11.7: Parallel and Perpendicular Lines
- Page ID
- 184026
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definitions and Theorems
Two distinct lines in the same plane are parallel if they do not intersect at any point.
Two lines are perpendicular if they intersect to form a right angle, that is, an angle of \(90^{\circ}\).
Two distinct non-vertical lines with slopes \(m_1\) and \(m_2\) are parallel if and only if their slopes are equal:\[m_1 = m_2.\nonumber\]In addition, any two distinct vertical lines are parallel.
Two non-vertical lines with slopes \(m_1\) and \(m_2\) are perpendicular if and only if the product of their slopes is \(-1\):\[m_1 m_2 = -1 \qquad \text{or, equivalently,} \qquad m_2 = -\dfrac{1}{m_1}.\nonumber\]In this form, the slope of each line is the negative reciprocal of the slope of the other. In addition, any horizontal line is perpendicular to any vertical line.
The product rule \(m_1 m_2 = -1\) requires that both slopes exist. A vertical line has undefined slope, so it cannot be tested with this rule—yet every vertical line is perpendicular to every horizontal line, and every pair of distinct vertical lines is parallel. Whenever a line in a problem is vertical, decide parallelism and perpendicularity from the definitions rather than from the slope formulas.
Examples
Two lines have the slopes given below. In each case, determine whether the lines are parallel, perpendicular, or neither.
- \(m_1 = 4\) and \(m_2 = 4\)
- \(m_1 = 3\) and \(m_2 = -\frac{1}{3}\)
- \(m_1 = 2\) and \(m_2 = -2\)
- Solutions
-
- The slopes are equal, \(m_1 = m_2 = 4\), so the lines are parallel.
- The product of the slopes is \(m_1 m_2 = (3)\left(-\frac{1}{3}\right) = -1\), so the lines are perpendicular.
- The slopes are not equal, and the product is \(m_1 m_2 = (2)(-2) = -4 \neq -1\). The lines are therefore neither parallel nor perpendicular. (Opposite slopes such as \(2\) and \(-2\) are a common trap: they are negatives, but not negative reciprocals.)
Determine whether the lines\[y = \dfrac{3}{4}x + 2 \qquad \text{and} \qquad y = -\dfrac{4}{3}x - 1\nonumber\]are parallel, perpendicular, or neither.
- Solution
- Each equation is in slope-intercept form \(y = mx + b\), so the slope is the coefficient of \(x\). The first line has slope \(m_1 = \frac{3}{4}\) and the second has slope \(m_2 = -\frac{4}{3}\). Their product is\[m_1 m_2 = \left(\dfrac{3}{4}\right)\left(-\dfrac{4}{3}\right) = -1,\nonumber\]so the lines are perpendicular.
Determine whether the lines\[3x - y = 4 \qquad \text{and} \qquad x + 3y = 9\nonumber\]are parallel, perpendicular, or neither.
- Solution
-
Solve each equation for \(y\) to read off its slope. For the first line,\[3x - y = 4 \;\Longrightarrow\; y = 3x - 4,\nonumber\]so \(m_1 = 3\). For the second line,\[x + 3y = 9 \;\Longrightarrow\; 3y = -x + 9 \;\Longrightarrow\; y = -\dfrac{1}{3}x + 3,\nonumber\]so \(m_2 = -\frac{1}{3}\).
The product of the slopes is \(m_1 m_2 = (3)\left(-\frac{1}{3}\right) = -1\), so the lines are perpendicular.
Find the equation of the line that passes through the point \((2,1)\) and is parallel to \(y = -3x + 4\).
- Solution
-
The given line has slope \(-3\). A line parallel to it has the same slope, so the required line also has slope \(m = -3\). Using the point-slope form with the point \((2,1)\),\[y - 1 = -3(x - 2).\nonumber\]Distribute and solve for \(y\):\[y - 1 = -3x + 6 \;\Longrightarrow\; y = -3x + 7.\nonumber\]The line \(y = -3x + 7\) passes through \((2,1)\) and is parallel to the given line.
Find the equation of the line that passes through the point \((4,-3)\) and is perpendicular to \(y = \dfrac{2}{5}x - 1\).
- Solution
-
The given line has slope \(\frac{2}{5}\). The slope of a perpendicular line is the negative reciprocal,\[m = -\dfrac{1}{\,2/5\,} = -\dfrac{5}{2}.\nonumber\]Using the point-slope form with the point \((4,-3)\),\[y - (-3) = -\dfrac{5}{2}(x - 4).\nonumber\]Distribute and solve for \(y\):\[y + 3 = -\dfrac{5}{2}x + 10 \;\Longrightarrow\; y = -\dfrac{5}{2}x + 7.\nonumber\]The line \(y = -\frac{5}{2}x + 7\) passes through \((4,-3)\) and is perpendicular to the given line.
Consider the lines\[ax + 4y = 7 \qquad \text{and} \qquad 2x - 3y = 5,\nonumber\]where \(a\) is a constant. Find the value of \(a\) that makes the lines
- parallel, and
- perpendicular.
- Solutions
-
First express each slope. Solving the first equation for \(y\),\[ax + 4y = 7 \;\Longrightarrow\; y = -\dfrac{a}{4}x + \dfrac{7}{4},\nonumber\]so its slope is \(m_1 = -\frac{a}{4}\). Solving the second equation for \(y\),\[2x - 3y = 5 \;\Longrightarrow\; y = \dfrac{2}{3}x - \dfrac{5}{3},\nonumber\]so its slope is \(m_2 = \frac{2}{3}\).
- The lines are parallel when \(m_1 = m_2\):\[-\dfrac{a}{4} = \dfrac{2}{3} \;\Longrightarrow\; a = -\dfrac{8}{3}.\nonumber\]
- The lines are perpendicular when \(m_1 m_2 = -1\):\[\left(-\dfrac{a}{4}\right)\left(\dfrac{2}{3}\right) = -1 \;\Longrightarrow\; -\dfrac{a}{6} = -1 \;\Longrightarrow\; a = 6.\nonumber\]
Sources
Several parts of this text use modifications from the following sources:
- Wikipedia article: "Slope"
- Wikipedia article: "Parallel (geometry)"
All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.