Chapter 25: distance from a point to a line

點到直線距離

Theorem 25.1 \[ \begin{array}{c} \begin{cases} P=P\left(x_{0},y_{0}\right)\\ L=L\left(x,y\right)=Ax+By+C=0,A^{2}+B^{2}\ne0 \end{cases}\\ \Downarrow\\ d\left(P,L\right)=\dfrac{\left|Ax_{0}+By_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} \end{array} \]

https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line

https://highscope.ch.ntu.edu.tw/wordpress/?p=47407

https://web.math.sinica.edu.tw/mathmedia/HTMLarticle18.jsp?mID=40312

Proofs:

25.1 by shortest \(\overline{PP^{\prime}}\)

\[ \begin{aligned} & P^{\prime}=P^{\prime}\left(x,y\right)\in L=Ax+By+C=0\\ \Rightarrow & y=\dfrac{-1}{B}\left(Ax+C\right) \end{aligned} \]

\[ \begin{aligned} \overline{PP^{\prime}}^{2}\left(x,y\right)= & \left(x_{0}-x\right)^{2}+\left(y_{0}-y\right)^{2}\\ = & \left(x_{0}-x\right)^{2}+\left(y_{0}-\dfrac{-1}{B}\left(Ax+C\right)\right)^{2}\\ = & \left(x-x_{0}\right)^{2}+\left(\dfrac{A}{B}x+\dfrac{C}{B}+y_{0}\right)^{2}=\overline{PP^{\prime}}^{2}\left(x\right) \end{aligned} \]

\[ \begin{aligned} 0=\dfrac{\partial}{\partial x}\overline{PP^{\prime}}^{2}\left(x\right)= & 2\left(x-x_{0}\right)+2\left(\dfrac{A}{B}x+\dfrac{C}{B}+y_{0}\right)\dfrac{A}{B}\\ = & \dfrac{2}{B^{2}}\left(B^{2}\left(x-x_{0}\right)+A^{2}x+AC+ABy_{0}\right)\\ = & \dfrac{2}{B^{2}}\left[\left(A^{2}+B^{2}\right)x-\left(B^{2}x_{0}-ABy_{0}-AC\right)\right]\\ x= & \dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}} \end{aligned} \] or by completing the square to find \(x\).

\[ \begin{aligned} & \overline{PP^{\prime}}^{2}\left(x=\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}}\right)\\ = & \left(\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}}-x_{0}\right)^{2}+\left(\dfrac{A}{B}\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}}+\dfrac{C}{B}+y_{0}\right)^{2}\\ = & \left(\dfrac{-A^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}}\right)^{2}+\left(\dfrac{A\left(B^{2}x_{0}-ABy_{0}-AC\right)+C\left(A^{2}+B^{2}\right)+B\left(A^{2}+B^{2}\right)y_{0}}{B\left(A^{2}+B^{2}\right)}\right)^{2}\\ = & \left(\dfrac{-A\left(Ax_{0}+By_{0}+C\right)}{A^{2}+B^{2}}\right)^{2}+\left(\dfrac{AB^{2}x_{0}+B^{3}y_{0}+B^{2}C}{B\left(A^{2}+B^{2}\right)}\right)^{2}\\ = & \dfrac{A^{2}\left(Ax_{0}+By_{0}+C\right)^{2}}{\left(A^{2}+B^{2}\right)^{2}}+\dfrac{B^{2}\left(Ax_{0}+By_{0}+C\right)^{2}}{\left(A^{2}+B^{2}\right)^{2}}\\ = & \dfrac{\left(Ax_{0}+By_{0}+C\right)^{2}}{A^{2}+B^{2}}\\ \overline{PP^{\prime}}= & \overline{PP^{\prime}}\left(x=\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}}\right)=\dfrac{\left|Ax_{0}+By_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} \end{aligned} \]

