Appendix A — Answer Key, Chapter 10: The Coordinate Plane

SOL 6.MG.3 · Covers textbook Chapter 10 and the companion workbook. Item numbers match the textbook; workbook items that repeat textbook problems share the same answers, and workbook-only items are keyed at the end. Reasoning answers show an acceptable response, not the only wording.

Lesson 10.1 — Parts of the Coordinate Plane

Guided practice

1. The horizontal axis is the $x$-axis; the vertical axis is the $y$-axis.
2. $(0, 0)$
3. Quadrant I
4. $-2$
5. The pair is ordered: the first number is horizontal and the second is vertical. For $(3, 4)$ you move 3 right and 4 up; for $(4, 3)$ you move 4 right and 3 up. Those are different locations.

Independent practice

6. Quadrant I $(+, +)$; Quadrant II $(-, +)$; Quadrant III $(-, -)$; Quadrant IV $(+, -)$.
7. It is $0$. No vertical movement means the point never leaves the $x$-axis.
8. a) Quadrant I b) Quadrant II c) Quadrant III d) Quadrant IV
9. Any pair of the form $(0, n)$ with $n \neq 0$, such as $(0, 4)$.
10. Quadrants are the four open regions between the axes. The origin sits on both axes, which are the boundaries, so it belongs to no region.
11. $(-5, 3)$; Quadrant II. West is the negative $x$-direction and north is the positive $y$-direction.
12. $(-1, 5)$ is in Quadrant II and $(4, -3)$ is in Quadrant IV, yet each has a negative coordinate. Quadrant III requires both coordinates to be negative.

Exit ticket 10.1

1. $(0, 0)$
2. Move 9 units to the left of the origin.
3. Quadrant IV
4. Its $y$-coordinate is $0$, so there is no vertical movement and the point stays on the horizontal axis.

Lesson 10.2 — Graphing and Naming Ordered Pairs

Guided practice

1. The $x$-coordinate is used first, and it moves you horizontally — right if positive, left if negative.
2. Right 6, then up 1.
3. Left 3, then down 5.
4. $C = (-2, -3)$
5. $(-7, 0)$

Independent practice

6. Plotted points, with locations for checking: $(2, 6)$ Quadrant I; $(-4, 5)$ Quadrant II; $(-6, -2)$ Quadrant III; $(3, -3)$ Quadrant IV; $(0, 5)$ on the $y$-axis; $(-1, 0)$ on the $x$-axis.
7. Start at the origin, move 8 units left, then 7 units up.
8. $(3, -9)$
9. Both points are in Quadrant I. $(5, 2)$ is far across and barely up; $(2, 5)$ is barely across and far up. They are mirror images of each other across the diagonal through the origin.
10. Any three pairs of the form $(0, n)$, such as $(0, 1)$, $(0, -2)$, and $(0, 6)$.
11. Tomatoes $(-4, -2)$, Quadrant III. Peppers $(3, 6)$, Quadrant I.
12. She used the $7$ as a horizontal move, but $7$ is the $y$-coordinate. The $x$-coordinate is $0$, so there is no horizontal move at all. The point is on the $y$-axis, 7 units above the origin.

Exit ticket 10.2

1. Start at the origin, move 6 units left, then 3 units up.
2. $(-2, -8)$
3. $(5, 0)$
4. Because the ordered pair lists the horizontal coordinate first. Reversing the order sends you to a different point — plotting $(2, 6)$ vertically first would land you on $(6, 2)$.

Lesson 10.3 — Quadrants and the Axes

Guided practice

1. Quadrant III
2. On the $y$-axis, 8 units below the origin.
3. Quadrant II
4. Any pair with both coordinates negative, such as $(-3, -5)$.
5. The $x$-axis.

Independent practice

6. a) Quadrant I b) on the $x$-axis c) Quadrant III d) on the $y$-axis e) Quadrant IV
7. Samples: I $(2, 5)$; II $(-2, 5)$; III $(-2, -5)$; IV $(2, -5)$.
8. Quadrant III. Only the $y$-coordinate changed sign, so the point moved from above the $x$-axis to below it while staying 3 units left.
9. Quadrant II points have a negative $x$-coordinate with a positive $y$-coordinate, and Quadrant IV points have a positive $x$-coordinate with a negative $y$-coordinate. A point on an axis can also have one negative coordinate. Quadrant III requires both coordinates negative.
10. Quadrant I: $(2, 2)$. Quadrant II: $(-5, 1)$, $(-1, 7)$. Quadrant III: $(-6, -6)$. Quadrant IV: $(9, -1)$. On an axis: $(0, -4)$ on the $y$-axis, $(3, 0)$ on the $x$-axis, $(0, 0)$ the origin.
11. The character is 12 units left and 5 units down from the starting point. Quadrant III.
12. Counterexample: $(0, 5)$ contains a $0$ but sits on the $y$-axis, 5 units above the origin. A point on an axis has exactly one coordinate equal to $0$; the origin has both equal to $0$.

