Appendix A — Answer Key, Chapter 11: Quadrilaterals in the Coordinate Plane

SOL G.PC.1 (a, b) · Covers textbook Chapter 11 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 112 across the chapter.

Conventions used in every answer below. Lengths are exact and in simplest radical form — $\sqrt{40}$ is written $2\sqrt{10}$, and two lengths are never compared as decimals. Slopes are exact fractions; a vertical segment has an undefined slope, not a slope of zero. A coordinate proof is finished only when the conclusion names the property, so every proof below ends with that line. Virginia's exclusive trapezoid definition holds throughout: exactly one pair of parallel sides, so every trapezoid proof carries two slope facts.

The single most useful habit in this chapter: ask "is it a parallelogram?" first. Two midpoints answer it, and the answer splits the six families in two.

Lesson 11.1 — The Three Formulas

Guided practice

1. The distance formula. Congruent means equal lengths.
2. The slope formula — the two slopes must be equal.
3. The midpoint formula, run on both diagonals. They bisect each other exactly when the two results are the same point.
4. Both $-\tfrac13$. Equal slopes prove $\overline{AB} \parallel \overline{DC}$.
5. $\tfrac13$ and $-3$, and $\left(\tfrac13\right)(-3) = -1$, which proves the two sides are perpendicular.
6. $M(-1, 0)$. One point proves a statement about two segments because it is the midpoint of both — and "each diagonal cuts the other in half" is exactly the claim that they share a midpoint.

Independent practice

7. $\dfrac{-3 - 1}{0 - (-4)} = \dfrac{-4}{4} = -1$.
8. $\dfrac{7 - 3}{2 - 6} = \dfrac{4}{-4} = -1$.
9. $AB = \sqrt{8^2 + 0^2} = 8$.
10. $CD = \sqrt{(-4)^2 + 0^2} = 4$.
11. $AD = \sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10}$.
12. $\left(\dfrac{0+7}{2}, \dfrac{0+6}{2}\right) = \left(\tfrac72, 3\right)$.
13. $\left(\dfrac{6+1}{2}, \dfrac{2+4}{2}\right) = \left(\tfrac72, 3\right)$.
14. Vertical: undefined. Horizontal: 0.
15. Distance.
16. Slope (show the product of the two slopes at $A$ is $-1$).
17. Midpoint.
18. Slope.
19. Slope. The north edge $\overline{DC}$ runs from $(-2,3)$ to $(2,3)$, slope $\dfrac{3-3}{2-(-2)} = 0$; the south edge $\overline{AB}$ runs from $(-4,-3)$ to $(4,-3)$, slope $0$. Equal slopes, so yes, they are parallel.
20. They actually proved something stronger than they wrote. $\sqrt{40}$ and $\sqrt{40}$ are equal, not approximately equal — the exact values are identical. Writing "$\approx$" claims only that two roundings agree, and two genuinely different lengths can round to the same two decimals. The correct line is $AB = CD = 2\sqrt{10}$.
21. Because the perpendicular test is an exact algebraic condition — the product of the slopes is $-1$ — and that condition characterises $90°$ and nothing else. There is no comparable slope condition for $60°$; expressing a general angle from coordinates needs trigonometry, which G.PC.1b does not include.
22. There is no error. A midpoint is an average, so it is a half-integer whenever the two coordinates being averaged have an odd sum. $\left(\tfrac72, 3\right)$ is the perfectly correct midpoint of $(0,0)$ and $(7,6)$.

Exit ticket 11.1

23. The slope formula; the two slopes must have a product of $-1$.
24. $\dfrac{2-1}{5-2} = \dfrac13$.
25. $AB = \sqrt{3^2 + 1^2} = \sqrt{10}$.
26. $\left(\dfrac{2+5}{2}, \dfrac{1+2}{2}\right) = \left(\tfrac72, \tfrac32\right)$.

