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Rigid Transformations and Symmetry: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. 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 move used twice

The same translation is applied twice to A. Its starting position is $A(8,-6)$ and its position after both applications is $A''(-6,4)$ .

Find the displacement of one application.

Find the intermediate point $A'$ .

Find the image of $B(-1,3)$ under one application.

Check your result by tracing A through both moves.
1. Hint 1
  Both applications use the same horizontal change and the same vertical change.
2. Hint 2
  The total change from A to A-double-prime contains two equal copies of each coordinate change.
3. Hint 3
  If one displacement is $(a,b)$ , the horizontal data give $8+2a=-6$ . Write the corresponding equation for height.
Answer

One displacement is $(-7,5)$ ; $A'(1,-1)$ ; $B'(-8,8)$ .

Full solution

The total horizontal change contains two equal additions.
Its equation is
$8+2a=-6.$
Add negative 8 to both sides to get $2a=-14$ , so $a=-7$ .

The height equation is $-6+2b=4$ .
Add 6 to get $2b=10$ , so $b=5$ .
The rule for one application is $(x,y)\mapsto(x-7,y+5)$ .

The first image of A is $(8-7,-6+5)$ , giving $A'(1,-1)$ .
A second application gives $(-6,4)$ , the required final point.

Apply just one move to B.
Its horizontal coordinate is $-1-7=-8$ and its height is $3+5=8$ , so $B'(-8,8)$ .

Answer

One displacement is $(-7,5)$ ; $A'(1,-1)$ ; $B'(-8,8)$ .

Key idea

Repeated uses of one translation repeat the same displacement, rather than creating a different instruction at each step.
Problem 2 Recover a crease on paper

P and P-prime are a reflection pair with $PP'=8.6$ cm. On paper draw an 8.6 cm segment for this pair, and place a point Q off its supporting line in roughly the position shown. The picture is not a centimeter scale.

Draw the only possible mirror for the pair with a ruler and protractor. Then construct the image Q-prime with a ruler, protractor, and compass. Keep the midpoint, perpendicular feet, and equal-distance marks visible.

Choose a point H on your mirror and give its image. Explain why your intended construction meets the reflection definition and what is approximate in the physical drawing.

Text description of this figure

P and P-prime are a given reflection pair joined by a segment labeled 8.6 centimeters. A separate point Q is shown. No mirror, midpoint, perpendicular foot, or image of Q is drawn. The picture is a placement model, not a centimeter scale.
1. Hint 1
  The reflection definition constrains where the mirror can cross the given nonzero connector.
2. Hint 2
  Locate the midpoint by measuring half the full length, then draw a right angle there with the protractor.
3. Hint 3
  Once the mirror is drawn, locate Q’s perpendicular foot and copy the distance to the opposite side with the compass. Treat H separately.
Answer

The mirror is perpendicular to PP-prime at its midpoint, 4.3 cm from each endpoint. Q-prime is across that mirror on Q's perpendicular with equal half-distances; $H'=H$ . Pencil lengths and angles are approximate.

Full solution

Half of the given connector length is
$8.6\div2=4.3.$
Use the ruler to mark M 4.3 cm from P along the segment.
It is also 4.3 cm from P-prime.

With the protractor centered at M, draw line m perpendicular to PP-prime.
This is the only possible mirror because the reflection definition requires both this midpoint and this perpendicular direction.
This is a ruler-and-protractor drawing, without using a later compass-and-straightedge bisector construction.

Through Q, use the protractor and ruler to draw a perpendicular to m, meeting it at K.
Set the compass to KQ, and copy that length from K onto the ray opposite Q to locate Q-prime.

Mark KQ and KQ-prime equal.
They lie on opposite sides of m along the same perpendicular, so m bisects their connector at a right angle.
The construction meets the reflection definition for Q.

H lies on m, so $H'=H$ .
Its connector is zero and needs no right-angle mark.

The exact intended midpoint, right angles, and copied distance justify the construction.
Ruler readings, protractor readings, compass settings, and pencil marks have limited precision, so checks on the actual drawing are approximate.

Single ticks pair the halves of PP-prime; double ticks pair the halves of QQ-prime.

Answer

The mirror is perpendicular to PP-prime at its midpoint, 4.3 cm from each endpoint. Q-prime is across that mirror on Q's perpendicular with equal half-distances; $H'=H$ . Pencil lengths and angles are approximate.

