Points, Lines, Planes, and Angle Relationships: practice solutions, Grade 10 – download the PDF

1 Point, line, or plane? ★★★

1. Point: it marks a location and has no size.
2. Line: straight and extending in both directions.
3. Plane: flat and extending in all directions.
4. Point: a single location.
5. Plane: a flat surface without edges.

2 Adding two segments ★★★

By the Segment Addition Postulate, \(PR = PQ + QR = 14.5 + 9.3 = 23.8\).

Answer: \(PR = 23.8\text{ cm}\).

3 Midpoint on a number line ★★★

Length: \(|10 - (-6)| = 16\). Midpoint: \(\dfrac{-6 + 10}{2} = 2\).

Answer: the length is 16 units and the midpoint is at \(2\). Check: \(2\) is 8 units from each endpoint.

4 Complement and supplement ★★★

Complement: \(90^\circ - 37^\circ = 53^\circ\). Supplement: \(180^\circ - 37^\circ = 143^\circ\).

5 Vertical angles ★★★

The vertical angle is congruent, so it measures \(48^\circ\). The two other angles each form a linear pair with it: \(180^\circ - 48^\circ = 132^\circ\).

Answer: \(48^\circ, 132^\circ, 132^\circ\).

6 A linear pair ★★★

A linear pair is supplementary: \(x + (x + 30) = 180\), so \(2x = 150\) and \(x = 75\).

Answer: the angles measure \(75^\circ\) and \(105^\circ\). Check: \(75 + 105 = 180\).

7 Angle addition ★★★

A right angle measures \(90^\circ\). By the Angle Addition Postulate, \(28 + m\angle BOC = 90\), so \(m\angle BOC = 62^\circ\).

8 Segment addition with algebra ★★★

\(LM + MN = LN\), so \((3x + 2) + (2x + 7) = 44\). That gives \(5x + 9 = 44\), \(5x = 35\), \(x = 7\).

\(LM = 3(7) + 2 = 23\) and \(MN = 2(7) + 7 = 21\). Check: \(23 + 21 = 44\).

Answer: \(x = 7\), \(LM = 23\text{ cm}\), \(MN = 21\text{ cm}\).

9 Midpoint and distance ★★★

Midpoint: \(\left(\dfrac{-3 + 9}{2}, \dfrac{8 + (-2)}{2}\right) = (3, 3)\).

Distance: \(AB = \sqrt{(9 - (-3))^2 + (-2 - 8)^2} = \sqrt{12^2 + (-10)^2} = \sqrt{244} = 2\sqrt{61} \approx 15.62\).

Answer: \(M(3, 3)\) and \(AB \approx 15.62\) units.

10 Find the other endpoint ★★★

Let \(B(x, y)\). Then \(\dfrac{7 + x}{2} = 4\), so \(x = 1\), and \(\dfrac{2 + y}{2} = -1\), so \(y = -4\).

Answer: \(B(1, -4)\). Check: the midpoint of \((7, 2)\) and \((1, -4)\) is \((4, -1)\).

11 Bisector equation ★★★

A bisector makes congruent angles: \(4x + 3 = 6x - 17\), so \(20 = 2x\) and \(x = 10\).

Each half measures \(4(10) + 3 = 43^\circ\), so \(m\angle ABC = 2 \times 43 = 86^\circ\).

12 Supplement equals three complements ★★★

Let the angle be \(x\). Then \(180 - x = 3(90 - x)\), so \(180 - x = 270 - 3x\), \(2x = 90\), \(x = 45\).

Check: the supplement is \(135^\circ\), the complement is \(45^\circ\), and \(3 \times 45 = 135\).

Answer: \(45^\circ\).

13 Vertical angles with variables ★★★

Vertical angles are congruent: \(5x - 12 = 3x + 20\), so \(2x = 32\) and \(x = 16\).

Each equals \(5(16) - 12 = 68^\circ\). The other two angles form linear pairs with them: \(180 - 68 = 112^\circ\).

14 True or false? ★★★

1. False. Angles of \(100^\circ\) and \(80^\circ\) far apart on a page are supplementary but are not adjacent, so they are not a linear pair.
2. True. This is the Vertical Angles Theorem: both angles are supplementary to the same neighboring angle.
3. True. Only one point splits a segment into two equal parts.