Lesson 11.2 — Proving a Parallelogram

Guided practice

27. Both pairs of opposite sides parallel (slope); both pairs of opposite sides congruent (distance); one pair both parallel and congruent (slope + distance); the diagonals bisect each other (midpoint).
28. The diagonal/midpoint route — two computations.
29. Because one pair of opposite sides parallel gives a trapezoid, not a parallelogram. It is the combination with congruence that works; either half alone is not enough.
30. $m_{AB} = m_{DC} = -\tfrac13$ and $m_{AD} = m_{BC} = \tfrac32$. Both pairs of opposite sides are parallel, which is the definition of a parallelogram.
31. Not a rectangle: $\left(-\tfrac13\right)\left(\tfrac32\right) = -\tfrac12 \ne -1$, so no angle is right. Not a rhombus: $AB = 2\sqrt{10}$ but $AD = 2\sqrt{13}$.
32. The diagonals bisect each other. $M$ is already fixed by $A$ and $C$, and $D$ then has to be the one point that makes $M$ the midpoint of $\overline{BD}$ — a midpoint determines the second endpoint uniquely, so there is exactly one such $D$.

Independent practice

33. Midpoint of $\overline{AC}$: $\left(\tfrac{-5+5}{2}, \tfrac{1+2}{2}\right) = \left(0, \tfrac32\right)$. Midpoint of $\overline{BD}$: $\left(\tfrac{-1+1}{2}, \tfrac{-2+5}{2}\right) = \left(0, \tfrac32\right)$. Same point, so the diagonals bisect each other, so yes, a parallelogram.

34. $m_{AB} = \tfrac{2-0}{6-0} = \tfrac13$ and $m_{DC} = \tfrac{6-4}{7-1} = \tfrac13$; $m_{AD} = \tfrac{4-0}{1-0} = 4$ and $m_{BC} = \tfrac{6-2}{7-6} = 4$. Both pairs parallel, so yes.

35. Midpoint of $\overline{AC}$: $\left(-\tfrac12, 0\right)$. Midpoint of $\overline{BD}$: $\left(\tfrac32, 0\right)$. Different, so the diagonals do not bisect each other and it is not a parallelogram.

36. Midpoint of $\overline{AC}$: $(1,1)$. Midpoint of $\overline{BD}$: $(1,1)$. Same, so yes.

37. $\left(0, \tfrac32\right)$.

38. $\left(0, \tfrac32\right)$.

39. $m_{AB} = \tfrac13$ and $m_{DC} = \tfrac13$.

40. $AB = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$, and $DC = \sqrt{6^2+2^2} = 2\sqrt{10}$.

41. Midpoint of $\overline{AC}$ is $\left(\tfrac72, \tfrac92\right)$, so $\dfrac{5+x}{2} = \tfrac72 \Rightarrow x = 2$ and $\dfrac{3+y}{2} = \tfrac92 \Rightarrow y = 6$. $D(2, 6)$.

42. Midpoint of $\overline{AC}$ is $(-1, 0)$, so $\dfrac{0+x}{2} = -1 \Rightarrow x = -2$ and $\dfrac{-4+y}{2} = 0 \Rightarrow y = 4$. $D(-2, 4)$.

43. $M$ is the midpoint of $\overline{AC}$, so $\dfrac{-3 + x}{2} = 2 \Rightarrow x = 7$ and $\dfrac{4 + y}{2} = -1 \Rightarrow y = -6$. $C(7, -6)$.

44. $\dfrac{4 + x}{2} = 0 \Rightarrow x = -4$ and $\dfrac{6 + y}{2} = 3 \Rightarrow y = 0$. $D(-4, 0)$.

45. Given: $A(-5,1)$, $B(-1,-2)$, $C(5,2)$, $D(1,5)$. Prove: $ABCD$ is a parallelogram.

    Midpoint of $\overline{AC} = \left(\dfrac{-5+5}{2}, \dfrac{1+2}{2}\right) = \left(0, \tfrac32\right)$. Midpoint of $\overline{BD} = \left(\dfrac{-1+1}{2}, \dfrac{-2+5}{2}\right) = \left(0, \tfrac32\right)$. The two midpoints are the same point, so $\overline{AC}$ and $\overline{BD}$ bisect each other. A quadrilateral whose diagonals bisect each other is a parallelogram. $\blacksquare$

46. $m_{AB} = \dfrac{-2-1}{-1-(-5)} = -\tfrac34$ and $m_{DC} = \dfrac{2-5}{5-1} = -\tfrac34$, so $\overline{AB} \parallel \overline{DC}$. $m_{AD} = \dfrac{5-1}{1-(-5)} = \tfrac23$ and $m_{BC} = \dfrac{2-(-2)}{5-(-1)} = \tfrac23$, so $\overline{AD} \parallel \overline{BC}$. Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram by definition. $\blacksquare$ — The midpoint proof is preferable: two computations instead of four, with the same force.

