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

1 Angles and algebra / 4 pts

(a) Alternate interior angles are congruent: \(7x - 4 = 5x + 16\) (1 pt), so \(2x = 20\) and \(x = 10\) (1 pt).

(b) \(7(10) - 4 = 66\); each angle measures \(66^\circ\) (1 pt). Check: \(5(10) + 16 = 66\).

(c) Same-side interior angles are supplementary: \(180^\circ - 66^\circ = 114^\circ\) (1 pt).

2 Prove parallel / 3 pts

(a) \(118 + 62 = 180\) (1 pt), so the same-side interior angles are supplementary and, by the converse theorem, \(p \parallel q\) (1 pt).

(b) \(118 + 64 = 182 \neq 180\), so the lines are not parallel (1 pt).

3 Slopes of three lines / 4 pts

\(m_1 = \dfrac{-2 - 4}{1 + 3} = -\dfrac{3}{2}\) (1 pt). \(m_2 = \dfrac{3 - 1}{3 - 0} = \dfrac{2}{3}\) (1 pt). \(m_3 = \dfrac{-1 - 5}{3 + 1} = -\dfrac{3}{2}\) (1 pt).

\(m_1 \cdot m_2 = -1\), so \(\ell_1 \perp \ell_2\). \(m_1 = m_3\), and \(\ell_1\) is \(y = -\dfrac{3}{2}x - \dfrac{1}{2}\), which gives \(y = 1\) at \(x = -1\), not \(5\): the lines are different, so \(\ell_1 \parallel \ell_3\) (1 pt).

4 Writing equations / 4 pts

(a) Slope \(-\dfrac{3}{4}\) (1 pt). \(y + 1 = -\dfrac{3}{4}(x - 8)\), so \(y = -\dfrac{3}{4}x + 5\) (1 pt). Check: \(-6 + 5 = -1\).

(b) Slope \(\dfrac{4}{3}\) (1 pt). \(y + 1 = \dfrac{4}{3}(x - 8)\), so \(y = \dfrac{4}{3}x - \dfrac{35}{3}\) (1 pt). Check: \(\dfrac{32}{3} - \dfrac{35}{3} = -1\).

5 Distance to a line / 3 pts

Formula: \(d = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\) (1 pt). \(d = \dfrac{|3(1) + 4(2) + 5|}{\sqrt{9 + 16}}\) (1 pt) \(= \dfrac{16}{5} = 3.2\) units (1 pt).

6 Perpendicular bisector / 2 pts

Midpoint \((2, 2)\) and slope of \(\overline{AB}\) equal to \(\dfrac{4}{6} = \dfrac{2}{3}\) (1 pt). The bisector has slope \(-\dfrac{3}{2}\): \(y - 2 = -\dfrac{3}{2}(x - 2)\), that is \(y = -\dfrac{3}{2}x + 5\) (1 pt).