Appendix A — Answer Key, Chapter 15: Dilations in the Coordinate Plane

SOL 7.MG.4 · Covers textbook Chapter 15 and the companion workbook. Item numbers match the textbook; workbook items are the same problems with the same numbers, so this key serves both. Every dilation in this chapter is centered at the origin, and every scale factor is $\tfrac14$, $\tfrac12$, $2$, $3$, or $4$. Sketching items are answered by listing the image vertices precisely, since a key cannot show a drawing. Reasoning answers show an acceptable response, not the only wording.

Lesson 15.1 — What a Dilation Does

Guided practice

1. $A'(2, 6)$, $B'(6, 2)$, $C'(4, 8)$
2. $P'(1, 1)$, $Q'(3, 1)$, $R'(3, 2)$, $S'(1, 2)$
3. a) enlargement b) reduction c) enlargement d) reduction
4. $40°$, $60°$, $80°$ — the same three measures. A dilation multiplies lengths but leaves angle measures unchanged, which is why the image is similar to the preimage.
5. $(kx, ky)$

Independent practice

6. a) $(6, 15)$ b) $(-3, 12)$ c) $(0, -9)$
7. a) $(2, 3)$ b) $(-1, 4)$ c) $(5, 0)$
8. $D'(-4, 6)$, $E'(8, 2)$, $F'(0, -10)$
9. $W'(2, 2)$, $X'(4, 2)$, $Y'(4, 6)$, $Z'(2, 4)$
10. $k = 3$. Divide each image coordinate by its partner: $15 \div 5 = 3$ and $6 \div 2 = 3$. Both give the same factor, so a single dilation did the work.
11. Image corners $(0, 0)$, $(12, 0)$, $(12, 8)$, $(0, 8)$. The decal is $12$ cm wide and $8$ cm tall.
12. Substituting the origin into the rule gives $(k \cdot 0, \; k \cdot 0) = (0, 0)$ for every value of $k$, so the origin maps to itself. For any other point, at least one coordinate is not zero, and $kx = x$ is false whenever $x \neq 0$ and $k \neq 1$ — so that point must move.

Exit ticket 15.1

13. $(8, -12)$
14. $J'(3, 1)$, $K'(-2, 4)$, $L'(0, -3)$
15. A reduction, because $\tfrac14$ is between $0$ and $1$. Every point's distance from the origin is multiplied by $\tfrac14$, so each point ends up one fourth as far out along the same ray.
16. Because a dilation preserves every angle measure and multiplies every side length by the same scale factor $k$. Congruent corresponding angles plus proportional corresponding sides is exactly the definition of similar, so $\triangle ABC \sim \triangle A'B'C'$. What stays the same: angle measures and shape. What changes: side lengths and distance from the origin, both multiplied by $k$.

Lesson 15.2 — Coordinates of a Dilated Polygon

Guided practice

17. $A(1, 4) \rightarrow (2 \cdot 1, \; 2 \cdot 4) = A'(2, 8)$; $B(5, 2) \rightarrow (10, 4)$; $C(3, -3) \rightarrow (6, -6)$
18. $P'(2, 3)$, $Q'(5, 1)$, $R'(-4, -2)$
19. $D'(-6, 3)$, $E'(0, 12)$, $F'(9, -6)$
20. $\dfrac{12}{3} = 4$ and $\dfrac{16}{4} = 4$, so $k = 4$.
21. $A(5, -3)$

Independent practice

22. a) $(8, 12)$ b) $(-4, 20)$ c) $(0, -8)$
23. $A'(1, 2)$, $B'(4, -1)$, $C'(-3, 0)$, $D'(5, 2)$
24. $E'(-2, -1)$, $F'(1, -1)$, $G'(1, 3)$, $H'(-2, 2)$
25. $R'(6, 3)$, $S'(15, 9)$, $T'(3, 12)$
26. $k = \tfrac12$, since $3 \div 6 = \tfrac12$ and $-2 \div (-4) = \tfrac12$. It is a reduction.
27. $(0, 0)$, $(12, 0)$, $(12, 12)$, $(0, 12)$. The side length went from $4$ units to $12$ units, three times as long, and the corner at the origin did not move.
28. The arithmetic is correct — $(3.5, 1.5)$ really is the image of $(7, 3)$ under $k = \tfrac12$. It is a perfectly valid point; it simply sits halfway between grid lines instead of on an intersection, which makes it awkward to plot and read. For $k = \tfrac12$ to produce whole-number image coordinates, both preimage coordinates must be even. (For $k = \tfrac14$, they must be multiples of $4$.)

