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Reflections: 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 One point, two axis mirrors

Start with $L(5,-7)$ separately in each case. Give the reflected image and name the coordinate that stays unchanged.

Case A. Reflect across the $x$ -axis.

Case B. Reflect across the $y$ -axis.
1. Hint 1
  Think about which direction crosses each mirror.
2. Hint 2
  A horizontal mirror changes height; a vertical mirror changes horizontal position.
Answer

Case A: $L'(5,7)$ ; horizontal coordinate unchanged. Case B: $L'(-5,-7)$ ; vertical coordinate unchanged.

Full solution

Case A.
The x-axis is horizontal, so keep the horizontal coordinate 5.
Reverse the sign of the height: $-(-7)=7$ .
The image is $L'(5,7)$ .

Case B.
The y-axis is vertical, so keep the height -7.
Reverse the sign of the horizontal coordinate: $-(5)=-5$ .
The image is $L'(-5,-7)$ .

Answer

Case A: $L'(5,7)$ ; horizontal coordinate unchanged. Case B: $L'(-5,-7)$ ; vertical coordinate unchanged.

Key idea

The coordinate across the mirror changes, while the coordinate along the mirror stays unchanged.
Problem 2 A reflected point and its mirror

Reflection across $m$ sends off-mirror point $R$ to $R'$ . The connector has length $RR'=11.4$ cm and meets the mirror at $H$ . Find the distance from $R$ to $m$ , then state the angle between the connector and the mirror.

The bracket labels the full given segment.

Text description of this figure

Line m meets segment RR-prime at H. The endpoints R and R-prime are a stated reflection pair across m. A bracket labels the whole segment length as 11.4 centimeters.
1. Hint 1
  The mirror crosses the point-image connector at its midpoint.
2. Hint 2
  The full length contains two equal perpendicular distances.
Answer

Distance $5.7$ cm; angle $90^\circ$ .

Full solution

The reflection definition gives $RH=HR'$ , so each half is half the full connector.

The distance from R to the mirror is the perpendicular half-length:
$RH=11.4\div2=5.7.$

The connector is perpendicular to the mirror by definition.
Their angle is $90^\circ$ .

The right angle and equal halves follow from the reflection definition.

Answer

Distance $5.7$ cm; angle $90^\circ$ .

Key idea

A point-to-image distance is twice the point-to-mirror distance for an off-mirror point.
Problem 3 Reflect a whole boundary

Quadrilateral $ABCD$ has vertices $A(-4,-2)$ , $B(-2,3)$ , $C(0,2)$ , and $D(-1,-3)$ , joined in that order. Reflect it across $x=1$ . Give every image vertex and draw the image boundary.

Explain how your rule supplies the perpendicular-midpoint condition for each off-mirror vertex.

Only the preimage and the chosen mirror are given.

Text description of this figure

A unit grid shows vertical mirror x equals 1 and solid quadrilateral ABCD with A at (negative 4, negative 2), B at (negative 2, 3), C at (0, 2), and D at (negative 1, negative 3). No image is drawn.
1. Hint 1
  Use one vertical mirror for all four vertices.
2. Hint 2
  Each height stays unchanged, and each starting and image horizontal coordinate must add to twice the mirror coordinate.
Answer

$A'(6,-2)$ , $B'(4,3)$ , $C'(2,2)$ , $D'(3,-3)$ ; join $A'B'C'D'$ in that order. Each connector is horizontal with midpoint on $x=1$ .

Full solution

Across $x=1$ , every point uses the rule $(x,y)\mapsto(2-x,y)$ .

For A, the new horizontal coordinate is $2-(-4)=6$ , giving $A'(6,-2)$ .

For B, it is $2-(-2)=4$ , giving $B'(4,3)$ .

For C, it is $2-0=2$ , giving $C'(2,2)$ .

For D, it is $2-(-1)=3$ , giving $D'(3,-3)$ .
Join the images in the given order.