Exit ticket 10.3

1. Quadrant II
2. On the $y$-axis, 9 units above the origin.
3. Any pair with a positive $x$-coordinate and a negative $y$-coordinate, such as $(2, -6)$.
4. A $y$-coordinate of $0$ means no vertical movement, so the point stays on the horizontal axis. The coordinate that is zero names the direction you did not travel, not the axis you land on.

Lesson 10.4 — Distance Between Points on a Horizontal or Vertical Line

Guided practice

1. $|8| = 8$ units
2. $|-2| = 2$ units
3. The same vertical line. Their $x$-coordinates are both $4$, so they are the same distance across and differ only up and down.
4. $|9 - 1| = 8$ units
5. $S = (2, -3)$

Independent practice

6. a) 5 from the $y$-axis, 12 from the $x$-axis b) 3 and 7 c) 0 and 9 d) 10 and 4
7. a) $|6 - (-2)| = 8$ b) $|9 - (-3)| = 12$ c) $|-4 - (-11)| = 7$
8. Any pair with $y = -5$, such as $(2, -5)$; then the distance is $|2 - (-8)| = 10$ units.
9. 9 units left of $M$: $(-10, -4)$. 5 units above $M$: $(-1, 1)$.
10. $(3, -8)$ is farther. Distance from the $x$-axis is the absolute value of the $y$-coordinate: $|-8| = 8$ and $|6| = 6$, and $8 > 6$.
11. $|9 - (-6)| = 15$ kilometers, heading east. Both points have $y$-coordinate $2$, so they lie on the same horizontal line and the path between them runs straight east and west.
12. She subtracted $4$ instead of $-4$, dropping the negative sign on the first $x$-coordinate. The correct work is $|6 - (-4)| = |6 + 4| = 10$ units. A quick check: the points are on opposite sides of the $y$-axis, so the two distances add — 4 units plus 6 units gives 10, not 2.

Exit ticket 10.4

1. $|-7| = 7$ units
2. $|4 - (-6)| = 10$ units
3. $(-2, -2)$
4. Their $y$-coordinates are equal. Equal height means the points differ only left and right, which is a horizontal line.

Lesson 10.5 — Polygons in the Coordinate Plane

Guided practice

1. Horizontal, because the $y$-coordinates match. Length $|9 - 2| = 7$ units.
2. Vertical, because the $x$-coordinates match. Length $|1 - (-6)| = 7$ units.
3. Sides are $6$ and $4$, so $P = 2(6) + 2(4) = 20$ units.
4. $A = 6 \times 4 = 24$ square units.
5. $(1, 2)$

Independent practice

6. Horizontal side $|2 - (-6)| = 8$ units; vertical side $|3 - (-4)| = 7$ units.
7. $P = 2(8) + 2(7) = 30$ units; $A = 8 \times 7 = 56$ square units.
8. Vertical side from $(-3, 6)$ to $(-3, -2)$: $|6 - (-2)| = 8$ units. Horizontal side from $(-3, -2)$ to $(4, -2)$: $|4 - (-3)| = 7$ units. The third side joins $(-3, 6)$ and $(4, -2)$, which differ in both coordinates, so it is slanted and its length cannot be found by subtracting coordinates.
9. Fourth vertex $(0, -1)$. Sides are $|5 - (-1)| = 6$ and $|0 - (-8)| = 8$, so $P = 2(6) + 2(8) = 28$ units.
10. Side length $|4 - (-3)| = 7$ units. The other two vertices are $(9, -3)$ and $(9, 4)$, or $(-5, -3)$ and $(-5, 4)$.
11. Horizontal side $|5 - (-7)| = 12$ yards; vertical side $|2 - (-3)| = 5$ yards. $P = 2(12) + 2(5) = 34$ yards. Cost $= 34 \times 4 = \$136$.
12. The two points differ in both coordinates — the $x$-coordinates are $-2$ and $4$, and the $y$-coordinates are $5$ and $1$ — so the segment joining them is slanted, not horizontal. Subtracting $x$-coordinates gives only the horizontal gap, not the actual distance. The method works only when the two points share a $y$-coordinate.

Exit ticket 10.5

1. $|6 - (-5)| = 11$ units
2. Sides $9$ and $5$, so $A = 45$ square units.
3. $(-2, 8)$
4. Both endpoints have the same $x$-coordinate, so the side runs straight up and down.