47. $m_{AB} = \tfrac13 = m_{DC}$, so $\overline{AB} \parallel \overline{DC}$; and $AB = DC = 2\sqrt{10}$, so $\overline{AB} \cong \overline{DC}$. One pair of opposite sides both parallel and congruent makes $ABCD$ a parallelogram.

48. $m_{AB} = m_{DC} = -\tfrac34$ and $m_{AD} = m_{BC} = \tfrac23$, so both pairs of opposite edges are parallel — the claim is verified. The edges are $AB = \sqrt{4^2+3^2} = 5$ m and $AD = \sqrt{6^2+4^2} = \sqrt{52} = 2\sqrt{13} \approx 7.21$ m.

49. It does not follow: $AB = CD$ is one pair of opposite sides congruent, and that alone is not one of the four tests. The isosceles trapezoid of item 94 is the counterexample — its opposite sides $\overline{BC}$ and $\overline{DA}$ are both $2\sqrt{10}$, and it is not a parallelogram. The fix: show that the same pair is also parallel, which is the third test.

50. They took the midpoints of two sides, not of the two diagonals. The test is about $\overline{AC}$ and $\overline{BD}$. (The midpoints of two opposite sides of a genuine parallelogram are different points anyway, so the computation they did could never have supported either conclusion.)

51. Because a quadrilateral has two diagonals but four sides. The diagonal test compares two objects and needs two computations; the slope test compares four sides and needs four.

52. The definition of a parallelogram is only both pairs of opposite sides are parallel. That opposite sides are congruent is a consequence, proved in Chapter 10 — and the statement being used here is its converse, which also had to be proved. So citing it means citing a theorem, and you are entitled to do so only because that theorem exists.

Exit ticket 11.2

53. The diagonals bisect each other, tested with the midpoint formula — two computations.
54. Midpoint of $\overline{AC} = (1,1)$; midpoint of $\overline{BD} = (1,1)$. Same point, so yes.
55. Midpoint of $\overline{AC}$ is $\left(\tfrac72, 3\right)$, so $\dfrac{6+x}{2} = \tfrac72 \Rightarrow x = 1$ and $\dfrac{2+y}{2} = 3 \Rightarrow y = 4$. $D(1,4)$.
56. The conclusion that names the property — for example, "a quadrilateral whose diagonals bisect each other is a parallelogram." Two matching midpoints are a computation, not yet a conclusion.

Lesson 11.3 — Rectangles, Rhombi, and Squares

Guided practice

57. They all start with the parallelogram test — the midpoints of the two diagonals.
58. Rectangle: the diagonals are congruent. Rhombus: two adjacent sides are congruent.
59. Slope: $m_{AB} = -\tfrac12$ and $m_{AD} = 2$, and $\left(-\tfrac12\right)(2) = -1$, so the angle at $A$ is right. Distance: $AC = \sqrt{8^2+6^2} = 10$ and $BD = \sqrt{0^2+10^2} = 10$, so the diagonals are congruent.
60. $\overline{AC}$ runs from $(1,0)$ to $(1,6)$ — both $x$-coordinates are $1$, so it is vertical and its slope is undefined. $\overline{BD}$ runs from $(5,3)$ to $(-3,3)$ — both $y$-coordinates are $3$, so it is horizontal. A vertical segment and a horizontal one are perpendicular, and that is stated in words because there is no number to multiply.
61. $AC = 6$ and $BD = 8$. Unequal diagonals rule out a rectangle, and therefore rule out a square.
62. (1) the diagonal midpoints are both $(3,3)$ → parallelogram; (2) $AB = AD = \sqrt{10}$ → rhombus; (3) $AC = BD = 2\sqrt{5}$ → rectangle; (4) a figure that is both is a square.