25.2 by perpendicular foot

\[ y=\dfrac{-A}{B}x-\dfrac{C}{B}=\dfrac{-1}{B}\left(Ax+C\right),\text{ if }B\ne0 \]

\[ L_{\perp}:\left(y=\dfrac{B}{A}x+K\right)\perp\left(y=\dfrac{-A}{B}x-\dfrac{C}{B}\right):L \]

\[ L_{\perp}=L_{\perp}\left(x,y\right)=Bx-Ay+K=0 \] \[ P=P\left(x_{0},y_{0}\right)\in L_{\perp}=B\left(x-x_{0}\right)-A\left(y-y_{0}\right)=0 \]

\[ L_{\perp}=Bx-Ay-\left(Bx_{0}-Ay_{0}\right)=0 \] perpendicular foot = foot of the perpendicular \(P^{\prime}\)

\[ \begin{aligned} P^{\prime}\in\left(L_{\perp}\cap L\right)= & \begin{cases} L=Ax+By+C=0\\ L_{\perp}=Bx-Ay-\left(Bx_{0}-Ay_{0}\right)=0 \end{cases}\\ = & \begin{cases} Ax+By=-C\\ Bx-Ay=Bx_{0}-Ay_{0} \end{cases}\\ P^{\prime}=P^{\prime}\left(x,y\right)= & \left(\dfrac{\begin{vmatrix}-C & B\\ Bx_{0}-Ay_{0} & -A \end{vmatrix}}{\begin{vmatrix}A & B\\ B & -A \end{vmatrix}},\dfrac{\begin{vmatrix}A & -C\\ B & Bx_{0}-Ay_{0} \end{vmatrix}}{\begin{vmatrix}A & B\\ B & -A \end{vmatrix}}\right)\\ = & \left(\dfrac{\begin{vmatrix}C & B\\ -Bx_{0}+Ay_{0} & -A \end{vmatrix}}{\begin{vmatrix}A & -B\\ B & A \end{vmatrix}},\dfrac{\begin{vmatrix}A & C\\ B & -Bx_{0}+Ay_{0} \end{vmatrix}}{\begin{vmatrix}A & -B\\ B & A \end{vmatrix}}\right)\\ = & \left(\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}},\dfrac{-ABx_{0}+A^{2}y_{0}-BC}{A^{2}+B^{2}}\right) \end{aligned} \]

\[ \begin{aligned} & d\left(P,L\right)=\overline{PP^{\prime}}\\ = & \left\Vert \left(x_{0},y_{0}\right)-\left(\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}},\dfrac{-ABx_{0}+A^{2}y_{0}-BC}{A^{2}+B^{2}}\right)\right\Vert \\ = & \sqrt{\left(x_{0}-\dfrac{B^{2}x_{0}-ABy_{0}-AC}{A^{2}+B^{2}}\right)^{2}+\left(y_{0}-\dfrac{-ABx_{0}+A^{2}y_{0}-BC}{A^{2}+B^{2}}\right)^{2}}\\ = & \sqrt{\left(\dfrac{A^{2}x_{0}+ABy_{0}+AC}{A^{2}+B^{2}}\right)^{2}+\left(\dfrac{ABx_{0}+B^{2}y_{0}+BC}{A^{2}+B^{2}}\right)^{2}}\\ = & \sqrt{\dfrac{A^{2}\left(Ax_{0}+By_{0}+C\right)^{2}+B^{2}\left(Ax_{0}+By_{0}+C\right)^{2}}{\left(A^{2}+B^{2}\right)^{2}}}=\sqrt{\dfrac{\left(Ax_{0}+By_{0}+C\right)^{2}}{A^{2}+B^{2}}}\\ = & \dfrac{\left|Ax_{0}+By_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} \end{aligned} \]

25.3 by normal vector

\[ \begin{cases} \overset{\rightharpoonup}{n}=\left(A,B\right)\perp L=Ax+By+C=0\\ \overset{\rightharpoonup}{PP^{\prime}}=P^{\prime}-P=\left(x-x_{0},y-y_{0}\right) \end{cases} \]

