Appendix A — Answer Key, Chapter 13: Translations in the Coordinate Plane

SOL 8.MG.3 (a, d) · Covers textbook Chapter 13 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 88 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Drawing convention used throughout: the preimage is dashed and the image is solid, and image vertices carry prime marks.

Two habits are worth grading for, because they catch nearly every error in this chapter. The same trip check: $x' - x$ must be the same number at every vertex, and so must $y' - y$. The congruence check: a horizontal or vertical side must have the same length in the image as in the preimage.

Sign conventions, restated: right is $x +$, left is $x -$, up is $y +$, down is $y -$.

Lesson 13.1 — What a Translation Does

Guided practice

1. $\triangle ABC$ is the preimage; $\triangle A'B'C'$ is the image. The prime marks identify the image.
2. $(x, y) \to (x + 6, y)$
3. $(x, y) \to (x, y - 5)$
4. $3$ units left and $7$ units up.
5. $2$ units right and $9$ units down.
6. False. A translation slides the figure without resizing it, so the image is congruent to the preimage — same side lengths, same angle measures, same size and shape. Only the position changes.

Independent practice

7. a) $(x, y) \to (x - 8, y)$ b) $(x, y) \to (x, y + 4)$ c) $(x, y) \to (x + 7, y - 2)$ d) $(x, y) \to (x - 3, y + 6)$
8. a) $9$ units right b) $1$ unit down c) $4$ units left and $4$ units down d) $2$ units left and $5$ units up
9. a) horizontal b) vertical c) combination d) combination
10. $P'Q' = 6$ units and $m\angle R' = 115^\circ$. A translation preserves side lengths and angle measures, so no arithmetic is needed: the numbers are copied from the preimage.
11. $(x, y) \to (x + 5, y + 3)$. Vertex $C(3, 4)$ traveled $5$ units right and $3$ units up, landing at $C'(8, 7)$ — the same trip every other vertex made.
12. If every vertex makes the same trip, the distances between vertices are unchanged, so the image is congruent to the preimage and the motion is a slide. If one vertex moved farther than the others, the sides touching that vertex would change length, so the figure would be stretched or distorted rather than slid, and the motion would not be a translation at all.
13. Rule: $(x, y) \to (x + 4, y + 1)$. New location: $(7, 3)$.
14. Left is the negative $x$-direction, so it decreases the $x$-coordinate; the student used the sign for "right." Correct rule: $(x, y) \to (x - 3, y)$.

Exit ticket 13.1

15. $(x, y) \to (x + 5, y - 8)$
16. $6$ units left and $2$ units down.
17. Preserved (any two): side lengths, angle measures, size, shape, orientation. Changed: position.
18. The preimage is the figure you start with and the image is the figure it becomes after the transformation. The prime mark is a label that keeps the two apart, so $A$ and $A'$ can be discussed in the same sentence: $A'$ means "the point that $A$ became."

Lesson 13.2 — Coordinates of a Translated Polygon

Guided practice

19. $(2 + 3,\; 5) = (5, 5)$
20. $(-1,\; 3 - 4) = (-1, -1)$
21. $(-4 + 6,\; 1 - 2) = (2, -1)$
22. $J'(1, -2)$, $K'(4, -2)$, $L'(2, 1)$. Every $x$ rose by $5$ and every $y$ fell by $4$.
23. $W'(-5, 3)$, $X'(-2, 3)$, $Y'(-2, 6)$, $Z'(-5, 5)$
24. Rule: $(x, y) \to (x - 5, y - 5)$. $A'(-5, -5)$, $B'(-2, -5)$, $C'(-5, -1)$. The vertex at the origin moved like every other vertex.
25. $x' - x = 1 - 6 = -5$ and $y' - y = 0$, so $(x, y) \to (x - 5, y)$: $5$ units left.
26. $x' - x = 0$ and $y' - y = -1 - 4 = -5$, so $(x, y) \to (x, y - 5)$: $5$ units down.