Key idea

A given reflection pair determines its mirror through the midpoint and perpendicular conditions, before that mirror is used for another point.
Problem 3 Two points turn around one pin

A segment has endpoints $A(-5,-1)$ and $B(-2,-6)$ . Turn it $90^\circ$ clockwise about $C(-2,-1)$ . Give A-prime, B-prime, and the image of C, then draw the image on the original axes.

Give the radius of each endpoint’s circular path. Do the different radii make the distance between the endpoints change? Explain without calculating that slanted segment length.

Text description of this figure

On fixed unit axes, A is at (-5,-1), B at (-2,-6), and center C at (-2,-1). Only the original segment AB and the center are shown. The task specifies a clockwise quarter-turn.
1. Hint 1
  Locate each endpoint relative to the actual center, rather than the origin.
2. Hint 2
  For a clockwise quarter-turn, left becomes up and down becomes left.
3. Hint 3
  Path radii concern each point and C; distance preservation also applies to the pair A,B.
Answer

$A'(-2,2)$ , $B'(-7,-1)$ , $C'=C(-2,-1)$ . The path radii are 3 and 5 units; the distance between A and B is unchanged.

Full solution

A starts 3 units left of C.
The clockwise turn places its image 3 units above C.
Its height is
$-1+3=2.$
Its horizontal coordinate is the center's, so $A'(-2,2)$ .

B starts 5 units below C.
Its image lies 5 units left of C, with horizontal coordinate
$-2-5=-7.$
Its height remains the center's height, so $B'(-7,-1)$ .

The center is fixed by definition: $C'=C$ .
Plot both endpoint images and join them; the turn maps the original straight segment onto that image segment.
Keep the coordinate axes fixed.

A follows a circle of radius 3 centered at C, while B follows a circle of radius 5 centered there.
Both turn through the same clockwise quarter-turn.

The accepted distance preservation of a rotation gives $A'B'=AB$ .
Different path radii do not change this internal distance; no formula for the slanted length is needed.

Answer

$A'(-2,2)$ , $B'(-7,-1)$ , $C'=C(-2,-1)$ . The path radii are 3 and 5 units; the distance between A and B is unchanged.

Key idea

A rotation preserves each radius and every pair distance, even when different points follow different circles.
Problem 4 Locate the second mirror

A device first reflects across $x=2$ and then across a vertical line $x=b$ . The intended whole-plane result is translation 13 units left. Find b.

Case A. Follow $P(0,-3)$ through those two reflections and give both images.

Case B. Restart at P and reverse the mirror order. Give the final image and the resulting whole-plane motion.

Explain why using the same mirror twice could not give the intended nonzero slide.
1. Hint 1
  The single motion depends on the directed perpendicular gap between the mirrors.
2. Hint 2
  The second mirror must be to the left of the first. Their gap is half the intended travel distance.
3. Hint 3
  The gap equation is $2-b=13/2$ . After locating b, apply each reflection to the preceding image.
Answer

$b=-4.5$ (or $-9/2$ ). Case A: $P'(4,-3)$ , $P''(-13,-3)$ . Case B: final image $(13,-3)$ , translation 13 units right. Repeating one mirror gives identity.

Full solution

The gap must be half the travel distance.
$13\div2=6.5.$
The directed slide is left, so the second mirror is 6.5 units left of x=2.

Its coordinate satisfies $b=2-6.5$ , giving $b=-4.5$ .
Thus the two mirrors are distinct.

Case A.
Reflect P across x=2.
The horizontal coordinate is
$2(2)-0=4.$
Height stays at $-3$ , giving $P'(4,-3)$ .

Case A.
Reflect that image across x=-4.5.
Its new horizontal coordinate is
$2(-4.5)-4=-13.$
Hence $P''(-13,-3)$ , as required.

Case B.
The reversed first reflection gives horizontal coordinate
$2(-4.5)-0=-9.$
The next gives
$2(2)-(-9)=13.$
The final point is $(13,-3)$ .

Case B.
Reversing the mirrors reverses the directed gap, so the whole plane now translates 13 units right.
If the mirrors coincided, the second fold would undo the first and give identity, or zero translation.

Answer

$b=-4.5$ (or $-9/2$ ). Case A: $P'(4,-3)$ , $P''(-13,-3)$ . Case B: final image $(13,-3)$ , translation 13 units right. Repeating one mirror gives identity.