Each point keeps its height, so its connector is horizontal and perpendicular to the vertical mirror.
Each starting and image horizontal coordinate has average 1, placing the midpoint on that mirror.

Solid preimage and dashed image share the same vertical mirror.

Answer

$A'(6,-2)$ , $B'(4,3)$ , $C'(2,2)$ , $D'(3,-3)$ ; join $A'B'C'D'$ in that order. Each connector is horizontal with midpoint on $x=1$ .

Key idea

Checking both the unchanged height and the midpoint checks a vertical-mirror image.
Problem 4 Compare the diagonal rules

Start with $S(-3,8)$ separately in each case. Find its image.

Case A. Reflect across $y=x$ .

Case B. Reflect across $y=-x$ .

Case C. Among $P(2,2)$ , $Q(-2,2)$ , and $O(0,0)$ , identify those fixed by both reflections.
1. Hint 1
  The rising and falling diagonal folds exchange coordinate roles in different ways.
2. Hint 2
  For a point to be fixed by both reflections, it must lie on both mirrors.
Answer

Case A: $S'(8,-3)$ . Case B: $S'(-8,3)$ . Case C: $O(0,0)$ only.

Full solution

Case A.
Across $y=x$ , exchange the coordinates without reversing their signs.
This gives $S'(8,-3)$ .

Case B.
Across $y=-x$ , exchange the coordinates and reverse both signs.
The new first coordinate is $-8$ , and the new second is $-(-3)=3$ , giving $S'(-8,3)$ .

Case C.
P lies on $y=x$ but not $y=-x$ .
Q lies on $y=-x$ but not $y=x$ .
O lies on both lines, so it is the only listed point fixed by both reflections.

Answer

Case A: $S'(8,-3)$ . Case B: $S'(-8,3)$ . Case C: $O(0,0)$ only.

Key idea

A fixed-point check can distinguish the two diagonal mirror rules.
Problem 5 Follow the correspondence around a triangle

A reflection maps noncollinear $X,Y,Z$ to $X',Y',Z'$ respectively. Given $XY=12$ cm, $YZ=7$ cm, and $m\angle XYZ=48^\circ$ , find the length of the two-segment path $X'Y'Z'$ and the measure of the angle at $Y'$ .

The order $X,Y,Z$ is clockwise. State the orientation of $X',Y',Z'$ and explain whether that changes the angle measure.
1. Hint 1
  Pair the image endpoints with the original endpoints before using preservation.
2. Hint 2
  The path contains two corresponding sides; orientation concerns the order around the triangle.
Answer

Path length $19$ cm; $m\angle X'Y'Z'=48^\circ$ ; counterclockwise orientation. The orientation reversal does not change the angle measure.

Full solution

Distance preservation gives $X'Y'=12$ cm and $Y'Z'=7$ cm.
Add the two path pieces:
$12+7=19.$

The vertex Y corresponds to Y-prime.
Angle preservation gives
$m\angle X'Y'Z'=48^\circ.$

Reflection reverses the orientation of this noncollinear triangle, so $X',Y',Z'$ is counterclockwise.
Reversal changes the boundary order without changing the opening at the corresponding vertex.

Answer

Path length $19$ cm; $m\angle X'Y'Z'=48^\circ$ ; counterclockwise orientation. The orientation reversal does not change the angle measure.

Key idea

A reflection reverses labeled boundary order while preserving corresponding measurements.
Problem 6 Recover the mirror, then compare journeys

A reflection in a vertical line sends $P(12,1)$ to $P'(-2,1)$ . Find the mirror and the image of $Q(3,-6)$ .

Calculate the travel distances $PP'$ and $QQ'$ . Explain why different travel distances are possible even though a reflection preserves distances.
1. Hint 1
  The first pair locates the midpoint of its horizontal connector.
2. Hint 2
  Distance preservation compares distances between two original points with distances between their images; it does not require equal travel for all points.
Answer

Mirror $x=5$ ; $Q'(7,-6)$ ; $PP'=14$ units and $QQ'=4$ units. Preservation compares distances within the preimage and image, not the travel distances of different points.

