4.6: Distance from a Point to a Line
- Page ID
- 34140
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Suppose we are given a point \(P\) and a line \(\overleftrightarrow{AB}\) as in Figure \(\PageIndex{1}\). We would like to find the shortest line segment that can be drawn from \(P\) to \(\overleftrightarrow{AB}\).

First we will prove a theorem:
In a right triangle, the hypotenuse is larger than either leg. In Figure \(\PageIndex{1}\), \(c>a\) and \(c>b\). (The symbol ">" means "is greater than.")

- Proof
-
By the Pythagorean Theorem,
\(c = \sqrt{a^2 + b^2} > \sqrt{a^2} = a.\)
\(c = \sqrt{a^2 + b^2} > \sqrt{b^2} = b.\)
Now we can give the answer to our question:
The perpendicular is the shortest line segment that can be drawn from a point to a straight line.
In Figure \(\PageIndex{3}\) the shortest line segment from \(P\) to \(\overleftrightarrow{AB}\) is \(PD\). Any other line segment, such as \(PC\), must be longer.

- Proof
-
\(PC\) is the hypotenuse of right triangle \(PCD\). Therefore by Theorem \(\PageIndex{1}\), \(PC > PD\).
We define the distance from a point to a line to be the length of the perpendicular.
Find the distance from \(P\) to \(\overleftrightarrow{AB}\):
Solution
Draw \(PD\) perpendicular to \(\overleftrightarrow{AB}\) (Figure \(\PageIndex{4}\)). \(\triangle PCD\) is a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle.

\(\begin{array} {rcl} {\text{hyp}} & = & {2s} \\ {8} & = & {2(CD)} \\ {4} & = & {CD} \\ {L} & = & {s\sqrt{3}} \\ {PD} & = & {4\sqrt{3}} \end{array}\)
Answer: \(4\sqrt{3}\)
Problems
1 - 6. Find the distance from \(P\) to \(\overleftrightarrow{AB}\):
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6.