Appendix A — Answer Key, Chapter 14: Reflections and Composite Transformations

SOL 8.MG.3 (b, c, e, f, g) · Covers textbook Chapter 14 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 128 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Rules used throughout:

• reflection over the x-axis: $(x, y) \to (x, -y)$
• reflection over the y-axis: $(x, y) \to (-x, y)$
• translation: $(x, y) \to (x + a, y + b)$

Drawing convention for every sketching item: the preimage is dashed, the image is solid, image vertices carry prime marks, and the intermediate figure of a composition is drawn lightly and labeled with single primes while the final image carries double primes.

Sketching items are scored on the coordinates listed here, on the image vertices being connected in the same order as the preimage, and on correct labeling.

Lesson 14.1 — Reflecting over the x- and y-Axis

Guided practice

1. $A'(3, -5)$
2. $B'(2, 4)$
3. $C'(-6, 1)$
4. $D'(-7, -3)$
5. $A'(1, -2)$, $B'(4, -2)$, $C'(4, -6)$
6. $A'(-1, 2)$, $B'(-4, 2)$, $C'(-4, 6)$
7. $E'(5, 0)$. The point already lies on the x-axis, and the rule gives $(5, -0) = (5, 0)$; a point on the mirror line is its own image.
8. The x-axis. The $x$-coordinates match and the $y$-coordinates are opposites, which is $(x, y) \to (x, -y)$.

Independent practice

9. a) $(2, -9)$ b) $(-2, 9)$ c) $(-5, 8)$ d) $(5, -8)$
10. $P'(-2, 1)$, $Q'(-6, 1)$, $R'(-6, 5)$, $S'(-3, 5)$
11. $J'(-4, -3)$, $K'(-1, -3)$, $L'(-1, -7)$
12. $(1, -2)$, $(3, -1)$, $(5, -3)$, $(4, -6)$, $(2, -5)$
13. Image 1 used the y-axis, with $A'(-2, 2)$: the $y$-coordinates match and the $x$-coordinates are opposites. Image 2 used the x-axis, with $A'(2, -2)$: the $x$-coordinates match and the $y$-coordinates are opposites.
14. In Quadrant III both coordinates are negative. Reflecting over the x-axis makes $y$ positive, giving Quadrant II. Reflecting over the y-axis makes $x$ positive, giving Quadrant IV.
15. Over the y-axis: $(6, 0)$. Over the x-axis: $(-6, 0)$, unchanged. The x-axis reflection left the point fixed, because $M$ lies on the x-axis and $-0 = 0$.
16. The y-axis. The $y$-coordinates are identical and the $x$-coordinates are opposites, which is exactly what $(x, y) \to (-x, y)$ does; no graph is needed to see it.
17. $R'(-2, -5)$, $S'(-5, -1)$, $T'(-1, -1)$. Only the $y$-coordinate of each vertex changed; every $x$-coordinate stayed the same.
18. The $y$-coordinate measures how far a point is from the x-axis, so it is the coordinate that must switch sides of the mirror. The $x$-coordinate measures position along the mirror, and a reflection does not slide points along the mirror line — the image stays in the same vertical line as the preimage, so $x$ is unchanged.
19. $(-1, 1)$, $(-5, 2)$, $(-3, 4)$. The two ships are congruent: a reflection preserves side lengths and angle measures, so only the direction the ship faces has changed.
20. The student negated the $x$-coordinate instead of the $y$-coordinate, which is the rule for a reflection over the y-axis, not the x-axis. Correct image: $(5, -2)$.

Exit ticket 14.1

21. $(-8, -3)$
22. $(8, 3)$
23. $D'(-2, 3)$, $E'(-5, 4)$, $F'(-3, 7)$
24. Negate the coordinate that measures distance from the mirror: reflecting over the x-axis negates $y$, and reflecting over the y-axis negates $x$.

