel with these correspondences.

Each specified step is rigid and takes straight segments to straight segments.
Thus all points on AB reach DE, all on BC reach EF, and all on CA reach FD.
The interior travels with its boundary under the turn and slide.

The whole-figure match establishes $\triangle ABC\cong\triangle DEF$ .
The ordered names record exactly the three verified matches.

Another valid order slides two down first, giving intermediate vertices $(-3,-1)$ , $(-1,-1)$ , and $(-2,2)$ , then turns counterclockwise 90 degrees about the origin.
Its final matches are the same.

Answer

One valid sequence: $90^\circ$ counterclockwise about origin, then two right; $A'(-1,-3)$ , $B'(-1,-1)$ , $C'(-4,-2)$ ; $A''=D$ , $B''=E$ , $C''=F$ ; $\triangle ABC\cong\triangle DEF$ .

Key idea

A congruence storyboard names the motions, retains the intermediate image, and verifies every correspondence.

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