Slopes of Parallel Lines – Proofs and Questions

Find which of the following lines are parallel:

\[ \begin{aligned} \text{a)}\ & y = 2x - 3 \\ \text{b)}\ & 2x - y = 2 \\ \text{c)}\ & -4y + 2x = 0 \\ \text{d)}\ & -4y + 8x = 9 \end{aligned} \]

Solution

Write each equation in slope–intercept form \(y = mx + b\):

\[ \begin{aligned} \text{a)}\ & y = 2x - 3 \implies m = 2 \\ \text{b)}\ & 2x - y = 2 \implies y = 2x - 2 \implies m = 2 \\ \text{c)}\ & -4y + 2x = 0 \implies 4y = 2x \implies y = \tfrac{1}{2}x \implies m = \tfrac{1}{2} \\ \text{d)}\ & -4y + 8x = 9 \implies 4y = 8x - 9 \implies y = 2x - \tfrac{9}{4} \implies m = 2 \end{aligned} \]

The lines in parts (a), (b), and (d) have equal slopes (\(m = 2\)) and are therefore parallel.

Find \(k\) so that the lines \[ 6kx - 3y = 9 \quad \text{and} \quad -4x + 5y = 7 \] are parallel.

Solution

\[ \begin{aligned} 6kx - 3y &= 9 \implies 3y = 6kx - 9 \implies y = 2kx - 3 \implies m_1 = 2k \\ -4x + 5y &= 7 \implies 5y = 4x + 7 \implies y = \tfrac{4}{5}x + \tfrac{7}{5} \implies m_2 = \tfrac{4}{5} \end{aligned} \]

Parallel lines have equal slopes: \[ 2k = \tfrac{4}{5} \] \[ k = \tfrac{2}{5} \]

Find \(a\) so that the line through \((1,2)\) and \((0,3)\) is parallel to the line through \((a,2)\) and \((-2,7)\).

Solution

\[ m_1 = \frac{3 - 2}{0 - 1} = \frac{1}{-1} = -1 \] \[ m_2 = \frac{7 - 2}{-2 - a} = \frac{5}{-2 - a} \]

Set the slopes equal: \[ \frac{5}{-2 - a} = -1 \] \[ 5 = -1(-2 - a) \implies 5 = 2 + a \implies a = 3 \]

Find the coordinates of point \(D (a,b)\) so that \(A(0,2)\), \(B(2,6)\), \(C(8,8)\), and \(D\) form a parallelogram.

Solution

Opposite sides of a parallelogram are parallel, meaning their slopes are equal (\(m_{AB} = m_{DC}\) and \(m_{BC} = m_{AD}\)):

\[ m_{AB} = \frac{6 - 2}{2 - 0} = \frac{4}{2} = 2 \] \[ m_{DC} = \frac{8 - b}{8 - a} \implies 2 = \frac{8 - b}{8 - a} \implies 8 - b = 16 - 2a \implies 2a - b = 8 \]

\[ m_{BC} = \frac{8 - 6}{8 - 2} = \frac{2}{6} = \tfrac{1}{3} \] \[ m_{AD} = \frac{b - 2}{a - 0} \implies \tfrac{1}{3} = \frac{b - 2}{a} \implies a = 3b - 6 \]

Substituting \(a = 3b - 6\) into \(2a - b = 8\): \[ 2(3b - 6) - b = 8 \implies 6b - 12 - b = 8 \implies 5b = 20 \implies b = 4 \] \[ a = 3(4) - 6 = 12 - 6 = 6 \]

Thus, the coordinates are \(D(6,4)\).

Find the equation of a line through \((2,1)\) parallel to the line through \((-1,2)\) and \((3,4)\).

Solution

\[ m = \frac{4 - 2}{3 - (-1)} = \frac{2}{4} = \tfrac{1}{2} \]

Using the point-slope form with point \((2,1)\) and slope \(m = \tfrac{1}{2}\): \[ y - 1 = \tfrac{1}{2}(x - 2) \implies y - 1 = \tfrac{1}{2}x - 1 \implies y = \tfrac{1}{2}x \]

Find the equation of a line through \((-1,1)\) parallel to \(2y + 4x = 2\).

Solution

Rewrite the given equation in slope-intercept form: \[ 2y + 4x = 2 \implies 2y = -4x + 2 \implies y = -2x + 1 \implies m = -2 \]

Using the point-slope form with point \((-1,1)\) and slope \(m = -2\): \[ y - 1 = -2(x - (-1)) \implies y - 1 = -2(x + 1) \implies y - 1 = -2x - 2 \implies y = -2x - 1 \]