Parallel and Perpendicular Lines: practice solutions, Grade 10 – download the PDF

1 Naming angle pairs ★★★

(a) Same corner at each crossing: corresponding angles.

(b) Between the lines, on opposite sides of \(t\): alternate interior angles.

(c) Between the lines, on the same side of \(t\): same-side interior angles.

(d) Outside the lines, on opposite sides of \(t\): alternate exterior angles.

2 Finding angles from one measure ★★★

\(\angle 6\) and \(\angle 3\) are alternate interior angles, so \(m\angle 6 = 65^\circ\).

\(\angle 7\) and \(\angle 3\) are corresponding angles, so \(m\angle 7 = 65^\circ\).

\(\angle 2\) and \(\angle 3\) are vertical angles, so \(m\angle 2 = 65^\circ\).

\(\angle 5\) and \(\angle 3\) are same-side interior angles, so \(m\angle 5 = 180^\circ - 65^\circ = 115^\circ\).

3 Crossing lines ★★★

The vertical angle also measures \(38^\circ\). Each of the two other angles forms a linear pair with the \(38^\circ\) angle, so it measures \(180^\circ - 38^\circ = 142^\circ\).

The four angles are \(38^\circ,\ 142^\circ,\ 38^\circ,\ 142^\circ\) and they add up to \(360^\circ\).

4 True or false? ★★★

False. Corresponding angles are congruent only when the two lines are parallel. For example, if two lines meet at a point and a transversal crosses them, the corresponding angles are different because the lines are tilted relative to each other. Conversely, if the corresponding angles are congruent, the lines are parallel.

5 Slope from two points ★★★

\(m = \dfrac{9 - 1}{4 - 0} = \dfrac{8}{4} = 2\).

(a) A parallel line has the same slope, \(2\).

(b) A perpendicular line has the opposite reciprocal slope, \(-\dfrac{1}{2}\). Check: \(2 \cdot \left(-\dfrac{1}{2}\right) = -1\).

6 Parallel in disguise ★★★

Solve the second equation for \(y\): \(-2y = -8x + 7\), so \(y = 4x - 3.5\).

Both slopes equal \(4\), but the \(y\)-intercepts are \(-5\) and \(-3.5\), so the lines are different. They are parallel.

7 Opposite reciprocals ★★★

(a) Flip and change the sign: \(\dfrac{3}{5}\).

(b) \(4 = \dfrac{4}{1}\), so the perpendicular slope is \(-\dfrac{1}{4}\).

(c) \(-2\).

Each product with the original slope equals \(-1\).

8 Alternate interior angles and algebra ★★★

Alternate interior angles are congruent: \(4x + 7 = 6x - 21\), so \(28 = 2x\) and \(x = 14\).

Each angle measures \(4(14) + 7 = 63^\circ\). Check: \(6(14) - 21 = 63\).

9 Same-side interior angles and algebra ★★★

Same-side interior angles are supplementary: \((2x + 15) + (3x + 25) = 180\), so \(5x + 40 = 180\) and \(x = 28\).

The angles are \(2(28) + 15 = 71^\circ\) and \(3(28) + 25 = 109^\circ\), and \(71 + 109 = 180\).

10 For which x are the lines parallel? ★★★

The lines are parallel exactly when the corresponding angles are congruent: \(5x + 8 = 7x - 12\), so \(20 = 2x\) and \(x = 10\).

Both angles measure \(5(10) + 8 = 58^\circ\). For any other value of \(x\) the corresponding angles differ and the lines are not parallel.

11 Parallel, perpendicular, or neither? ★★★

Slope of \(A\): \(\dfrac{5 - 2}{5 - 1} = \dfrac{3}{4}\). Slope of \(B\): \(\dfrac{-4 - 0}{3 - 0} = -\dfrac{4}{3}\). Slope of \(C\): \(\dfrac{4 - 1}{2 + 2} = \dfrac{3}{4}\).

\(A\) and \(B\): \(\dfrac{3}{4} \cdot \left(-\dfrac{4}{3}\right) = -1\), so they are perpendicular.

\(A\) and \(C\): equal slopes. Is \(C\) the same line as \(A\)? The line \(A\) is \(y = \dfrac{3}{4}x + \dfrac{5}{4}\); at \(x = -2\) it gives \(-\dfrac{1}{4}\), not \(1\). So they are different lines: parallel.

12 A parallel line in standard form ★★★

Solve for \(y\): \(5y = -2x + 10\), so \(y = -\dfrac{2}{5}x + 2\) and the slope is \(-\dfrac{2}{5}\).

Point-slope form: \(y + 3 = -\dfrac{2}{5}(x - 4)\), so \(y = -\dfrac{2}{5}x + \dfrac{8}{5} - 3 = -\dfrac{2}{5}x - \dfrac{7}{5}\).

Multiplying by 5: \(2x + 5y = -7\). Check: \(2(4) + 5(-3) = 8 - 15 = -7\).

13 A perpendicular line through a point ★★★

The given slope is \(-3\), so the perpendicular slope is \(\dfrac{1}{3}\).

\(y - 1 = \dfrac{1}{3}(x - 6)\), so \(y = \dfrac{1}{3}x - 2 + 1 = \dfrac{1}{3}x - 1\).