Exit ticket 15.2

29. $(-10, 6)$
30. $A'(2, 3)$, $B'(-4, 1)$, $C'(0, -2)$
31. $k = 3$, since $9 \div 3 = 3$ and $-6 \div (-2) = 3$.
32. Both coordinates must be multiplied by the same factor because the image point has to stay on the ray from the origin through the preimage point. Keeping the same direction means keeping the same ratio of rise to run, which only happens when both coordinates are scaled equally. If only the $x$-coordinate were multiplied, the figure would be stretched horizontally: the angle measures would change, the image would not be similar to the preimage, and the result would not be a dilation at all.

Lesson 15.3 — Sketching a Dilation

Guided practice

33. $A'(2, 2)$, $B'(8, 2)$, $C'(2, 6)$. Grid needs to reach $8$ horizontally and $6$ vertically.
34. $P'(-3, 2)$, $Q'(1, 2)$, $R'(1, -1)$, $S'(-3, -1)$
35. $D'(6, 3)$, $E'(9, 9)$, $F'(3, 6)$
36. $W'(-2, 1)$, $X'(1, 1)$, $Y'(1, -2)$, $Z'(-2, -2)$
37. Yes. $A(1, 1)$ and $A'(2, 2)$ both have a rise-to-run ratio of $1$ to $1$ measured from the origin, so both lie on the same ray. That confirms the image vertex was plotted in the right direction from the center, at twice the distance.

Independent practice

38. $J'(-4, 2)$, $K'(-2, 8)$, $L'(6, 4)$
39. $M'(-2, 3)$, $N'(2, 3)$, $P'(3, -1)$, $Q'(-3, -2)$
40. $A'(-3, -3)$, $B'(6, -6)$, $C'(0, 9)$
41. $R'(4, 8)$, $S'(8, -4)$, $T'(-4, 4)$
42. $E'(1, 2)$, $F'(3, 1)$, $G'(2, -2)$, $H'(-1, -1)$
43. Image vertices $(0, 0)$, $(9, 0)$, $(0, 12)$. New side lengths $3 \times 3 = 9$ cm, $4 \times 3 = 12$ cm, and $5 \times 3 = 15$ cm.
44. The extreme coordinates $-5$ and $5$ become $-20$ and $20$, so the grid must run from at least $-20$ to $20$ in both directions. Checking first saves time because a grid drawn only to $\pm 10$ cannot hold the image, and the whole sketch would have to be redrawn on a bigger grid.

Exit ticket 15.3

45. $A'(4, 6)$, $B'(8, 2)$, $C'(2, -4)$
46. $P'(-4, 1)$, $Q'(-1, 3)$, $R'(2, -2)$
47. $(3, 3)$, $(6, 3)$, $(6, 6)$, $(3, 6)$
48. The definition of a dilation says the image vertex lies on the ray from the center through the preimage vertex. So if you draw that ray and your plotted image point is not on it, the point is misplotted. The ray also shows the direction is right; comparing distances then confirms the scale factor.

Lesson 15.4 — Dilations in the World

Guided practice

49. $10$ in by $14$ in. An enlargement, since $2 > 1$.
50. $3$ px by $4$ px. A reduction, since $\tfrac14 < 1$.
51. a) $\dfrac{12}{6} = 2$ and $\dfrac{18}{9} = 2$. The ratios agree, so it is a dilation with scale factor $2$. b) $\dfrac{12}{6} = 2$ and $\dfrac{15}{9} = \tfrac53 \approx 1.7$. The ratios disagree, so it is not a dilation.
52. Image vertices $(0, 0)$, $(12, 0)$, $(0, 9)$. New side lengths $12$, $9$, and $15$.
53. Original perimeter $2(8 + 5) = 26$ units. The image is $16$ by $10$, so the new perimeter is $2(16 + 10) = 52$ units — exactly twice the original, because perimeter is multiplied by $k$.