Chapter 10 Review

Part A — Axes, origin, and quadrants (6.MG.3a)

1. Horizontal: the $x$-axis. Vertical: the $y$-axis. They meet at the origin, $(0, 0)$.
2. Quadrant I $(+, +)$, Quadrant II $(-, +)$, Quadrant III $(-, -)$, Quadrant IV $(+, -)$, numbered counterclockwise from the upper right.
3. The quadrants are the four regions between the axes. The origin lies on both axes, so it is on a boundary rather than inside a region.
4. Student sketch. Check that the $x$-axis is horizontal with values increasing right, the $y$-axis is vertical with values increasing up, the origin is at the crossing, and the Roman numerals run counterclockwise from the upper right.

Part B — Locating a point from its ordered pair (6.MG.3b)

5. a) Quadrant II b) Quadrant IV c) on the $y$-axis d) Quadrant III e) on the $x$-axis
6. Samples: Quadrant II $(-4, 7)$; Quadrant IV $(4, -7)$.
7. It lies on the $x$-axis, to the right of the origin, a number of units equal to its $x$-coordinate.

Part C — Graphing ordered pairs (6.MG.3c)

8. Plotted points, with locations for checking: $(4, 4)$ Quadrant I; $(-2, 6)$ Quadrant II; $(-5, -3)$ Quadrant III; $(1, -7)$ Quadrant IV; $(0, 3)$ on the $y$-axis; $(-6, 0)$ on the $x$-axis.
9. Start at the origin, move 9 units left, then 2 units down.

Part D — Identifying ordered pairs from points (6.MG.3d)

10. $A = (3, 5)$; $C = (-2, -3)$; $F = (-6, 0)$
11. $(-6, 4)$
12. $(10, 0)$

Part E — Distance from the axes and between points (6.MG.3e)

13. a) 8 units from the $y$-axis, 3 units from the $x$-axis b) 6 units from the $y$-axis, 15 units from the $x$-axis
14. a) $|9 - (-4)| = 13$ units b) $|5 - (-6)| = 11$ units
15. $(8, -8)$
16. Their $x$-coordinates are equal, so the points are the same distance across and differ only up and down.

Part F — Polygons in the coordinate plane (6.MG.3f)

17. Horizontal side $|4 - (-5)| = 9$ units; vertical side $|6 - (-2)| = 8$ units; $P = 2(9) + 2(8) = 34$ units; $A = 9 \times 8 = 72$ square units.
18. $(-6, -1)$
19. Vertical side $|9 - 2| = 7$ units; horizontal side $|10 - 1| = 9$ units.

Part G — Mixed application and reasoning

20. Gate to pond: $|0 - (-5)| = 5$ blocks. Pond to playground: $|2 - (-6)| = 8$ blocks. The playground $(2, -5)$ is in Quadrant IV.
21. Horizontal side $|6 - (-4)| = 10$ m; vertical side $|3 - (-4)| = 7$ m. Area $= 10 \times 7 = 70$ square meters. Railing $= 2(10) + 2(7) = 34$ meters.
22. The distance from the $x$-axis is $|y| = |-9| = 9$, so the point is 9 units below the $x$-axis. The distance from the $y$-axis is $|x| = |0| = 0$, so the point sits directly on the $y$-axis.
23. $(9, 3)$ and $(-19, 3)$. The shared $y$-coordinate must stay $3$, and the other $x$-coordinate is 14 units away from $-5$ in either direction: $-5 + 14 = 9$ and $-5 - 14 = -19$.

Workbook-only items

Page 2, fill in the blanks. The horizontal number line is the $x$-axis. The vertical number line is the $y$-axis. The axes cross at the origin, written $(0, 0)$. In $(x, y)$ the first number tells you how far right or left and the second tells you how far up or down.

Page 2, explain. They are ordered pairs, so position in the pair matters. $(2, 9)$ means 2 right and 9 up; $(9, 2)$ means 9 right and 2 up.

Page 3, quadrant table.

Page 3, true or false. False. The quadrants are the open regions between the axes, so a point on an axis is on a boundary and belongs to no quadrant.

Page 5, describe the moves. $(6, 1)$: right 6, up 1. $(-3, -5)$: left 3, down 5. $(-8, 7)$: left 8, up 7. $(0, -3)$: no horizontal move, down 3. $(-5, 0)$: left 5, no vertical move.

Page 5, steps fill-in. Start at the origin. Move horizontally first, then vertically.