Independent practice

27. a) $(1, 7)$ b) $(-7, 4)$ c) $(-4, 0)$ d) $(0, 0)$
28. $D'(-5, 5)$, $E'(-1, 5)$, $F'(-2, 2)$
29. Rule: $(x, y) \to (x + 8, y)$. $Q'(2, -1)$, $R'(6, -1)$, $S'(6, 2)$, $T'(3, 2)$. Every $y$-coordinate is unchanged, as it must be for a horizontal translation.
30. $A'(1, -5)$, $B'(3, -5)$, $C'(4, -3)$, $D'(2, -1)$, $E'(0, -3)$
31. Using $A$ and $A'$: $-1 - 2 = -3$ and $-2 - 3 = -5$, so $(x, y) \to (x - 3, y - 5)$, a translation $3$ units left and $5$ units down. Confirmed at $C$: $1 - 4 = -3$ and $2 - 7 = -5$.
32. The rule is $(x, y) \to (x, y + 5)$, since $4 - 4 = 0$ and $-2 - (-7) = 5$. So $N(-1, 3) \to N'(-1, 8)$. One pair was enough because a translation moves every point by the same amounts, so the trip measured at $M$ is the trip taken by every other point.
33. The translation moved nothing: $(x, y) \to (x, y)$, a translation of $0$ units. A translation adds the same numbers to every point's coordinates, so if the additions are $0$ for one point they are $0$ for all of them, and no point moves. A dilation multiplies instead of adding, and multiplying $0$ by any scale factor still gives $0$, which is why the center of a dilation can stay put while every other point moves.
34. A horizontal translation has rule $(x, y) \to (x + a, y)$ for some number $a$. The second entry is $y$ itself: nothing is added to it, so the output $y$-coordinate equals the input $y$-coordinate for every vertex.
35. East is right and south is down, so the rule is $(x, y) \to (x + 6, y - 4)$. Swing: $(3, -2)$. Slide: $(7, 1)$. Bench: $(1, -5)$.
36. Two sign errors. The student subtracted $2$ from $-5$ instead of adding, getting $-7$ rather than $-3$; and added $3$ to $4$ instead of subtracting, getting $7$ rather than $1$. Correct image: $(-3, 1)$.

Exit ticket 13.2

37. $(4 - 9,\; -6 + 1) = (-5, -5)$
38. $A'(3, -1)$, $B'(6, -1)$, $C'(3, 2)$
39. $x' - x = 2 - 7 = -5$ and $y' - y = -4 - 2 = -6$, so $(x, y) \to (x - 5, y - 6)$.
40. For each vertex compute $x' - x$ and $y' - y$. All the $x$-changes must be equal to one another, and all the $y$-changes must be equal to one another, because every point of a translated figure makes the same trip. A vertex whose changes differ from the rest is the one that is wrong.

Lesson 13.3 — Sketching a Translated Image

Sketches are graded on three things: the image has the same number of vertices as the preimage, the image vertices sit at the listed coordinates, and the vertices are connected in the same order as the preimage so the orientation is preserved.

Guided practice

41. $A'(-5, 2)$, $B'(-2, 2)$, $C'(-4, 5)$. Check: $AB = 3$ and $A'B' = 3$.
42. $D'(2, -4)$, $E'(5, -4)$, $F'(5, -1)$, $G'(2, -2)$
43. Rule: $(x, y) \to (x, y + 6)$. $H'(-3, 1)$, $J'(2, 1)$, $K'(1, 4)$, $L'(-2, 4)$. Every $x$-coordinate is unchanged, so the image sits directly above the preimage.
44. Connect the image vertices, in the same order in which the preimage vertices are connected. The order matters because a different order joins different pairs of points, producing a polygon with different side lengths — a figure that is not congruent to the preimage and so is not its image.
45. Rule: $(x, y) \to (x - 5, y)$. $P'(-5, 0)$, $Q'(-1, 0)$, $R'(-5, 3)$.
46. Rule: $(x, y) \to (x, y - 4)$. Image corners: $(1, -3)$, $(3, -3)$, $(3, -1)$, $(1, -1)$.
47. $(x, y) \to (x + 6, y - 4)$. Any two corresponding pairs confirm it; for example $R(-5, 3) \to R'(1, -1)$ gives right $6$ and down $4$, and $T(-4, 6) \to T'(2, 2)$ gives the same.
48. Check 1, count the vertices — catches a vertex left out or a point plotted twice. Check 2, compare a horizontal or vertical side in both figures — catches one vertex moved by the wrong amount, which shows up as a stretched or squashed image. Check 3, compare the segments joining corresponding vertices — catches a single misplaced vertex, whose segment will differ in length or direction from the others.