Full solution

The vertical mirror lies halfway between horizontal coordinates 12 and -2:
$h=\frac{12-2}{2}=5.$

Reflect Q across $x=5$ by keeping its height and calculating the new horizontal coordinate:
$x'=2(5)-3=7.$
The image is $Q'(7,-6)$ .

P and P-prime share a height, so their travel distance is the horizontal difference:
$PP'=12-(-2)=14.$

Q and Q-prime also share a height, giving
$QQ'=7-3=4.$

P is farther from the mirror than Q, so its travel is longer.
Distance preservation compares PQ with P-prime Q-prime; it does not compare PP-prime with QQ-prime.

Answer

Mirror $x=5$ ; $Q'(7,-6)$ ; $PP'=14$ units and $QQ'=4$ units. Preservation compares distances within the preimage and image, not the travel distances of different points.

Key idea

Points farther from the mirror travel farther, while the distances within the figure are preserved.
Problem 7 Track a segment and its points

Segment $AB$ joins $A(-6,-6)$ to $B(6,6)$ . Reflect it across $y=-x$ . Give the images of both endpoints, identify every fixed point on the segment, and decide whether its entire set of points occupies the same position afterward.

Given segment AB and mirror $y=-x$ .

Text description of this figure

A unit grid shows segment AB joining A at (negative 6, negative 6) and B at (6, 6), with the mirror line y equals negative x. No image is drawn.
1. Hint 1
  A point may move into another point’s old position.
2. Hint 2
  Apply the falling-diagonal rule to the endpoints, then find where the segment meets the mirror.
Answer

$A'(6,6)=B$ , $B'(-6,-6)=A$ ; the only fixed point is $(0,0)$ . The image segment occupies the same set of positions.

Full solution

Exchange coordinates and reverse their signs.
This gives $A'(6,6)$ and $B'(-6,-6)$ , so the endpoints exchange places.
The image joins the same two endpoint positions and therefore occupies the original segment.

Every point on this segment has equal coordinates, so write it as $(t,t)$ with
$-6\le t\le6.$
A point is also on the mirror only when $t=-t$ .

Solving gives $2t=0$ , so $t=0$ .
The only fixed point is $(0,0)$ .
Other points move to different places within the same segment.

Answer

$A'(6,6)=B$ , $B'(-6,-6)=A$ ; the only fixed point is $(0,0)$ . The image segment occupies the same set of positions.

Key idea

The image set staying in place does not mean that each of its points is fixed.
Problem 8 Check and repair a proposed boundary image

Triangle $ABC$ has $A(-5,-1)$ , $B(-2,4)$ , and $C(3,6)$ . A proposed reflection across $y=-x$ lists $A'(1,5)$ , $B'(-4,2)$ , and $C'(-6,3)$ .

Check every listed pair. Repair any incorrect image and state how to join the corrected boundary.
1. Hint 1
  Use the same falling-diagonal rule at each vertex.
2. Hint 2
  Both signs reverse after the coordinates exchange roles; check both image coordinates.
Answer

A and B are correct; repair C to $C'(-6,-3)$ . Join the corrected vertices as $A'B'C'$ .

Full solution

For A, the rule sends $(-5,-1)$ to $(1,5)$ , agreeing with the proposal.

For B, it sends $(-2,4)$ to $(-4,2)$ , also agreeing.

For C, the new horizontal coordinate is $-6$ and the new vertical coordinate is $-3$ .
The proposed height 3 has the wrong sign, so replace that vertex with $C'(-6,-3)$ .

Join $A'$ to $B'$ , $B'$ to the corrected $C'$ , and $C'$ to $A'$ .
The corrected boundary uses the same reflection for all vertices.