Independent practice

54. $12$ in by $20$ in
55. $5$ cm by $6$ cm
56. a) $\dfrac{5}{10} = \tfrac12$ and $\dfrac{7}{14} = \tfrac12$. A dilation, scale factor $\tfrac12$. b) $\dfrac{20}{10} = 2$ and $\dfrac{24}{14} = \tfrac{12}{7} \approx 1.7$. Not a dilation. c) $\dfrac{27}{9} = 3$ and $\dfrac{36}{12} = 3$. A dilation, scale factor $3$.
57. $(0, 0)$, $(3, 0)$, $(3, 2)$, $(0, 2)$. The logo went from $6$ units wide and $4$ units tall to $3$ units wide and $2$ units tall — half as long in every direction, with the same angles and the corner at the origin unmoved. It is a reduction.
58. $(0, 0)$, $(16, 0)$, $(16, 24)$, $(0, 24)$
59. New width $4 \times 3 = 12$ cm; new height $6 \times 3 = 18$ cm. Original perimeter $2(4 + 6) = 20$ cm; new perimeter $2(12 + 18) = 60$ cm. Original area $4 \times 6 = 24$ square centimeters; new area $12 \times 18 = 216$ square centimeters. The area was multiplied by $9$, which is $3^2$.
60. The ratios are $\dfrac{16}{8} = 2$ for the width and $\dfrac{10}{10} = 1$ for the height. They disagree, so it is not a dilation. Only one dimension was scaled, so the image is stretched horizontally: circles become ovals, squares become rectangles, and every angle that was not a right angle changes measure. The result is not similar to the original.

Exit ticket 15.4

61. $12$ in by $16$ in
62. $4$ px by $6$ px
63. $\dfrac{15}{5} = 3$ and $\dfrac{20}{8} = 2.5$. The ratios disagree, so it is not a dilation — the image was stretched more in one direction than the other.
64. Sample answer: a photo taken on a phone and printed as an $8$ in by $10$ in poster. The preimage is the $4$ in by $5$ in print and the image is the $8$ in by $10$ in poster, and because every length doubled it is an enlargement with scale factor $2$. (Other acceptable answers: a map or floor plan, which is a reduction of the real place; a model car, a reduction; a projected slide, an enlargement.)

Chapter 15 Review

Part A — Coordinates of a dilated image (7.MG.4a)

65. $(6, -10)$
66. $(-12, 6)$
67. $(5, -3)$
68. $(3, -2)$
69. $A'(4, -8)$, $B'(16, 0)$, $C'(-12, 20)$
70. $W'(2, 1)$, $X'(4, 2)$, $Y'(1, -3)$, $Z'(-2, 0)$
71. $k = 4$, since $8 \div 2 = 4$ and $28 \div 7 = 4$.

Part B — Sketching a dilation (7.MG.4b)

72. $A'(2, 4)$, $B'(6, -2)$, $C'(-4, 2)$
73. $P'(-3, 4)$, $Q'(2, 3)$, $R'(1, -2)$
74. $D'(6, 0)$, $E'(0, 6)$, $F'(-3, -3)$
75. $G'(-2, 3)$, $H'(2, 2)$, $J'(1, -1)$, $K'(-3, -2)$
76. $(4, 4)$, $(8, 4)$, $(8, 12)$, $(4, 12)$
77. Yes, it passes through $B'$. From the origin, $B(3, -1)$ has a rise-to-run ratio of $-1$ to $3$, and $B'(6, -2)$ has $-2$ to $6$, which is the same ratio. That tells you $B'$ was plotted in the correct direction from the center of dilation, at twice the distance, so the sketch is right.