Page 6, ordered pairs. A $(3, 5)$; B $(-4, 2)$; C $(-2, -3)$; D $(5, -4)$; E $(0, 4)$; F $(-6, 0)$.

Page 6, quadrant or axis. A Quadrant I; B Quadrant II; C Quadrant III; D Quadrant IV; E on the $y$-axis; F on the $x$-axis.

Page 6, write the ordered pair. $(3, -9)$; $(-7, 0)$.

Page 6, find the error. Priya treated the $7$ as a horizontal move, but it is the $y$-coordinate. The $x$-coordinate is $0$, so the point is on the $y$-axis, 7 units above the origin.

Page 8, first location table. $(8, 15)$ Quadrant I; $(-4, 0)$ on the $x$-axis; $(-9, -2)$ Quadrant III; $(0, 11)$ on the $y$-axis; $(6, -7)$ Quadrant IV.

Page 8, second location table. $(-2, -9)$ Quadrant III; $(0, -8)$ on the $y$-axis; $(-11, 4)$ Quadrant II; $(7, 0)$ on the $x$-axis; $(0, 0)$ the origin.

Page 8, your own pairs. Samples: I $(3, 1)$; II $(-3, 1)$; III $(-3, -1)$; IV $(3, -1)$; on the $x$-axis $(5, 0)$; on the $y$-axis $(0, 5)$.

Page 9, sort. Quadrant I: $(2, 2)$. Quadrant II: $(-5, 1)$, $(-1, 7)$. Quadrant III: $(-6, -6)$. Quadrant IV: $(9, -1)$. On an axis: $(0, -4)$, $(3, 0)$, $(0, 0)$.

Page 9, sign flip. $(-3, -6)$ is in Quadrant III, because the $y$-coordinate turned negative and the point dropped below the $x$-axis. $(3, 6)$ is in Quadrant I. $(3, -6)$ is in Quadrant IV.

Page 9, apply it. The character is 12 units left and 5 units down from the origin. Quadrant III.

Page 9, challenge. Counterexample: $(0, 5)$, which is on the $y$-axis, not at the origin. A point on an axis has one coordinate equal to $0$; only the origin has both equal to $0$.

Page 11, distance table. $(5, 12)$: 5 and 12. $(-3, -7)$: 3 and 7. $(0, 9)$: 0 and 9. $(-10, 4)$: 10 and 4. $(8, -2)$: 8 and 2.

Page 11, circle. $(3, -8)$, because $|-8| = 8$ is greater than $|6| = 6$.

Page 12, distance table. $(1, 5)$ and $(9, 5)$: horizontal, $|9 - 1| = 8$. $(-2, 8)$ and $(6, 8)$: horizontal, $|6 - (-2)| = 8$. $(-5, -3)$ and $(-5, 9)$: vertical, $|9 - (-3)| = 12$. $(7, -4)$ and $(7, -11)$: vertical, $|-4 - (-11)| = 7$. $(3, -6)$ and $(3, 4)$: vertical, $|4 - (-6)| = 10$.

Page 12, move from a point. $(2, -3)$; $(-10, -4)$; $(-1, 1)$; $(5, 6)$.

Page 12, find the error. She dropped the negative sign and subtracted $4$ instead of $-4$. The correct distance is $|6 - (-4)| = 10$ units.

Page 14, rectangle $ABCD$. $\overline{AB} = |4 - (-3)| = 7$; $\overline{BC} = |2 - (-1)| = 3$; perimeter $20$ units; area $21$ square units.

Page 14, second rectangle. Horizontal side $|2 - (-6)| = 8$; vertical side $|3 - (-4)| = 7$; perimeter $30$ units; area $56$ square units.

Page 15, missing vertices. $(1, 2)$; $(-4, -2)$; $(0, -1)$; $(-2, 8)$.

Page 15, right triangle. Vertical side $|6 - (-2)| = 8$; horizontal side $|4 - (-3)| = 7$. The third side joins points that differ in both coordinates, so it is slanted and cannot be measured by subtracting coordinates.

Page 15, apply it. Perimeter $= 2(12) + 2(5) = 34$ yards. Cost $= 34 \times 4 = \$136$.

Page 15, challenge. The points $(-2, 5)$ and $(4, 1)$ differ in both coordinates, so the segment between them is slanted. Subtracting the $x$-coordinates gives only how far apart they are horizontally, which is not the distance between the points.

Page 17, item 4. Student labeling. Check that the $x$-axis is horizontal, the $y$-axis vertical, the origin is at the crossing, and the quadrants run I, II, III, IV counterclockwise from the upper right.

Page 18, item 10. A $(3, 5)$; C $(-2, -3)$; F $(-6, 0)$.