Slopes of Perpendicular Lines

This page presents a detailed proof showing that the slopes of two perpendicular lines are negative reciprocals of each other. Several practice questions on perpendicular lines are also included, each with a complete step-by-step solution.

Proof: Slopes of Perpendicular Lines

Questions on Perpendicular Lines

Question 1

Which of the following lines are perpendicular?

\[ \text{a) } y = -2x - 3 \quad \text{b) } -4y + 5x = 4 \quad \text{c) } \frac{1}{2}x - y = 2 \quad \text{d) } -6y - 3x = 9 \]

Solution

\[ \begin{aligned} \text{a)}\;& m = -2 \\ \text{b)}\;& y = \frac{5}{4}x - 1 \Rightarrow m = \frac{5}{4} \\ \text{c)}\;& y = \frac{1}{2}x - 2 \Rightarrow m = \frac{1}{2} \\ \text{d)}\;& y = -\frac{1}{2}x - \frac{3}{2} \Rightarrow m = -\frac{1}{2} \end{aligned} \] \[ (-2)\left(\frac{1}{2}\right) = -1 \]

Therefore, the lines in parts a and c are perpendicular.

Question 2

Find \(k\) so that the line through \((1,-2)\) and \((0,3)\) is perpendicular to the line through \((3,-k)\) and \((2,-5)\).

Solution

\[ m_1 = \frac{3 - (-2)}{0 - 1} = -5 \] \[ m_2 = \frac{-5 - (-k)}{2 - 3} = 5 - k \] \[ -5(5 - k) = -1 \] \[ k = \frac{24}{5} \]

Question 3

Find the value of \(k\) such that the lines with equations

\[ - kx + 2x - y = -3 \quad \text{and} \quad -2x + y = 7 \]

are perpendicular.

Solution

First, write each equation in slope–intercept form \(y = mx + b\).

\[ - kx + 2x - y = -3 \quad \Rightarrow \quad y = (-k + 2)x + 3 \] \[ \text{Slope of first line: } m_1 = -k + 2 \] \[ -2x + y = 7 \quad \Rightarrow \quad y = 2x + 7 \] \[ \text{Slope of second line: } m_2 = 2 \]

Since the lines are perpendicular, the product of their slopes is \(-1\):

\[ m_1 m_2 = -1 \] \[ 2(-k + 2) = -1 \] \[ -2k + 4 = -1 \] \[ k = \frac{5}{2} \]

Question 4

Find the coordinates of point \(C\) such that the points

\[ A(0,2), \quad B(2,6), \quad C(2k, k) \]

are the vertices of a right-angled triangle with hypotenuse \(AC\).

Solution

Since \(AC\) is the hypotenuse, segments \(AB\) and \(BC\) must be perpendicular.

Slope of segment \(AB\):

\[ m_{AB} = \frac{6 - 2}{2 - 0} = 2 \]

Slope of segment \(BC\):

\[ m_{BC} = \frac{k - 6}{2k - 2} \]

Because the segments are perpendicular, the product of their slopes is \(-1\):

\[ 2 \cdot \frac{k - 6}{2k - 2} = -1 \] \[ 2(k - 6) = - (2k - 2) \] \[ 2k - 12 = -2k + 2 \] \[ 4k = 14 \] \[ k = \frac{7}{2} \]

Therefore, the coordinates of point \(C\) are:

\[ C(2k, k) = \left(7, \frac{7}{2}\right) \]

Question 5

Find the equation of a line \(L_1\) that passes through the point \((3,1)\) and is perpendicular to the line \(L_2\) with equation

\[ - x + 2y - 6 = 0 \]

Solution

First, find the slope of line \(L_2\) by writing its equation in slope–intercept form.

\[ - x + 2y - 6 = 0 \quad \Rightarrow \quad 2y = x + 6 \] \[ y = \frac{1}{2}x + 3 \] \[ m_2 = \frac{1}{2} \]

Since the lines are perpendicular, the slope \(m_1\) of \(L_1\) satisfies:

\[ m_1 \cdot \frac{1}{2} = -1 \] \[ m_1 = -2 \]

Using point–slope form with point \((3,1)\):

\[ y - 1 = -2(x - 3) \] \[ y = -2x + 7 \]

Question 6

Find the equation of a line \(L_1\) that passes through the point \((-2,-1)\) and is perpendicular to the line \(L_2\) passing through \((0,-2)\) and \((2,4)\).

Solution

First, find the slope of line \(L_2\):

\[ m_2 = \frac{4 - (-2)}{2 - 0} = 3 \]

Since the lines are perpendicular:

\[ m_1 \cdot 3 = -1 \] \[ m_1 = -\frac{1}{3} \]

Using point–slope form with point \((-2,-1)\):

\[ y - (-1) = -\frac{1}{3}(x - (-2)) \] \[ y + 1 = -\frac{1}{3}(x + 2) \] \[ y = -\frac{1}{3}x - \frac{5}{3} \]

More References and Links

Equations of Lines in Different Forms
Slope of a Line
Tutorial on Equations of Lines
Line Problems
Lines Through Two Points – Parallel & Perpendicular