Part C — Dilations in context (7.MG.4c)

78. $12$ in by $15$ in. An enlargement.
79. $6$ mm by $8$ mm. A reduction.
80. $\dfrac{14}{7} = 2$ and $\dfrac{16}{9} \approx 1.8$. The ratios disagree, so it is not a dilation.
81. $(0, 0)$, $(3, 0)$, $(3, 1)$, $(0, 1)$. The logo went from $6$ units by $2$ units to $3$ units by $1$ unit — every length halved, angles unchanged, and the corner at the origin fixed. It is a reduction.
82. A scale drawing is similar to the real object it represents, and a dilation is the transformation that produces a similar figure. If you place the real object's outline on a coordinate grid with the origin as the center, the drawing is exactly the image of a dilation. The scale of the drawing — the number relating drawing lengths to actual lengths — plays the role of the scale factor $k$. A drawing smaller than the object corresponds to $k$ between $0$ and $1$, a reduction.

Part D — Mixed application and reasoning

83. New dimensions $36$ in by $48$ in. Original perimeter $2(9 + 12) = 42$ in; new perimeter $2(36 + 48) = 168$ in, which is $42 \times 4$.
84. $(0, 0)$, $(3, 0)$, $(3, 2)$, $(0, 2)$. A reduction, since $\tfrac14 < 1$.
85. Draw the ray from the origin through a preimage vertex. Moving from the origin to $(x, y)$ means going a run of $x$ and a rise of $y$. Moving from the origin to $(kx, ky)$ means going a run of $kx$ and a rise of $ky$, which is the same direction — the ratio of rise to run is unchanged — but a trip $k$ times as long. So the image point stays on the ray and its distance from the center is multiplied by $k$, which is exactly what the definition of a dilation requires.
86. Substituting into the rule gives $(k \cdot 0, \; k \cdot 0) = (0, 0)$, so the origin maps to itself no matter what $k$ is. Every other point has at least one nonzero coordinate, and multiplying a nonzero number by a factor other than $1$ always changes it, so every other point moves.
87. No error. $\tfrac14 \times 5 = 1.25$ and $\tfrac14 \times 7 = 1.75$, so $(1.25, 1.75)$ is the correct image, and it is a genuine point of the plane. It just does not fall on a grid intersection, which makes it hard to plot precisely. For the image to land on grid intersections under $k = \tfrac14$, both preimage coordinates would have to be multiples of $4$ — for example, $(4, 8)$ instead of $(5, 7)$.
88. Angle measures: unchanged — all four are still right angles, because dilations preserve angle measure. Side lengths: $4 \times 2 = 8$ and $6 \times 2 = 12$, so each side is twice as long. Perimeter: $2(4 + 6) = 20$ becomes $2(8 + 12) = 40$, multiplied by $k = 2$. Area: $4 \times 6 = 24$ becomes $8 \times 12 = 96$, multiplied by $k^2 = 4$. The image is similar to the preimage with scale factor $2$.

Workbook-only items

Page 2, fill in the blanks. The figure you start with is the preimage. The figure you end with is the image. The fixed point everything is measured from is the center of dilation; in this chapter it is always the origin. A dilation centered at the origin sends $(x, y)$ to $(kx, ky)$. A scale factor greater than 1 produces an enlargement; a scale factor between 0 and 1 produces a reduction. A dilation changes side lengths but leaves angle measures alone, so the image is similar to the preimage.

Page 6, complete the sentence. The image vertex lies on the ray from the origin through the preimage vertex, and its distance from the origin is multiplied by the scale factor.

Page 6, rise-over-run table.

Page 6, explain. Multiplying only $x$ would change the ratio of rise to run, so the image point would fall off the ray from the origin. The figure would be stretched horizontally instead of resized, its angle measures would change, and the result would not be similar to the preimage — so it would not be a dilation.

Page 15, item 44 blanks. The grid must run from $-20$ to $20$ in both directions.

Page 24, item 88 table. Angle measures: four right angles, unchanged. Side lengths: $4$ and $6$ become $8$ and $12$. Perimeter: $20$ becomes $40$. Area: $24$ becomes $96$.