Lesson 14.2 — Sketching a Reflection

Guided practice

25. $A'(1, -1)$, $B'(5, -2)$, $C'(2, -4)$
26. $A'(-1, 1)$, $B'(-5, 2)$, $C'(-2, 4)$
27. $(-2, 1)$, $(-6, 1)$, $(-6, 3)$, $(-2, 3)$
28. $(-1, -2)$, $(-4, -2)$, $(-5, -5)$, $(-2, -6)$
29. $(0, 3)$, $(-4, 1)$, $(-3, 5)$. The vertex $(0, 3)$ did not move, because it lies on the y-axis and $-0 = 0$.
30. $(0, 3)$ is $0$ units from the y-axis and stays put; $(4, 1)$ is $4$ units right and its image is $4$ units left at the same height; $(3, 5)$ is $3$ units right and its image is $3$ units left at the same height.

Independent practice

31. $(1, -1)$, $(4, -1)$, $(5, -3)$, $(3, -5)$, $(1, -4)$
32. $(-1, 1)$, $(-4, 1)$, $(-5, 3)$, $(-3, 5)$, $(-1, 4)$
33. $(-2, 1)$, $(-6, 1)$, $(-4, 5)$. The image lies in Quadrant II.
34. $(-2, 2)$, $(-7, 2)$, $(-6, 5)$, $(-3, 5)$
35. $(2, 1)$, $(-3, 1)$, $(-1, 4)$. The preimage crosses the y-axis, so the image crosses it too and the two triangles overlap near the axis; the vertex that was $2$ units left is now $2$ units right, and the vertex that was $3$ units right is now $3$ units left.
36. $(1, -1)$, $(4, -1)$, $(1, -3)$. Reading $A' \to B' \to C'$ runs clockwise: a reflection reverses orientation.
37. $(1, -3)$, $(4, -2)$, $(6, -5)$, $(2, -7)$
38. Folding the paper along the mirror line carries every point to the point the same distance away on the other side, which is exactly what the reflection rule does. So in a correct sketch the image lands precisely on the preimage when the paper is folded; if any vertex misses its partner, that vertex was plotted at the wrong distance from the axis or on the wrong side.
39. In Quadrant I both coordinates are positive. Reflecting over the x-axis makes $y$ negative, giving Quadrant IV. Reflecting over the y-axis makes $x$ negative, giving Quadrant II.
40. Second wing: $(0, 0)$, $(-2, 1)$, $(-5, 2)$, $(-3, 5)$, $(0, 4)$. The vertices $(0, 0)$ and $(0, 4)$ lie on the y-axis and do not move, so the two wings join along the body line.
41. The student drew a translation $5$ units down: every point moved down $5$ with no flip. The orientation check exposes it — the student's figure runs the same way around as the preimage, but a reflection must reverse it. Correct image: $(1, -2)$, $(4, -2)$, $(3, -5)$.
42. Any triangle symmetric about the y-axis works, for example $(-3, 1)$, $(3, 1)$, $(0, 5)$. Reflecting negates each $x$-coordinate and returns the same three points, so the image lands on the preimage. What is special is that the y-axis is a line of symmetry of the figure.

Exit ticket 14.2

43. $(2, -1)$, $(6, -3)$, $(3, -6)$
44. $(-2, 1)$, $(-6, 3)$, $(-3, 6)$
45. It stays at $(0, -4)$. The point is on the y-axis, and negating $0$ gives $0$.
46. Fold the paper along the mirror line; the image should land exactly on the preimage. If you had reflected over the wrong axis, the two figures would not meet at all when folded along the axis named in the problem.

Lesson 14.3 — Composite Transformations and the Order of the Moves

Guided practice

47. $A'(-2, 2)$, $A''(-2, -2)$
48. $A'(3, -2)$, $A''(-2, -2)$. The final point matches item 47: the translation is horizontal, so it runs parallel to the x-axis mirror and the order can be swapped.
49. $B'(-4, 4)$, $B''(4, 4)$
50. $B'(4, 1)$, $B''(4, 4)$. The final point matches item 49: the translation is vertical, so it runs parallel to the y-axis mirror.
51. $C'(2, 9)$, $C''(2, -9)$
52. $C'(2, -5)$, $C''(2, -1)$. This time the final points differ, because an up-and-down translation is perpendicular to the x-axis mirror. The gap of $8$ is twice the $4$-unit translation.
53. Preimage $A(1, 2)$, $B(4, 2)$, $C(4, 5)$; after the translation $A'(3, 1)$, $B'(6, 1)$, $C'(6, 4)$; after the reflection $A''(-3, 1)$, $B''(-6, 1)$, $C''(-6, 4)$.
54. After the reflection $A'(-1, 2)$, $B'(-4, 2)$, $C'(-4, 5)$; after the translation $A''(1, 1)$, $B''(-2, 1)$, $C''(-2, 4)$. Different from item 53, because the translation has a horizontal component and the mirror is the y-axis.