Independent practice

49. a) $(x, y) \to (x + 5, y)$: $A'(1, 1)$, $B'(4, 1)$, $C'(2, 4)$. b) $(x, y) \to (x, y - 3)$: $A'(-4, -2)$, $B'(-1, -2)$, $C'(-3, 1)$. c) $(x, y) \to (x + 5, y - 3)$: $A'(1, -2)$, $B'(4, -2)$, $C'(2, 1)$. Note that the image in part c) is the image of part a) dropped $3$, and the image of part b) pushed right $5$: the horizontal and vertical parts of a rule act independently.
50. $A'(-6, 4)$, $B'(-2, 4)$, $C'(-3, 1)$, $D'(-5, 1)$
51. $P'(-5, -5)$, $Q'(-2, -5)$, $R'(-1, -3)$, $S'(-3, -1)$, $T'(-5, -3)$
52. $A'(6, -5)$, $B'(6, -2)$, $C'(2, -5)$
53. Rule: $(x, y) \to (x - 2, y + 5)$. Image: $(-3, 4)$, $(0, 2)$, $(2, 5)$, $(-1, 7)$.
54. The picture shows congruence because corresponding sides are the same length. For example, in item 52 the vertical side $AB$ runs from $y = 2$ to $y = 5$, a length of $3$, and $A'B'$ runs from $y = -5$ to $y = -2$, also $3$. Comparing any one horizontal or vertical side is enough to catch a misplaced vertex; comparing all of them confirms the whole figure.
55. Step 2 — connecting the image vertices in the same order as the preimage. Done carelessly, it joins the wrong pairs of points, which changes the side lengths and can reverse the way the vertices run around the figure. The plotted points would be right and the drawn polygon still wrong.
56. Rule: $(x, y) \to (x + 5, y)$. New corners: $(5, 0)$, $(9, 0)$, $(9, 4)$, $(5, 4)$. The block is a $4$ by $4$ square in both positions.
57. Rule: $(x, y) \to (x - 3, y - 2)$. New corners: $(-1, -1)$, $(3, -1)$, $(3, 1)$, $(-1, 1)$. The riser is still $4$ units by $2$ units, as it must be — moving a riser does not resize it.
58. A translation moves every point of the figure by the same amount. Moving one vertex and leaving two behind changes the lengths of the two sides that meet at that vertex, so the drawn triangle is not congruent to the preimage and is not the image of anything. Correct procedure: apply $(x, y) \to (x + 3, y - 2)$ to all three vertices, plot $A'$, $B'$, and $C'$, then connect them in the order $A'B'$, $B'C'$, $C'A'$.

Exit ticket 13.3

59. $A'(-6, -2)$, $B'(-3, -2)$, $C'(-3, 2)$
60. Rule: $(x, y) \to (x + 7, y - 6)$. Image corners: $(2, -5)$, $(5, -5)$, $(5, -2)$, $(2, -2)$.
61. $x' - x = 4 - (-3) = 7$ and $y' - y = 1 - 5 = -4$, so $(x, y) \to (x + 7, y - 4)$.
62. Acceptable response: first apply the rule to every vertex and plot the image points, labeling each with a prime mark. Then connect those points in the same order as the preimage vertices are connected, and check that a horizontal or vertical side came out the same length in both figures.

Chapter 13 Review

Part A — Identifying the coordinates of a translated image (8.MG.3a)

63. a) $(7, -3)$ b) $(2, -1)$ c) $(11, -5)$
64. $A'(3, -5)$, $B'(6, -5)$, $C'(4, -2)$
65. $K'(-5, -2)$, $L'(-1, -1)$, $M'(-2, 2)$, $N'(-6, 1)$
66. Rule: $(x, y) \to (x - 10, y)$, so the image is $(-7, -4)$.
67. $x' - x = 2 - (-4) = 6$ and $y' - y = 1 - (-6) = 7$, so $(x, y) \to (x + 6, y + 7)$: $6$ units right and $7$ units up.
68. The translation of $0$ units, $(x, y) \to (x, y)$. Since $x' - x = 0$ and $y' - y = 0$ and every point makes the same trip, nothing moved anywhere. This is the only translation that leaves any point fixed, and it leaves all of them fixed.
69. Reverse each operation: $(3 + 4,\; -5 - 2) = (7, -7)$. Forward check: $(7 - 4,\; -7 + 2) = (3, -5)$, which is $A'$.
70. $V'(2, 2)$, $W'(5, 2)$, $X'(6, 4)$, $Y'(4, 6)$, $Z'(1, 4)$
71. The image vertex corresponding to $C$ is $C'$. $B'C' = 9$ units, because a translation preserves side lengths — the image is congruent to the preimage.
72. The rule is $(x, y) \to (x + a, y + b)$ with $a$ and $b$ integers. Each image coordinate is an integer plus an integer, and a sum of two integers is an integer, so every image coordinate is an integer and every image vertex lands on a grid intersection.