Independent practice

63. Rectangle. Midpoint of $\overline{AC}$ = midpoint of $\overline{BD}$ = $(1,2)$ → parallelogram. $AC = \sqrt{10^2+2^2} = \sqrt{104} = 2\sqrt{26}$ and $BD = \sqrt{2^2+10^2} = 2\sqrt{26}$ → rectangle. Not a square: $AB = 4\sqrt{2}$ and $AD = 6\sqrt{2}$ differ.
64. Rhombus. Midpoints both $(0,0)$ → parallelogram. $AB = \sqrt{6^2+3^2} = \sqrt{45} = 3\sqrt{5}$ and $AD = \sqrt{6^2+3^2} = 3\sqrt{5}$ → rhombus. Not a square: $AC = 12$ and $BD = 6$ differ.
65. Square. Midpoints both $(1,1)$ → parallelogram. $AB = AD = \sqrt{6^2+2^2} = 2\sqrt{10}$ → rhombus. $AC = BD = \sqrt{8^2+4^2} = \sqrt{80} = 4\sqrt{5}$ → rectangle. Both → square.
66. Parallelogram, and no more. Midpoints both $\left(\tfrac72, 3\right)$. $AB = 2\sqrt{10}$ but $AD = \sqrt{17}$, so not a rhombus; $AC = \sqrt{85}$ but $BD = \sqrt{29}$, so not a rectangle.
67. $AC = BD = 2\sqrt{26}$ — congruent diagonals, which in a parallelogram proves a rectangle.
68. $AB = 4\sqrt{2}$ and $AD = 6\sqrt{2}$ — adjacent sides differ, which rules out a rhombus and therefore a square.
69. $AB = 3\sqrt{5}$ and $AD = 3\sqrt{5}$ — congruent adjacent sides, which in a parallelogram proves a rhombus.
70. $\overline{AC}$ is vertical (slope undefined) and $\overline{BD}$ is horizontal (slope $0$). A vertical and a horizontal segment are perpendicular — stated in words, since the product test needs two numbers and "undefined" is not one.
71. $AC = 6$ and $BD = 8$ — unequal, which rules out a rectangle and therefore a square.
72. $AB = \sqrt{3^2+1^2} = \sqrt{10}$ and $AD = \sqrt{1^2+3^2} = \sqrt{10}$.
73. $AC = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$ and $BD = \sqrt{4^2+2^2} = 2\sqrt{5}$.
74. A rectangle.
75. A rhombus.
76. A square.
77. Midpoint of $\overline{AC}$ = midpoint of $\overline{BD}$ = $(3,3)$ → parallelogram. $AB = AD = \sqrt{10}$ → rhombus. $AC = BD = 2\sqrt{5}$ → rectangle. Both → square. Six computations, and the side is $\sqrt{10}$ inches.
78. Square-cornered: $m_{AB} = \dfrac{-3-1}{0-(-4)} = -1$ and $m_{AD} = \dfrac{7-1}{2-(-4)} = 1$, and $(-1)(1) = -1$, so the corner at $A$ is right. (Equivalently $AC = BD = 2\sqrt{26}$.) Not equal-sided: $AB = 4\sqrt{2}$ and $AD = 6\sqrt{2}$. Both requirements are met.
79. Four congruent sides makes it a rhombus, not a square. A square additionally needs a right angle — equivalently, congruent diagonals — and nothing here shows that. A rhombus with $50°$ and $130°$ angles also has four sides of $5$.
80. Missing: that the figure is a parallelogram in the first place. "A parallelogram with a right angle is a rectangle" — the hypothesis is doing real work. Any number of quadrilaterals have one right angle and belong to no family at all.
81. Because in a parallelogram, opposite sides are already congruent — that is a theorem you have before you start. So $AB = AD$ forces $AB = BC = CD = DA$ all at once, and computing $BC$ and $CD$ returns information you already had.
82. Because the slope of a vertical segment is undefined, so there is no number to put into the product, and $(\text{undefined}) \times 0$ is meaningless. Write instead: one diagonal is vertical and the other is horizontal, so they are perpendicular.

Exit ticket 11.3

83. The diagonals are congruent.
84. Two adjacent sides are congruent.
85. No. $AC = \sqrt{12^2 + 0^2} = 12$ and $BD = \sqrt{0^2 + 6^2} = 6$. The diagonals are unequal, so it is not a rectangle and therefore not a square. (It is a rhombus.)
86. Six. Two midpoints (parallelogram), two adjacent sides (rhombus), and two diagonals (rectangle).