Independent practice

55. $P'(-6, -3)$, $P''(-6, -7)$
56. $P'(6, -7)$, $P''(-6, -7)$ — same final point as item 55, since the translation is vertical and the mirror is the y-axis.
57. $Q'(1, 8)$, $Q''(1, -8)$
58. $Q'(-2, -7)$, $Q''(1, -6)$ — different from item 57, because the translation has a vertical component and the mirror is the x-axis.
59. Preimage $D(-5, 1)$, $E(-2, 1)$, $F(-2, 4)$; after the reflection $D'(-5, -1)$, $E'(-2, -1)$, $F'(-2, -4)$; after the translation $D''(1, -1)$, $E''(4, -1)$, $F''(4, -4)$.
60. After the translation $W'(1, 3)$, $X'(5, 3)$, $Y'(5, 5)$, $Z'(2, 5)$; after the reflection $W''(1, -3)$, $X''(5, -3)$, $Y''(5, -5)$, $Z''(2, -5)$.
61. After the reflection $W'(1, -1)$, $X'(5, -1)$, $Y'(5, -3)$, $Z'(2, -3)$; after the translation $W''(1, 1)$, $X''(5, 1)$, $Y''(5, -1)$, $Z''(2, -1)$.
62. The two final images have the same $x$-coordinates, and every $y$-coordinate differs by $4$. The translation moves the figure up $2$. In item 60 that rise happens first and the mirror then turns it into a $2$-unit drop; in item 61 the rise happens after the mirror and stays a rise. Reversing a $2$-unit rise into a $2$-unit drop accounts for the gap of $2 \times 2 = 4$.
63. $A'(-5, 3)$. Translating $(2, 3)$ seven units left gives $(-5, 3)$, and reflecting that over the x-axis gives the stated $(-5, -3)$.
64. $B(3, 6)$. Undo the reflection first: reflecting $(4, -6)$ over the x-axis gives $(4, 6)$. Then undo the translation by moving $1$ unit left: $(3, 6)$. Forward check: $(3, 6) \to (4, 6) \to (4, -6)$.
65. After the conveyor moves it $8$ units left: $(-7, 2)$, $(-4, 2)$, $(-4, 6)$. After the flip over the x-axis: $(-7, -2)$, $(-4, -2)$, $(-4, -6)$.
66. The intermediate point $R'(4, 6)$ is right, but the student then reflected the original point $R(4, 1)$ instead of $R'$. The second transformation must be applied to the output of the first. Correct: $R''(4, -6)$.

Exit ticket 14.3

67. $M'(-3, -1)$, $M''(3, -1)$
68. After the translation $(1, 6)$, $(3, 6)$, $(1, 9)$; after the reflection $(1, -6)$, $(3, -6)$, $(1, -9)$.
69. After the reflection $(1, -1)$, $(3, -1)$, $(1, -4)$; after the translation $(1, 4)$, $(3, 4)$, $(1, 1)$. Different from item 68.
70. Order does not matter when the translation runs parallel to the mirror line — a horizontal slide with an x-axis reflection, or a vertical slide with a y-axis reflection — because the mirror has nothing in that direction to reverse. Order does matter whenever the translation has a component perpendicular to the mirror, since the reflection reverses that part of the slide.