Part B — Sketching a translated image (8.MG.3d)

73. $A'(-6, 1)$, $B'(-3, 1)$, $C'(-3, 5)$
74. $A'(-6, -4)$, $B'(-2, -4)$, $C'(-2, -1)$, $D'(-5, -1)$
75. $(2, 5)$, $(6, 5)$, $(5, 2)$, $(3, 2)$
76. Rule: $(x, y) \to (x, y + 8)$. Image: $D'(1, 2)$, $E'(5, 2)$, $F'(3, 5)$.
77. $(-7, -2)$, $(-7, 2)$, $(-3, 2)$, $(-3, -2)$
78. $(1, -4)$, $(4, -1)$, $(7, -4)$, $(4, -7)$
79. $A'(-6, 5)$, $B'(-3, 5)$, $C'(-3, 2)$
80. $J'(2, -5)$, $K'(5, -5)$, $L'(3, -1)$. $JK = 3$ and $J'K' = 3$. The two are equal, which is the congruence check passing: the translation preserved that side length, as it preserves every length, so the image is congruent to the preimage.
81. Rule: $(x, y) \to (x + 2, y + 3)$. New corners: $(3, 4)$, $(7, 4)$, $(5, 7)$. The design is the same triangle, three feet higher and two feet to the right.
82. The image sides must join the same pairs of vertices as the preimage sides. Connecting $A'$ to $C'$ and $C'$ to $B'$ draws segments that correspond to no side of the preimage, so the drawn quadrilateral has different side lengths and is not congruent to the preimage. Rule for connecting: join the image vertices in the same order as the preimage vertices — $A'B'$, $B'C'$, $C'D'$, $D'A'$.

Part C — Mixed practice and reasoning

83. Rule: $(x, y) \to (x + 4, y + 5)$. Image: $A'(-1, 3)$, $B'(3, 3)$, $C'(3, 8)$. Check: $AB$ runs from $x = -5$ to $x = -1$, a length of $4$, and $A'B'$ runs from $x = -1$ to $x = 3$, also $4$.
84. $P(-4, 1) \to P'(1, -4)$ gives $x' - x = 5$ and $y' - y = -5$, so the rule is $(x, y) \to (x + 5, y - 5)$: $5$ units right and $5$ units down. Confirming pair: $R(-2, 4) \to R'(3, -1)$, again right $5$ and down $5$. (The pair $Q(-1, 1) \to Q'(4, -4)$ works equally well.)
85. From $(6, 2)$ to $(1, 7)$ is $x' - x = -5$ and $y' - y = 5$, so the rule is $(x, y) \to (x - 5, y + 5)$. The other corners become $(9, 2) \to (4, 7)$, $(9, 4) \to (4, 9)$, and $(6, 4) \to (1, 9)$.
86. Left decreases $x$, so the student should have computed $-3 - 4 = -7$ rather than $-3 + 4 = 1$. Down decreases $y$, so the student should have computed $5 - 2 = 3$ rather than $5 + 2 = 7$. Correct image: $(-7, 3)$.
87. Every point of the figure is moved by the same rule, adding the same number to each $x$-coordinate and the same number to each $y$-coordinate. So each connecting segment has the same horizontal run and the same vertical rise, which makes them all the same length and all the same direction — in fact they are parallel. If that were not true, different vertices would have moved by different amounts, and the resulting figure would be distorted rather than congruent, so the motion would not be a translation.
88. Answers vary. A correct response names a triangle with integer vertices, a rule with a nonzero horizontal part and a nonzero vertical part, image coordinates obtained by applying the rule to each vertex, a sketch with the preimage dashed and the image solid, and a check. Sample: $\triangle ABC$ with $A(1, 1)$, $B(5, 1)$, $C(1, 4)$ under $(x, y) \to (x - 6, y + 3)$ gives $A'(-5, 4)$, $B'(-1, 4)$, $C'(-5, 7)$. Check: $AB = 4$ and $A'B' = 4$; $AC = 3$ and $A'C' = 3$; the same trip check gives $-6$ and $+3$ at all three vertices.

Coverage note

Every item in this chapter serves 8.MG.3a, 8.MG.3d, or the vocabulary and sign work that both depend on. Reflections over the $x$- or $y$-axis, combinations of a translation with a reflection, and describing transformations in context — bullets (b), (c), (e), (f), and (g) — are answered in the Chapter 14 key.