Answer

A and B are correct; repair C to $C'(-6,-3)$ . Join the corrected vertices as $A'B'C'$ .

Key idea

A correct match at one vertex does not excuse a different rule at another.
Problem 9 Two fixed points reveal the mirror

A reflection fixes distinct points $A(1,1)$ and $B(3,3)$ . Which mirror is possible among $x=2$ , $y=2$ , $y=x$ , and $y=-x$ ? Justify your choice, then find the image of $P(-2,5)$ .
1. Hint 1
  A fixed point lies on the reflection’s mirror.
2. Hint 2
  Test which listed line contains both fixed points; two distinct points determine one line.
Answer

Mirror $y=x$ ; $P'(5,-2)$ .

Full solution

The mirror must contain both A and B, because both are fixed.
Of the listed lines, only y=x contains both points.
Two distinct points determine one line, so another mirror through both is impossible.

Across $y=x$ , exchange the coordinates of P:
$(-2,5)\mapsto(5,-2).$

Answer

Mirror $y=x$ ; $P'(5,-2)$ .

Key idea

Two distinct fixed points determine the mirror, then its rule determines every image.
Problem 10 Build an image across an oblique mirror

On paper, draw an oblique line $m$ and a noncollinear triangle $ABC$ , with $A,B$ off $m$ on the same side and $C$ on $m$ . Use the picture as a model; no exact scale is required.

Construct its reflected image with a ruler, protractor, and compass. Keep the perpendicular feet and equal-distance marks visible. Explain why the intended construction meets the reflection definition, identify the fixed vertex, and state what is approximate in a real pencil drawing.

Choose a similar noncollinear triangle on your own paper.

Text description of this figure

An oblique mirror m rises toward the right. Triangle ABC lies on one side, with A and B off the mirror and C on it. No perpendiculars, feet, copied lengths, or image triangle are shown.
1. Hint 1
  An off-mirror point needs a perpendicular connector with equal halves.
2. Hint 2
  Use the protractor to draw the right angle, then use the compass to transfer a distance across the foot.
3. Hint 3
  Treat the vertex on the mirror separately before joining the image vertices.
Answer

A reflected triangle $A'B'C'$ with perpendicular connectors bisected by $m$ ; $C'=C$ is the fixed vertex. Pencil angles and lengths are approximate.

Full solution

Through A, use a ruler and protractor to draw a line perpendicular to m.
Label its intersection with the mirror H.
The existence and uniqueness of this perpendicular foot were proved in Proving Lines Parallel or Perpendicular.

Set the compass to HA.
On the ray from H opposite A, mark A-prime at that compass distance.
Thus $HA'=HA$ , with A and A-prime on opposite sides.
The connector is perpendicular to m, and H is its midpoint.
Mark these equal halves with single ticks.

Repeat for B, labeling its perpendicular foot K.
Copy KB to the opposite ray to obtain B-prime.
Mark the right angle at K and these equal halves with double ticks, using a different family from the A connector.

C lies on m, so set $C'=C$ .
It has no nonzero connector and needs no perpendicular mark.
Join the three image vertices in corresponding order.

Each off-mirror vertex now meets the opposite-side, perpendicular, and midpoint conditions.
The fixed vertex meets the on-mirror clause.
Reflection takes the straight sides to the joined image sides.

The ideal angle-drawing foundation and compass length-copy fact justify the intended conditions.
Actual pencil marks and protractor readings have limited precision, so measurement checks give approximate evidence about the physical drawing.

Single and double tick families distinguish the equal halves within each connector.

Answer

A reflected triangle $A'B'C'$ with perpendicular connectors bisected by $m$ ; $C'=C$ is the fixed vertex. Pencil angles and lengths are approximate.

Key idea

For an oblique mirror, construct each off-mirror image from a perpendicular and an equal distance, while keeping mirror points fixed.

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