\(P\)到\(L\)的距離\(d\left(P,L\right)\)即為\(L\)線上一點\(P^{\prime}\)對應之\(\overset{\rightharpoonup}{PP^{\prime}}\)在\(L\)法向量\(\overset{\rightharpoonup}{n}\)方向上的投影長

\[ \begin{aligned} \overset{\rightharpoonup}{PP^{\prime}}\cdot\overset{\rightharpoonup}{n}= & \left\Vert \overset{\rightharpoonup}{PP^{\prime}}\right\Vert \left\Vert \overset{\rightharpoonup}{n}\right\Vert \cos\theta\\ \left|\overset{\rightharpoonup}{PP^{\prime}}\cdot\overset{\rightharpoonup}{n}\right|= & \left\Vert \overset{\rightharpoonup}{PP^{\prime}}\right\Vert \left\Vert \overset{\rightharpoonup}{n}\right\Vert \left|\cos\theta\right|\\ \left\Vert \overset{\rightharpoonup}{PP^{\prime}}\right\Vert \left|\cos\theta\right|= & \left|\overset{\rightharpoonup}{PP^{\prime}}\cdot\widehat{\boldsymbol{n}}\right|=\dfrac{\left|\overset{\rightharpoonup}{PP^{\prime}}\cdot\overset{\rightharpoonup}{n}\right|}{\left\Vert \overset{\rightharpoonup}{n}\right\Vert }=\dfrac{\left|\left(x-x_{0},y-y_{0}\right)\cdot\left(A,B\right)\right|}{\left\Vert \left(A,B\right)\right\Vert }\\ = & \dfrac{\left|A\left(x-x_{0}\right)+B\left(y-y_{0}\right)\right|}{\sqrt{A^{2}+B^{2}}}=\dfrac{\left|-Ax_{0}-By_{0}+Ax+By\right|}{\sqrt{A^{2}+B^{2}}}\\ \begin{subarray}{c} Ax+By+C=0\\ =\\ Ax+By=-C \end{subarray} & \dfrac{\left|-Ax_{0}-By_{0}-C\right|}{\sqrt{A^{2}+B^{2}}}=\dfrac{\left|Ax_{0}+By_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} \end{aligned} \]

PDF LaTeX \usepackage{fdsymbol} to have \overrightharpoon vector; however, there are too many side effects, including ugly mathptmx \(\sum\), …

\usepackage{fdsymbol} % vector over accent, but will use mathptmx
% replace the rather ugly mathptmx \sum operator with the equivalent Computer Modern one
\let\sum\relax
\DeclareSymbolFont{CMlargesymbols}{OMX}{cmex}{m}{n}
\DeclareMathSymbol{\sum}{\mathop}{CMlargesymbols}{"50}

25.4 by Cauchy inequality

\[ \begin{aligned} Ax+By+C= & 0\\ Ax+By= & -C\\ \left(Ax+By\right)-\left(Ax_{0}+By_{0}\right)= & -C-\left(Ax_{0}+By_{0}\right)\\ A\left(x-x_{0}\right)+B\left(y-y_{0}\right)= & -\left(Ax_{0}+By_{0}+C\right) \end{aligned} \]

\[ \begin{aligned} \overline{PP^{\prime}}^{2}= & \left(x_{0}-x\right)^{2}+\left(y_{0}-y\right)^{2}\\ \left[A^{2}+B^{2}\right]\overline{PP^{\prime}}^{2}= & \left[A^{2}+B^{2}\right]\left[\left(x_{0}-x\right)^{2}+\left(y_{0}-y\right)^{2}\right]\\ \ge & \left[A\left(x-x_{0}\right)+B\left(y-y_{0}\right)\right]^{2}\\ = & \left[-\left(Ax_{0}+By_{0}+C\right)\right]^{2}=\left(Ax_{0}+By_{0}+C\right)^{2}\\ \overline{PP^{\prime}}^{2}\ge & \dfrac{\left(Ax_{0}+By_{0}+C\right)^{2}}{A^{2}+B^{2}}\\ \overline{PP^{\prime}}\ge & \dfrac{\left|Ax_{0}+By_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} \end{aligned} \]