Lesson 14.4 — Sketching Compositions and Describing Transformations in Context

Guided practice

71. After the translation $A'(-5, 1)$, $B'(-2, 1)$, $C'(-2, 3)$; after the reflection $A''(-5, -1)$, $B''(-2, -1)$, $C''(-2, -3)$.
72. After the reflection $A'(1, -1)$, $B'(4, -1)$, $C'(4, -3)$; after the translation $A''(-5, -1)$, $B''(-2, -1)$, $C''(-2, -3)$. The final image is the same as in item 71, because the translation is horizontal and therefore parallel to the x-axis mirror. The intermediate figures are in different places, but the two routes end together.
73. After the translation $(2, 3)$, $(5, 3)$, $(5, 6)$, $(3, 6)$; after the reflection $(2, -3)$, $(5, -3)$, $(5, -6)$, $(3, -6)$.
74. After the reflection $(2, -1)$, $(5, -1)$, $(5, -4)$, $(3, -4)$; after the translation $(2, 1)$, $(5, 1)$, $(5, -2)$, $(3, -2)$. Different from item 73: every $y$-coordinate differs by $4$, twice the $2$-unit vertical translation, because that translation is perpendicular to the x-axis mirror.
75. A reflection over the y-axis. Every $y$-coordinate is unchanged, every $x$-coordinate is negated, and the orientation is reversed. Rule: $(x, y) \to (-x, y)$.
76. A reflection. Size and shape are preserved — the printed design is congruent to the design cut into the stamp — but the orientation is reversed, which is why any lettering on the stamp must be carved backwards to print forwards.

Independent practice

77. After the reflection $(1, 2)$, $(4, 2)$, $(4, 6)$; after the translation $(1, -1)$, $(4, -1)$, $(4, 3)$.
78. After the translation $(3, 1)$, $(6, 2)$, $(7, 5)$, $(4, 4)$; after the reflection $(-3, 1)$, $(-6, 2)$, $(-7, 5)$, $(-4, 4)$.
79. After the reflection $(2, -2)$, $(6, -2)$, $(6, -5)$; after the translation $(-1, -1)$, $(3, -1)$, $(3, -4)$.
80. After the translation $(-1, 3)$, $(3, 3)$, $(3, 6)$; after the reflection $(-1, -3)$, $(3, -3)$, $(3, -6)$. Different from item 79: every $y$-coordinate differs by $2$, twice the $1$-unit vertical part of the translation. The horizontal part of the translation runs parallel to the mirror and causes no difference at all.
81. The preimage $(1, 1)$, $(4, 2)$, $(2, 4)$ has image $(4, -1)$, $(7, -2)$, $(5, -4)$. The orientation is reversed and every $y$-coordinate is negated, so the mirror is the x-axis; every $x$-coordinate is $3$ larger, so there is also a translation $3$ units right. Rule: $(x, y) \to (x + 3, -y)$. Because the translation is parallel to the mirror, describing it in either order is correct.
82. A translation $3$ units left and $4$ units down: the image is $(-2, -3)$, $(1, -2)$, $(-1, 0)$, the orientation is unchanged, and every vertex moved by the same amounts. Rule: $(x, y) \to (x - 3, y - 4)$.
83. The reflection appears at $(-4, 2)$. The mirror reverses orientation, so left and right are exchanged: the hand on the dancer's right is the hand nearest the left side of the reflected figure, which reads as the reflection's left hand.
84. $(x, y) \to (x + 5, -y)$. The corner at $(-2, 3)$ moves to $(3, 3)$ on the conveyor and then to $(3, -3)$ after the flip.
85. The first border uses a translation $4$ units right, repeated. The second border uses a composition of that translation with a reflection over the horizontal center line for every other copy.
86. A translation preserves orientation and a reflection reverses it, so reading the labeled vertices around each figure and comparing the direction of travel decides the question by itself. Size tells you nothing here because both transformations produce a congruent image — every figure in this chapter is the same size as its preimage.
87. It is safe exactly when the translation runs parallel to the mirror line, since then the two moves can be performed in either order and give the same image. It gives the wrong image whenever the translation has a component perpendicular to the mirror: performing the slide after the reflection leaves that component pointing the original way, while performing it before the reflection has the mirror reverse it.
88. Every transformation in this chapter produces a congruent image, so "same size and shape" is true of translations, reflections, and their compositions alike, and rules nothing out. The student skipped the orientation check: reading the vertices in order around both figures shows whether the direction of travel reversed, which is what identifies a reflection.