Key idea

Two parallel mirrors need half the intended travel as their gap, with the first-to-second direction determining the sign.
Problem 5 The correspondence on a cutout

A specified rigid-motion sequence carries the whole triangle ABC exactly onto triangle PQR, with $A\mapsto R$ , $B\mapsto P$ , and $C\mapsto Q$ . Write the congruence statement in this correspondence and find each parameter and common measurement below.

Length case. $AB=4t-9$ cm and $RP=2t+17$ cm.

Angle case. $m\angle ABC=(4u-5)^\circ$ and $m\angle RPQ=(2u+47)^\circ$ .
1. Hint 1
  Follow the endpoint names for a side and the middle letter for an angle.
2. Hint 2
  The rigid sequence preserves each matching measurement separately.
3. Hint 3
  The length equation is $4t-9=2t+17$ . Write a separate equation for the angle case before solving.
Answer

$\triangle ABC\cong\triangle RPQ$ . Length case: $t=13$ , common length 43 cm. Angle case: $u=26$ , common measure $99^\circ$ .

Full solution

The target order corresponding to A,B,C is R,P,Q.
The given whole-figure rigid match establishes $\triangle ABC\cong\triangle RPQ$ .

Length case.
AB corresponds to RP, so
$4t-9=2t+17.$
Subtract 2t and add 9 to get $2t=26$ , hence $t=13$ .

Length case.
Substitute in the first expression:
$4(13)-9=43.$
The other gives $2(13)+17=43$ too.
The common length is 43 cm.

Angle case.
B maps to P, so the stated image angle has the correct vertex.
Angle preservation gives
$4u-5=2u+47.$
Subtract 2u and add 5 to get $2u=52$ , hence $u=26$ .

Angle case.
Substitute to find
$4(26)-5=99.$
The other expression gives
$2(26)+47=99.$
The common angle is $99^\circ$ , a valid triangle angle.

Answer

$\triangle ABC\cong\triangle RPQ$ . Length case: $t=13$ , common length 43 cm. Angle case: $u=26$ , common measure $99^\circ$ .

Key idea

A rigid sequence establishes the match; ordered correspondence tells you which lengths and angle measures to equate.
Problem 6 Recover the starting triangle

A triangular cutout ABC first reflects across $y=-x$ , then translates by $(-5,4)$ , then makes a half-turn about the origin.

The final match is $A\mapsto D(4,-2)$ .

The final match is $B\mapsto E(4,2)$ .

The final match is $C\mapsto F(1,-1)$ .

Recover the three starting vertices. Show the reverse motions in their required order, then check all three final matches by following the original forward sequence.

Text description of this figure

The given final triangle has D(4,-2), E(4,2), and F(1,-1) on fixed coordinate axes. No earlier image or starting triangle is shown.
1. Hint 1
  The final operation must be undone before the earlier ones.
2. Hint 2
  A half-turn undoes itself. The inverse slide has the opposite displacement. The same diagonal fold undoes the fold.
3. Hint 3
  The first reverse step sends D to $(-4,2)$ . Continue from that point, rather than restarting from D.
Answer

$A(2,-1)$ , $B(6,-1)$ , $C(3,-4)$ . Reverse order: half-turn about origin, translate by $(5,-4)$ , reflect across $y=-x$ .

Full solution

Undo the last motion with another half-turn about the origin.
The recovered second-stage vertices are $A''(-4,2)$ , $B''(-4,-2)$ , and $C''(-1,1)$ .

Undo the slide by adding 5 horizontally and subtracting 4 vertically.
This gives $A'(1,-2)$ , $B'(1,-6)$ , and $C'(4,-3)$ .

Undo the falling-diagonal reflection with that same reflection.
Its rule $(x,y)\mapsto(-y,-x)$ gives $A(2,-1)$ , $B(6,-1)$ , and $C(3,-4)$ .

Forward check for A.
The fold gives $(1,-2)$ .
The slide gives $(-4,2)$ .
The half-turn gives $(4,-2)=D$ .

Forward check for B.
The fold gives $(1,-6)$ .
The slide gives $(-4,-2)$ .
The half-turn gives $(4,2)=E$ .

Forward check for C.
The fold gives $(4,-3)$ .
The slide gives $(-1,1)$ .
The half-turn gives $(1,-1)=F$ .