Check at \(x = 6\): \(2 - 1 = 1\).

14 The sprinkler pipe ★★★

Write the pipe as \(5x + 12y - 39 = 0\). Then \(d = \dfrac{|5(0) + 12(0) - 39|}{\sqrt{5^2 + 12^2}} = \dfrac{39}{13} = 3\).

The shortest distance is 3 meters, about \(3 \times 3.28 \approx 9.8\) feet.

15 Perpendicular bisector equation ★★★

Midpoint: \(\left(\dfrac{-2 + 4}{2}, \dfrac{3 + 7}{2}\right) = (1, 5)\).

Slope of \(\overline{AB}\): \(\dfrac{7 - 3}{4 + 2} = \dfrac{2}{3}\), so the bisector has slope \(-\dfrac{3}{2}\).

\(y - 5 = -\dfrac{3}{2}(x - 1)\), so \(y = -\dfrac{3}{2}x + \dfrac{13}{2}\).

16 Is the triangle a right triangle? ★★★

(a) Slope of \(\overline{AB}\): \(\dfrac{3 - 1}{5 - 1} = \dfrac{1}{2}\). Slope of \(\overline{BC}\): \(\dfrac{7 - 3}{3 - 5} = -2\). Since \(\dfrac{1}{2} \cdot (-2) = -1\), \(\overline{AB} \perp \overline{BC}\): the right angle is at \(B\).

(b) \(AB = \sqrt{4^2 + 2^2} = \sqrt{20}\) and \(BC = \sqrt{(-2)^2 + 4^2} = \sqrt{20}\). The legs are the two perpendicular sides, so the area is \(\dfrac{1}{2} \cdot \sqrt{20} \cdot \sqrt{20} = \dfrac{20}{2} = 10\) square units.

17 Is PQRS a rectangle? ★★★

\(\overline{PQ}\): \(\dfrac{2 - 0}{4 - 0} = \dfrac{1}{2}\). \(\overline{QR}\): \(\dfrac{6 - 2}{2 - 4} = -2\). \(\overline{RS}\): \(\dfrac{4 - 6}{-2 - 2} = \dfrac{1}{2}\). \(\overline{SP}\): \(\dfrac{0 - 4}{0 + 2} = -2\).

Opposite sides have equal slopes, so they are parallel: \(PQ \parallel RS\) and \(QR \parallel SP\). Adjacent sides have slopes with product \(-1\), so the angles are right angles. The quadrilateral has four right angles: it is a rectangle.

\(PQ = \sqrt{20}\) and \(QR = \sqrt{4 + 16} = \sqrt{20}\), so the area is \(20\) square units (it is actually a square).

18 Distance between parallel lines ★★★

The lines have the same slope \(2\), so they are parallel. Choose the point \((0, 3)\) on the first line.

Write the second line as \(2x - y - 7 = 0\). Then \(d = \dfrac{|2(0) - 3 - 7|}{\sqrt{2^2 + (-1)^2}} = \dfrac{10}{\sqrt{5}} = 2\sqrt{5} \approx 4.47\).

The distance between the lines is \(2\sqrt{5} \approx 4.47\) units.

19 Equidistant point on the x-axis ★★★

Let \(P = (x, 0)\). \(PA^2 = (x - 1)^2 + 16\) and \(PB^2 = (x - 5)^2 + 4\).

Set them equal: \(x^2 - 2x + 17 = x^2 - 10x + 29\), so \(8x = 12\) and \(x = 1.5\). So \(P(1.5, 0)\).

Check: \(PA^2 = 0.25 + 16 = 16.25\) and \(PB^2 = 12.25 + 4 = 16.25\). By the converse of the perpendicular bisector theorem, \(P\) lies on the perpendicular bisector of \(\overline{AB}\), which crosses the \(x\)-axis at \(P\).

20 Two perpendiculars to a third line ★★★

Angle proof. Let \(c\) be a transversal of \(a\) and \(b\). Since \(a \perp c\) and \(b \perp c\), the two corresponding angles formed with \(c\) both measure \(90^\circ\). Corresponding angles are congruent, so \(a \parallel b\) by the converse of the corresponding angles theorem.

Slope proof. The slope of \(c\) is \(\dfrac{2}{5}\), so each line perpendicular to \(c\) has slope \(-\dfrac{5}{2}\). Lines \(a\) and \(b\) have the same slope, so they are parallel (or the same line).

21 Foot of the perpendicular ★★★

(a) The perpendicular through \(P\) has slope \(-\dfrac{1}{2}\): \(y = -\dfrac{1}{2}x + 6\). Intersect with \(\ell\): \(2x - 4 = -\dfrac{1}{2}x + 6\), so \(\dfrac{5}{2}x = 10\) and \(x = 4\). Then \(y = 2(4) - 4 = 4\). So \(H(4, 4)\).

(b) \(PH = \sqrt{(4 - 0)^2 + (4 - 6)^2} = \sqrt{20} = 2\sqrt{5}\). With the formula, \(\ell\) is \(2x - y - 4 = 0\), so \(d = \dfrac{|0 - 6 - 4|}{\sqrt{5}} = \dfrac{10}{\sqrt{5}} = 2\sqrt{5}\). The two results agree.