Exit ticket 14.4

89. After the translation $(1, -2)$, $(3, -4)$, $(5, -1)$; after the reflection $(-1, -2)$, $(-3, -4)$, $(-5, -1)$.
90. After the reflection $(-1, 3)$, $(-3, 1)$, $(-5, 4)$; after the translation $(-1, -2)$, $(-3, -4)$, $(-5, -1)$. The final image is the same as in item 89, because a vertical translation runs parallel to the y-axis mirror, so the reflection has nothing in that direction to reverse.
91. $(x, y) \to (-x, y + 4)$. Applying it to $(3, -1)$ gives the intermediate point $(-3, -1)$ and the final point $(-3, 3)$.
92. A translation carries the figure along without turning it over, so the vertices still run the same way around; a reflection flips the figure, so the vertices run the opposite way around. Reversed orientation therefore means a reflection was part of what happened.

Chapter 14 Review

Part A — Coordinates of a reflection over the x- or y-axis (8.MG.3b)

93. $(7, 2)$
94. $(-7, -2)$
95. $A'(-3, 1)$, $B'(-6, 1)$, $C'(-6, 5)$
96. $D'(-2, -2)$, $E'(-5, -2)$, $F'(-6, -5)$, $G'(-3, -6)$
97. The x-axis. The $x$-coordinates are identical and the $y$-coordinates are opposites, which is $(x, y) \to (x, -y)$.
98. $(0, -6)$, unchanged. The point lies on the y-axis, and negating $0$ gives $0$, so a point on the mirror line is its own image.
99. Over the x-axis: $(5, -5)$, in Quadrant IV. Over the y-axis: $(-5, 5)$, in Quadrant II.

Part B — Coordinates of a composition (8.MG.3c)

100. Intermediate $(5, 4)$; final $(-5, 4)$.
101. Intermediate $(-1, 4)$; final $(3, 4)$. This differs from item 100 because the translation is horizontal and the mirror is the y-axis, so the reflection reverses the slide.
102. After the translation $P'(2, 4)$, $Q'(5, 4)$, $R'(5, 6)$; after the reflection $P''(2, -4)$, $Q''(5, -4)$, $R''(5, -6)$.
103. After the reflection $P'(2, -1)$, $Q'(5, -1)$, $R'(5, -3)$; after the translation $P''(2, 2)$, $Q''(5, 2)$, $R''(5, 0)$.
104. Each final $y$-coordinate differs by $6$. The translation is a $3$-unit rise. Doing it first lets the mirror turn it into a $3$-unit drop; doing it last leaves it a $3$-unit rise. Changing $+3$ into $-3$ is a change of $2 \times 3 = 6$.
105. $(6, -3)$. Undo the translation by moving down $5$, giving $(-6, -3)$; undo the y-axis reflection by negating $x$, giving $(6, -3)$. Forward check: $(6, -3) \to (-6, -3) \to (-6, 2)$.
106. Intermediate $(4, -4)$; final $(4, 4)$.

Part C — Sketching a reflection (8.MG.3e)

107. $(1, -2)$, $(5, -2)$, $(3, -6)$
108. $(-2, 1)$, $(-6, 2)$, $(-5, 5)$, $(-2, 4)$
109. $(-3, 2)$, $(-7, 2)$, $(-5, 6)$. The image lies in Quadrant II.
110. $(0, 4)$, $(-3, 2)$, $(-2, 6)$. The vertex $(0, 4)$ did not move, because it lies on the y-axis.
111. $(1, 2)$ is $2$ units above the x-axis and its image is $2$ units below at $x = 1$; $(5, 2)$ is $2$ above and its image $2$ below at $x = 5$; $(3, 6)$ is $6$ above and its image $6$ below at $x = 3$.

Part D — Sketching a composition (8.MG.3f)

112. After the translation $(1, -3)$, $(4, -2)$, $(2, 1)$; after the reflection $(-1, -3)$, $(-4, -2)$, $(-2, 1)$.
113. After the reflection $(-1, 1)$, $(-4, 2)$, $(-2, 5)$; after the translation $(-1, -3)$, $(-4, -2)$, $(-2, 1)$. The final image matches item 112, because the translation is vertical and therefore parallel to the y-axis mirror.
114. After the translation $(2, 3)$, $(5, 3)$, $(5, 5)$, $(3, 5)$; after the reflection $(2, -3)$, $(5, -3)$, $(5, -5)$, $(3, -5)$.
115. After the reflection $(2, -2)$, $(5, -2)$, $(5, -4)$, $(3, -4)$; after the translation $(2, -1)$, $(5, -1)$, $(5, -3)$, $(3, -3)$. Every $y$-coordinate differs from item 114 by $2$, twice the $1$-unit vertical translation, because that translation is perpendicular to the x-axis mirror.
116. After the reflection $(2, 1)$, $(2, 5)$, $(6, 1)$; after the translation $(3, -1)$, $(3, 3)$, $(7, -1)$.