Each forward step acts on the preceding image, verifying every specified correspondence.

The recovered starting triangle is plotted on fixed unit axes.

Answer

$A(2,-1)$ , $B(6,-1)$ , $C(3,-4)$ . Reverse order: half-turn about origin, translate by $(5,-4)$ , reflect across $y=-x$ .

Key idea

Reverse both the order and the individual motions, then check the recovered data in the original order.
Problem 7 Two established folds

A whole figure has line symmetry across m and also across n. The mirrors meet at O, with the chosen ray of n $36^\circ$ clockwise from the chosen ray of m. The figure has finite rotational order about O.

No counterclockwise angle strictly between $0^\circ$ and $72^\circ$ matches the figure about O. Determine its least positive matching angle, its order including identity once, and its number of nonidentity rotations. Does it have point symmetry about O? Explain using the given whole-figure matches.

Text description of this figure

Lines m and n meet at O. The chosen ray of m points right; the chosen ray of n is 36 degrees clockwise from it. The given mirror angle is marked. These lines are symmetry guides for the figure described in the task; no particular shape is assumed.
1. Hint 1
  A motion that matches the whole figure can be followed by another such motion.
2. Hint 2
  Relate the ordered pair of mirror reflections to a turn about their intersection. The inverse of a matching turn matches too.
3. Hint 3
  The first rotation to examine has magnitude $2\times36^\circ$ . Use the no-smaller-angle condition before treating any angle as least.
Answer

Least positive angle $72^\circ$ ; order 5; 4 nonidentity rotations; no point symmetry about O.

Full solution

Reflect first in m, then in n.
The first reflection leaves the whole figure occupying its original positions, and the second also matches that entire figure.
Their composition therefore matches.

The directed double-mirror turn is clockwise, with magnitude
$2\times36^\circ=72^\circ.$
Its center is O.

The inverse is a 72-degree counterclockwise turn about the same O, which also matches.
The given absence of smaller positive counterclockwise matches makes 72 degrees the least positive angle.

The finite order is
$360\div72=5.$
Count identity once; the number of nonidentity matches is $5-1=4$ .

The matching angles are $0^\circ$ , $72^\circ$ , $144^\circ$ , $216^\circ$ , and $288^\circ$ .
A half-turn is not on this list, so the figure has no point symmetry about O.

Answer

Least positive angle $72^\circ$ ; order 5; 4 nonidentity rotations; no point symmetry about O.

Key idea

Compose established whole-figure matches, then use an actual minimality condition before applying the finite-order formula.
Problem 8 Investigate the paired data

Case A. The proposed pairs are $A(-1,0)\mapsto A'(-1,4)$ and $B(0,2)\mapsto B'(-3,5)$ . On graph paper plot the pairs and draw their two connectors. With a ruler and protractor, investigate their perpendicular bisectors and find their intersection C. Treat this as an exploratory search, then check equal radii and one common directed turn at C.

Case B. Keep the rotation verified in Case A. An added record claims $D(-7,2)\mapsto D'(-3,6)$ . Does that same rotation make this additional match? Check its radius and directed angle. If it fails, give the corrected image of D under the Case A rotation.

Explain why an intersection of search lines alone would not have been a complete check for arbitrary proposed data.

Text description of this figure

A unit grid gives A(-1,0), A-prime (-1,4), B(0,2), and B-prime (-3,5). The two connectors are drawn. No candidate center, perpendicular bisector, or angle arc is shown.
1. Hint 1
  Use nonzero connectors whose perpendicular bisectors are distinct. Return to the rotation definition after locating a candidate.
2. Hint 2
  One connector is vertical. The other runs diagonally up and left; draw each midpoint perpendicular with your ruler and protractor.
3. Hint 3
  Compare the positions of each point and its image relative to C. A matching radius does not choose the turning direction.
Answer

Case A: candidate $C(-3,2)$ is verified by a $90^\circ$ counterclockwise turn. Case B: the added record needs a clockwise quarter-turn and fails for that rotation; the corrected image is $D'(-3,-2)$ .

Full solution

Case A.
The nonzero vertical connector AA-prime has midpoint $(-1,2)$ .
Its midpoint perpendicular is the horizontal line $y=2$ .