15 Intersecting lines ★★★

A linear pair adds up to \(180^\circ\): \((2x + 10) + (3x - 5) = 180\), so \(5x + 5 = 180\) and \(x = 35\).

\(m\angle 1 = 2(35) + 10 = 80^\circ\) and \(m\angle 2 = 3(35) - 5 = 100^\circ\). Vertical angles are congruent, so \(m\angle 3 = 80^\circ\) and \(m\angle 4 = 100^\circ\).

Check: \(80 + 100 + 80 + 100 = 360\).

16 A hiking map ★★★

1. \(\left(\dfrac{2 + 14}{2}, \dfrac{3 + 12}{2}\right) = (8, 7.5)\).
2. \(TC = \sqrt{(14 - 2)^2 + (12 - 3)^2} = \sqrt{144 + 81} = \sqrt{225} = 15\) miles. In kilometers: \(15 \times 1.609 \approx 24.1\) km.

Answer: the rest stop is at \((8, 7.5)\) and the cabin is 15 miles (about 24.1 km) away.

17 Four points on a fence ★★★

1. \(AB + BC + CD = AD\): \(2x + (x + 3) + (3x - 1) = 50\), so \(6x + 2 = 50\) and \(x = 8\). Then \(AB = 16\), \(BC = 11\), \(CD = 23\). Check: \(16 + 11 + 23 = 50\).
2. The midpoint is 25 feet from \(A\). \(B\) is 16 ft from \(A\) and \(C\) is \(16 + 11 = 27\) ft from \(A\), and \(16 < 25 < 27\), so it lies on \(\overline{BC}\), \(25 - 16 = 9\) feet from \(B\).

18 Three adjacent angles ★★★

1. By angle addition, \(x + (2x + 10) + (3x - 4) = 150\), so \(6x + 6 = 150\) and \(x = 24\). The angles are \(24^\circ\), \(58^\circ\), and \(68^\circ\). Check: \(24 + 58 + 68 = 150\).
2. The bisector makes \(75^\circ\) with \(OA\). \(m\angle AOB = 24^\circ\) and \(m\angle AOC = 24 + 58 = 82^\circ\). Since \(24 < 75 < 82\), the bisector lies between \(OB\) and \(OC\), \(51^\circ\) from \(OB\).

19 Proof: vertical angles ★★★

1. \(\angle 1\) and \(\angle 2\) form a linear pair, so \(m\angle 1 + m\angle 2 = 180^\circ\) (Linear Pair Postulate).
2. \(\angle 2\) and \(\angle 3\) form a linear pair, so \(m\angle 2 + m\angle 3 = 180^\circ\) (Linear Pair Postulate).
3. \(m\angle 1 + m\angle 2 = m\angle 2 + m\angle 3\) (substitution, both sums equal \(180^\circ\)).
4. \(m\angle 1 = m\angle 3\) (subtraction property of equality).

20 Bisectors of a linear pair ★★★

Let the angles be \(a\) and \(b\) with \(a + b = 180\). The bisectors make angles \(\dfrac{a}{2}\) and \(\dfrac{b}{2}\), which are adjacent, so the angle between the bisectors is \(\dfrac{a}{2} + \dfrac{b}{2} = \dfrac{180}{2} = 90^\circ\). So they are perpendicular.

Check: \(\dfrac{50}{2} + \dfrac{130}{2} = 25 + 65 = 90\).

21 Endpoint and length ★★★

\(x_Q = 2(-1.5) - (-5) = 2\) and \(y_Q = 2(4) - 7 = 1\), so \(Q(2, 1)\).

\(PQ = \sqrt{(2 - (-5))^2 + (1 - 7)^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22\).

22 Why does the construction work? ★★★

1. Draw an arc centered at \(B\) crossing the two sides at \(P\) and \(Q\), so \(BP = BQ\).
2. With the same opening, more than half of \(PQ\), draw arcs centered at \(P\) and at \(Q\). They meet at \(R\), so \(PR = QR\).
3. Draw ray \(BR\).

Justification: triangles \(BPR\) and \(BQR\) have \(BP = BQ\), \(PR = QR\), and the shared side \(BR\). They are congruent by SSS, so the corresponding angles \(\angle PBR\) and \(\angle QBR\) are equal.