Lesson 11.4 — Trapezoids, and Choosing the Proof

Guided practice

87. $m_{AB} = 0$ and $m_{DC} = 0$, so the bases are parallel; $m_{AD} = 3$ and $m_{BC} = -2$, so the legs are not parallel.
88. Because every parallelogram also has a pair of parallel sides. Without the second fact you have not excluded a parallelogram — and under Virginia's exclusive definition a parallelogram is not a trapezoid, so the proof would be incomplete.
89. $AD = BC = 2\sqrt{10}$ — congruent legs, so it is an isosceles trapezoid.
90. $AC = BD = 10$; the midpoints are $(-1,1)$ and $(1,1)$. Together: the diagonals are congruent but do not bisect each other — the first fact is shared with a rectangle, the second is what keeps it out of the parallelogram family.
91. "Is it a parallelogram?" It costs only two computations, and its answer splits the six families into two groups, so every later question depends on it.
92. Square: six ($2$ midpoints $+ 2$ sides $+ 2$ diagonals). Isosceles trapezoid: six ($4$ slopes $+ 2$ legs).

Independent practice

93. Trapezoid. $m_{AB} = 0 = m_{DC}$, so one pair is parallel; $m_{AD} = 2$ and $m_{BC} = -6$, so the other is not. Not isosceles: $AD = \sqrt{3^2+6^2} = 3\sqrt{5}$ and $BC = \sqrt{1^2+6^2} = \sqrt{37}$.
94. Isosceles trapezoid. $m_{AB} = 0 = m_{DC}$; $m_{AD} = 3$ and $m_{BC} = -3$, so exactly one pair is parallel. $AD = BC = \sqrt{2^2+6^2} = 2\sqrt{10}$ → isosceles.
95. Trapezoid. $m_{AB} = 0 = m_{DC}$; $m_{AD} = 3$ and $m_{BC} = -2$. Not isosceles: $AD = 2\sqrt{10}$ and $BC = 3\sqrt{5}$.
96. Isosceles trapezoid. $m_{AB} = 0 = m_{DC}$; $m_{AD} = 3$ and $m_{BC} = -3$. $AD = BC = 2\sqrt{10}$ → isosceles.
97. $m_{AB} = 0$, $m_{BC} = -3$, $m_{CD} = 0$, $m_{DA} = 3$. The two facts: $\overline{AB} \parallel \overline{DC}$ (both $0$), and $\overline{AD} \not\parallel \overline{BC}$ ($3 \ne -3$) — so exactly one pair is parallel.
98. $AD = BC = 2\sqrt{10}$, so it is an isosceles trapezoid.
99. $AC = \sqrt{6^2+6^2} = \sqrt{72} = 6\sqrt{2}$ and $BD = \sqrt{6^2+6^2} = 6\sqrt{2}$.
100. $(-1, 0)$ and $(1, 0)$. They are different, so the diagonals do not bisect each other, so the figure is not a parallelogram.
101. $AD = 3\sqrt{5}$ and $BC = \sqrt{37}$.
102. A trapezoid.
103. Because Virginia's definition is exclusive — exactly one pair. Two pairs of parallel sides makes the figure a parallelogram, and in this course no parallelogram is a trapezoid.
104. $m_{AB} = 0 = m_{DC}$, so the bases are parallel; $m_{AD} = 3$ and $m_{BC} = -3$, so the legs are not — a trapezoid. Then $AD = BC = 2\sqrt{10}$, so it is isosceles. Each leg is $2\sqrt{10}$ ft $\approx 6.32$ ft.
105. Compute the two diagonal midpoints first: both are $\left(0, \tfrac32\right)$, so it is a parallelogram. Then compute the two diagonals: $AC = \sqrt{10^2+1^2} = \sqrt{101}$ and $BD = \sqrt{2^2+7^2} = \sqrt{53}$. Unequal, so it is not a rectangle — four computations in total.
106. Missing: the second slope fact, that the other pair of sides is not parallel. Without it the figure could be a parallelogram, which under the exclusive definition is not a trapezoid — so the proof does not yet establish the family it claims.
107. Item 94's figure has $AC = BD = 6\sqrt{2}$, and its diagonals have slopes $\dfrac{3-(-3)}{2-(-4)} = 1$ and $\dfrac{3-(-3)}{-2-4} = -1$, whose product is $-1$ — so the diagonals genuinely are both congruent and perpendicular. But it is an isosceles trapezoid, not a square: its diagonal midpoints are $(-1,0)$ and $(1,0)$, which differ. The skipped check is the parallelogram test. Congruent-and-perpendicular diagonals name a square only within the parallelogram family.
108. Because it costs two computations and it partitions every possibility in two — parallelogram or not. Every later question ("congruent diagonals?", "congruent legs?") means something different depending on the answer, so asking it later risks computing things that turn out to be irrelevant.