Case A.
The midpoint of BB-prime is $(-1.5,3.5)$ .
Drawing its midpoint perpendicular on the grid gives the rising diagonal $y=x+5$ .
This is a distinct search line.

Case A.
At the intersection, $2=x+5$ , so $x=-3$ .
The candidate is $C(-3,2)$ .
The general justification of this center-search construction is deferred; now test the candidate directly.

Case A.
Relative to C, A is 2 right and 2 down, while A-prime is 2 right and 2 up.
A counterclockwise quarter-turn changes the first relative position into the second.
This quarter-turn keeps the radius, without needing to calculate its slanted length.

Case A.
B is 3 right of C and B-prime is 3 above C.
Both radii are 3, and the same counterclockwise quarter-turn matches them.
Thus that rotation verifies both given matches.

Case B.
D is 4 left of C, while its proposed image is 4 above C.
The radii are equal, but this pair needs a clockwise quarter-turn.
It disagrees with the Case A directed turn.

Case B.
The verified counterclockwise turn puts D 4 units below C instead.
Its image height is
$2-4=-2.$
The corrected image is $D'(-3,-2)$ .

An intersection only supplies a candidate in this exploration.
The equal-radius and common directed-angle checks are separate requirements, as the added record demonstrates.

Answer

Case A: candidate $C(-3,2)$ is verified by a $90^\circ$ counterclockwise turn. Case B: the added record needs a clockwise quarter-turn and fails for that rotation; the corrected image is $D'(-3,-2)$ .

Key idea

A candidate center must pass every radius and directed-angle check; equal radii alone do not verify one common rotation.
Problem 9 A letter changes position

The exact line letter A consists only of segments LT, TR, and UV, where $L(-4,-4)$ , $T(0,4)$ , $R(4,-4)$ , $U(-2,0)$ , and $V(2,0)$ . There is no bottom segment or other decoration. The names only identify points.

Translate the whole letter by $(-2,3)$ . Give the images of all five named points and draw the translated figure.

Test the entire translated figure for reflection across $x=-2$ and for a half-turn about $C(-2,3)$ . Give a complete match or a specific failed point for each test.

Text description of this figure

The given letter A consists of segments LT, TR, and UV only. L(-4,-4), T(0,4), R(4,-4), U(-2,0), and V(2,0) locate its endpoints. The bottom is open, and there are no other strokes or decorations. Names only identify points.
1. Hint 1
  Every endpoint receives the same coordinate changes, and the original segment connections remain.
2. Hint 2
  Track how the complete two legs and the crossbar behave under the proposed fold.
3. Hint 3
  For the half-turn, compare the image of the top with all three segments, including the open gap at the bottom.
Answer

$L'(-6,-1)$ , $T'(-2,7)$ , $R'(2,-1)$ , $U'(-4,3)$ , $V'(0,3)$ . The $x=-2$ reflection matches; the half-turn about C fails because $T'$ maps to the absent point $(-2,-1)$ .

Full solution

Add displacement $(-2,3)$ at every endpoint.
L goes to $L'(-6,-1)$ and T to $T'(-2,7)$ .

The other images are $R'(2,-1)$ , $U'(-4,3)$ , and $V'(0,3)$ .
Join L-prime to T-prime, T-prime to R-prime, and U-prime to V-prime.

For reflection across x=-2, L-prime and R-prime exchange, T-prime is fixed, and U-prime and V-prime exchange.
Their matching heights and horizontal midpoints at $x=-2$ verify those endpoint pairs.

The two complete legs therefore exchange, while the crossbar covers itself.
Reflection maps every straight segment between corresponding endpoints, so the whole translated letter matches.

For the half-turn, T-prime is 4 units above C.
Its image would be 4 units below C, with height
$3-4=-1.$

That image is $(-2,-1)$ .
The bottom endpoints are $(-6,-1)$ and $(2,-1)$ , and there is no bottom stroke between them.
The point is absent from both legs and the crossbar, so the half-turn fails.

Answer

$L'(-6,-1)$ , $T'(-2,7)$ , $R'(2,-1)$ , $U'(-4,3)$ , $V'(0,3)$ . The $x=-2$ reflection matches; the half-turn about C fails because $T'$ maps to the absent point $(-2,-1)$ .

