7.3: Perpendicular Lines
- Page ID
- 45197
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Two distinct lines \(l\) and \(q\) are perpendicular, written \(l ⊥ q\), if their intersection form four right angles or angles with measure \(90^{\circ}\). The slopes of the perpendicular lines \(l\) and \(q\) are negative reciprocals. That is,
\[m_l = −\dfrac{1}{m_q} \nonumber \]
and
\[m_q = − \dfrac{1}{m_l} \nonumber \]
Determine if the given lines are perpendicular. The line \(l\) that passes through the points \((0, 1)\) and \((1, 3)\), and the line \(q\) that passes through the points \((−1, 4)\) and \((5, 1)\).
Solution
To determine if the lines are perpendicular, first find their slopes using the slope of the line formula. The slope of line \(l\), \(m_l\), that passes through the points \((0, 1)\) and \((1, 3)\) is,
\(\begin{array}s m_l &= \dfrac{3 − 1}{1 − 0} \\ &= \dfrac{2}{1} \\ &= 2 \end{array}\)
The slope of line \(q\), \(m_q\), that passes through the points \((−1, 4)\) and \((5, 1)\), is
\(\begin{array}s m_q &= \dfrac{1 − 4}{5 − (-1)} \\ &= \dfrac{-3}{6} \\ &= \dfrac{-1}{2} \end{array}\)
Now, lines \(l\) and \(q\) are perpendicular if and only if:
\(m_l = −\dfrac{1}{m_q} \text{ and } m_q = −\dfrac{1}{m_l}\)
\(m_l = 2\) and \(m_q = −\dfrac{1}{m_l} = −\dfrac{1}{2}\). Hence, the slopes of the lines are negative reciprocals so it can be concluded that lines \(l\) and \(q\) are perpendicular lines.
Find the slope of a line perpendicular to line \(l\) that passes through the points \((−3, 0)\) and \((3, 4)\).
Solution
Start by finding the slope of line \(l\) that passes through the points \((−3, 0)\) and \((3, 4)\), using the slope of the line formula. Thus,
\(\begin{array} s m_l &= \dfrac{y_2 − y_1}{x_2 − x_1} \\ &= \dfrac{4 − 0}{3 − (−3)} \\ &= \dfrac{4}{6} \\ &= \dfrac{2}{3} \end{array}\)
Any line perpendicular to line \(l\) must have a slope that is negative reciprocal of its slope. Since \(m_l = \dfrac{2}{3}\) then the slope of the line perpendicular to line \(l\) must be \(m = −\dfrac{3}{2}\)
Determine if the given lines are perpendicular.
- The line \(l\) that passes through the points \((0, 4)\) and \((5, 3)\) and the line \(q\) that passes through the points \((1, 5)\) and \((−1, −5)\).
- The line \(l\) that passes through the points \((−2, −5)\) and \((1, 7)\) and the line \(q\) that passes through the points \((−4, 1)\) and \((−3, −3)\).
Find the slope of a line perpendicular to:
- Line \(l\) that passes through the points \((4, 2)\) and \((−1, −2)\).
- Line \(q\) that passes through the points \((7, −8)\) and \((9, 1)\).