Exit ticket 11.4

109. One pair of opposite sides has equal slopes, and the other pair does not.
110. The legs are congruent — one distance computation on each.
111. No. Midpoint of $\overline{AC}$ is $(-1, 0)$ and midpoint of $\overline{BD}$ is $(1, 0)$ — different points, so the diagonals do not bisect each other.
112. "Is it a parallelogram?", answered by the midpoint formula run on both diagonals.

Chapter 11 Review — answers

Review 1 (G.PC.1b). $A(-3,-1)$, $B(3,-3)$, $C(5,3)$, $D(-1,5)$.

• Midpoint of $\overline{AC} = \left(\tfrac{-3+5}{2}, \tfrac{-1+3}{2}\right) = (1,1)$; midpoint of $\overline{BD} = \left(\tfrac{3+(-1)}{2}, \tfrac{-3+5}{2}\right) = (1,1)$. Same point, so the diagonals bisect each other and $ABCD$ is a parallelogram.
• $AB = \sqrt{6^2+(-2)^2} = \sqrt{40} = 2\sqrt{10}$ and $AD = \sqrt{2^2+6^2} = 2\sqrt{10}$. Two adjacent sides are congruent, so it is a rhombus.
• $AC = \sqrt{8^2+4^2} = \sqrt{80} = 4\sqrt{5}$ and $BD = \sqrt{(-4)^2+8^2} = 4\sqrt{5}$. The diagonals are congruent, so it is a rectangle.
• A square, in six computations. Computing all four sides, all four slopes, and both diagonals would be ten — and would prove exactly the same thing.

Review 2 (G.PC.1 a, b).

• $A(0,0)$, $B(6,2)$, $C(7,6)$, $D(1,4)$ → parallelogram. Decided by $AB = 2\sqrt{10}$ against $AD = \sqrt{17}$ (not a rhombus); the diagonals $\sqrt{85}$ and $\sqrt{29}$ also differ, so not a rectangle either.
• $A(-6,0)$, $B(0,-3)$, $C(6,0)$, $D(0,3)$ → rhombus. Decided by $AB = AD = 3\sqrt{5}$; and $AC = 12 \ne BD = 6$ stops it short of a square.
• $A(-4,1)$, $B(0,-3)$, $C(6,3)$, $D(2,7)$ → rectangle. Decided by $AC = BD = 2\sqrt{26}$; and $AB = 4\sqrt{2} \ne AD = 6\sqrt{2}$ stops it short of a square.
• $A(-4,-3)$, $B(4,-3)$, $C(3,3)$, $D(-1,3)$ → trapezoid. Decided by $AD = 3\sqrt{5} \ne BC = \sqrt{37}$ — not isosceles.
• $A(-4,-3)$, $B(4,-3)$, $C(2,3)$, $D(-2,3)$ → isosceles trapezoid. Decided by $AD = BC = 2\sqrt{10}$.

Review 3 (G.PC.1b). $A(-6,-2)$, $B(0,-4)$, $C(4,2)$.

• $M$, the midpoint of $\overline{AC}$, is $\left(\tfrac{-6+4}{2}, \tfrac{-2+2}{2}\right) = (-1, 0)$. In a parallelogram $M$ is also the midpoint of $\overline{BD}$, so $\dfrac{0+x}{2} = -1 \Rightarrow x = -2$ and $\dfrac{-4+y}{2} = 0 \Rightarrow y = 4$. $D(-2, 4)$.
• $AB = \sqrt{6^2+(-2)^2} = 2\sqrt{10}$ cm and $AD = \sqrt{4^2+6^2} = \sqrt{52} = 2\sqrt{13}$ cm. Adjacent sides differ, so it is not a rhombus.
• $AC = \sqrt{10^2+4^2} = \sqrt{116} = 2\sqrt{29}$ cm and $BD = \sqrt{(-2)^2+8^2} = \sqrt{68} = 2\sqrt{17}$ cm. Unequal, so it is not a rectangle.
• No choice of $D$ can make it a square, and one computation already made proves it. $\overline{AB}$ and $\overline{BC}$ are two adjacent sides of the figure, and both are fixed by the three given points alone — $AB = 2\sqrt{10}$ and $BC = \sqrt{4^2+6^2} = 2\sqrt{13}$. A square needs adjacent sides congruent, and these two already differ before $D$ is placed at all. (There is no freedom in $D$ in any case: requiring a parallelogram pins it to $(-2,4)$ uniquely.)