Key idea

A translated symmetry must be tested at the translated location, and an exact letter includes its missing strokes as well as its present ones.
Problem 10 The mark inside the cutout

The starting figure consists of the triangular boundary joining $A(-2,-1)$ , $B(3,-1)$ , $C(-2,4)$ , together with a single filled dot $P(-1,1)$ .

The target consists of the boundary joining $D(5,0)$ , $E(5,5)$ , $F(10,0)$ , together with a single filled dot $Q(8,1)$ .

The dots belong to the figures; the point names do not.

First reflect across either $y=x$ or $y=-x$ , then apply one translation.

Choose the reflection and determine the translation for the required matches $A\mapsto D$ , $B\mapsto E$ , and $C\mapsto F$ . Test both permitted mirrors and verify all three final matches for your chosen sequence.

Does your chosen sequence also send P to Q and map the complete starting figure exactly onto the target?

If it fails, keep the target boundary and move only its dot to repair the match. Give the repaired dot coordinates and explain why your chosen sequence then establishes congruence of the two complete figures.

Text description of this figure

Separate panels use the same length scale. Start: triangle ABC has A(-2,-1), B(3,-1), C(-2,4), and an included dot P(-1,1). Target: triangle DEF has D(5,0), E(5,5), F(10,0), and an included dot Q(8,1). Only the given boundaries and dots are shown. Names are labels, not printed decoration.
1. Hint 1
  A valid translation after the reflection must use one common displacement for all three required vertex matches.
2. Hint 2
  Test both permitted mirrors. For each, the A-to-D match determines the translation; check B and C with that same displacement.
3. Hint 3
  Across $y=-x$ , A becomes $(1,2)$ . Compare that intermediate point with D before checking B. After finding a sequence that matches the vertices, follow the included dot through it too.
Answer

Reflect across $y=x$ , then translate by $(6,2)$ ; all three vertex matches work. The $y=-x$ choice fails. The complete match fails: P maps to $(7,1)$ . Move Q to $(7,1)$ ; the repaired figures are congruent.

Full solution

Choice $y=x$ .
Reflecting A gives $A'(-1,-2)$ .
Its required target is $D(5,0)$ .

Choice $y=x$ .
The horizontal change from A-prime to D is
$5-(-1)=6.$

Choice $y=x$ .
The vertical change is
$0-(-2)=2.$
Thus the required translation is $(6,2)$ , which sends A-prime to D.

Choice $y=x$ .
B first becomes $B'(-1,3)$ .
Applying that same slide gives $(5,5)=E$ .

Choice $y=x$ .
C first becomes $C'(4,-2)$ .
The same slide gives $(10,0)=F$ .
All three required vertex matches work.

Choice $y=-x$ .
Reflecting A gives $(1,2)$ .
The required horizontal change to D is $5-1=4$ , and the required vertical change is $0-2=-2$ .
This forces translation $(4,-2)$ for this mirror choice.

Choice $y=-x$ .
B first becomes $(1,-3)$ .
The forced slide sends it to $(5,-5)$ , which is not $E(5,5)$ .
No different translation can fix this B match while retaining the required A-to-D match, so this mirror choice fails.

Choose reflection across $y=x$ followed by translation $(6,2)$ .
Each rigid step takes a straight segment to the segment between its endpoint images.
The three verified vertex matches therefore send AB to DE, BC to EF, and CA to FD, covering every boundary side.

Dot check.
P first becomes $P'(1,-1)$ .
Its final horizontal coordinate is $1+6=7$ and its final height is $-1+2=1$ .
Hence $P''(7,1)$ .

Dot check.
The given target dot is at (8,1), a different position.
Both dot positions are strictly inside the target triangle and off its boundary, so the unmatched image cannot be supplied by a boundary point.
The chosen sequence fails for the complete figures.

Repair.
Move only Q to (7,1).
Now the chosen sequence sends A to D, B to E, C to F, and P to the repaired Q.
It covers each target side and its one dot, so the entire starting figure maps exactly onto the repaired target.

Repair.
Both steps are rigid.
This complete mapping establishes congruence of the repaired marked figures, with all four correspondences specified.

Answer

Reflect across $y=x$ , then translate by $(6,2)$ ; all three vertex matches work. The $y=-x$ choice fails. The complete match fails: P maps to $(7,1)$ . Move Q to $(7,1)$ ; the repaired figures are congruent.

Key idea

A valid sequence must carry every included component, even when the boundary vertices already match.

Multiple choice

Core practice