Part E — Identifying and describing transformations in context (8.MG.3g)

117. A reflection over the y-axis. The image is $(-1, 1)$, $(-4, 2)$, $(-2, 4)$: the $y$-coordinates are unchanged, the $x$-coordinates are negated, and the orientation is reversed. Rule: $(x, y) \to (-x, y)$.
118. A translation $3$ units left and $4$ units down. The image is $(-2, -3)$, $(1, -2)$, $(-1, 0)$, every vertex moved by the same amounts, and the orientation is unchanged, so no reflection occurred. Rule: $(x, y) \to (x - 3, y - 4)$.
119. A composition: a reflection over the x-axis together with a translation $3$ units right, giving the image $(4, -1)$, $(7, -2)$, $(5, -4)$. Rule: $(x, y) \to (x + 3, -y)$. Either order is a correct description, because the translation runs parallel to the mirror line.
120. A reflection over the y-axis, $(x, y) \to (-x, y)$. The side lengths, angle measures, and the size of the icon all stayed the same; only the direction it points, and therefore its orientation, changed.
121. A reflection over the y-axis. A spectator can tell it is not a slide because the mirrored formation is reversed: a member who was on the outside edge of the first shape is on the inside edge of the second. A slide would have kept every part of the shape facing the same way.
122. a) translation b) reflection c) composition — a translation $5$ units right followed by a reflection d) translation e) reflection

Part F — Mixed reasoning and application

123. Label a triangle $A$, $B$, $C$ so that $A \to B \to C$ runs counterclockwise. A translation moves all three vertices the same distance in the same direction, so their positions relative to one another are untouched and the trip around the image still runs counterclockwise. A reflection sends each vertex to the opposite side of the mirror, which exchanges the two sides of the figure — what was on the left of the path is now on the right — so the trip around the image runs clockwise. Size and shape survive both; only the sense of the loop is reversed, and only by the reflection.
124. The student negated the $y$-coordinate instead of the $x$-coordinate, which is the rule for a reflection over the x-axis. Over the y-axis the rule is $(x, y) \to (-x, y)$, so the correct image is $(-6, -2)$.
125. The second transformation must be applied to the image of the first, not to the original point. The intermediate point is $(1, 7)$, and reflecting that over the x-axis gives the correct final point $(1, -7)$.
126. With a reflection over the x-axis, the order does not matter for a purely horizontal translation, because left-and-right movement is parallel to the mirror and an up-down flip does not disturb it. With a reflection over the y-axis, the order does not matter for a purely vertical translation, for the mirror-image reason. In every other case the slide has a component pointing straight at the mirror, the reflection reverses that component, and doing the slide before or after the flip gives different results.
127. Reflected half: $(0, 0)$, $(-2, 1)$, $(-5, 2)$, $(-3, 5)$, $(0, 4)$. After sliding that half up $3$: $(0, 3)$, $(-2, 4)$, $(-5, 5)$, $(-3, 8)$, $(0, 7)$.
128. Answers vary. A complete response states a triangle with integer coordinates, a translation, and a reflection over the x- or y-axis; lists the intermediate coordinates and the final coordinates, each obtained by applying the correct rule; and settles the order question correctly — the order changes the answer unless the translation runs parallel to the mirror line. Sample: triangle $(1, 1)$, $(4, 1)$, $(4, 3)$; translate up $2$, then reflect over the x-axis. Intermediate $(1, 3)$, $(4, 3)$, $(4, 5)$; final $(1, -3)$, $(4, -3)$, $(4, -5)$. Reversing the order would change the answer, because the translation is perpendicular to the x-axis mirror; the other order gives $(1, 1)$, $(4, 1)$, $(4, -1)$.