点と直線の距離 (直交する直線の式を経由して求める)

直交する直線の式より交点を求め，距離を求める方法

$\mathrm{点}P(x_p,y_p)$と$\mathrm{直}\mathrm{線}ax+by+c=0$の距離の公式を導く． $$\begin{array}{rcl}ax+by+c& =& 0\\ y& =& -\frac{a}{b}x-\frac{c}{b}\end{array}$$ 傾き$\alpha$の直線に直交する直線の傾きは$-\frac{1}{\alpha }$となるので上記の直線の場合，直交する直線の傾きは以下のようになる． $$\begin{array}{rcl}-\frac{1}{\alpha }& =& -\frac{1}{-\frac{a}{b}}\\ & =& \frac{b}{a}\end{array}$$ 上記傾きの直線が$\mathrm{点}P(x_p,y_p)$を通るのでこの直線の式は以下のようになる． $$\begin{array}{rcl}y-y_p& =& \frac{b}{a}(x-x_p)\\ y& =& \frac{b}{a}(x-x_p)+y_p\\ & =& \frac{b}{a}x-\frac{b}{a}{x}_p+y_p\end{array}$$ もとの$\mathrm{直}\mathrm{線}y=-\frac{a}{b}x-\frac{c}{b}$と上記の直交する$\mathrm{直}\mathrm{線}y=\frac{b}{a}x-\frac{b}{a}{x}_p+y_p$との$\mathrm{交}\mathrm{点}Q(x_q,y_q)$を求める． $$\begin{array}{r}\\ -\frac{a}{b}{x}_q-\frac{c}{b}& =& \frac{b}{a}{x}_q-\frac{b}{a}{x}_p+y_p\\ -\frac{a}{b}{x}_q-\frac{b}{a}{x}_q& =& -\frac{b}{a}{x}_p+y_p+\frac{c}{b}\\ (-\frac{a}{b}-\frac{b}{a})x_q& =& -\frac{b}{a}{x}_p+y_p+\frac{c}{b}\\ -\frac{a^2+b^2}{ab}{x}_q& =& -\frac{b}{a}{x}_p+y_p+\frac{c}{b}\\ x_q& =& \frac{-ab}{a^2+b^2}(-\frac{b}{a}{x}_p+y_p+\frac{c}{b})\\ & =& \frac{1}{a^2+b^2}(b^2{x}_p-ab{y}_p-ac)\\ \\ y_q& =& -\frac{a}{b}{x}_q-\frac{c}{b}\\ & =& -\frac{a}{b}\left\{\frac{1}{a^2+b^2}(b^2{x}_p-ab{y}_p-ac)\right\}-\frac{c}{b}\\ & =& \frac{1}{a^2+b^2}(-ab{x}_p+a^2{y}_p+\frac{a^2}{b}c)-\frac{c}{b}\\ & =& \frac{1}{a^2+b^2}\{-ab{x}_p+a^2{y}_p+\frac{a^2}{b}c-(a^2+b^2)\frac{c}{b}\}\\ & =& \frac{1}{a^2+b^2}\{-ab{x}_p+a^2{y}_p+\frac{c}{b}(a^2-a^2-b^2)\}\\ & =& \frac{1}{a^2+b^2}(-ab{x}_p+a^2{y}_p-\frac{b^{2}c}{b})\\ & =& \frac{1}{a^2+b^2}(-ab{x}_p+a^2{y}_p-bc)\end{array}$$ 以上より$\mathrm{点}P(x_p,y_p)$と$\mathrm{点}Q(x_q,y_q)$の距離を求める． $$\begin{array}{r}\\ x_q-x_p& =& \frac{1}{a^2+b^2}(b^2{x}_p-ab{y}_p-ac)-x_p\\ & =& \frac{1}{a^2+b^2}\{b^2{x}_p-ab{y}_p-ac-(a^2+b^2)x_p\}\\ & =& \frac{1}{a^2+b^2}(b^2{x}_p-ab{y}_p-ac-a^2{x}_p-b^2{x}_p)\\ & =& \frac{1}{a^2+b^2}(-a^2{x}_p-ab{y}_p-ac)\\ & =& \frac{-a}{a^2+b^2}(a{x}_p+b{y}_p+c)\\ \\ {(x_q-x_p)}^2& =& {\left\{\frac{-a}{a^2+b^2}(a{x}_p+b{y}_p+c)\right\}}^2\\ & =& {\left(\frac{-a}{a^2+b^2}\right)}^2{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{{(-a)}^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{a^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2\\ \\ y_q-y_p& =& \frac{1}{a^2+b^2}(-ab{x}_p+a^2{y}_p-bc)-y_p\\ & =& \frac{1}{a^2+b^2}\{-ab{x}_p+a^2{y}_p-bc-(a^2+b^2)y_p\}\\ & =& \frac{1}{a^2+b^2}(-ab{x}_p+a^2{y}_p-bc-a^2{y}_p-b^2{y}_p)\\ & =& \frac{1}{a^2+b^2}(-ab{x}_p-b^2{y}_p-bc)\\ & =& \frac{-b}{a^2+b^2}(a{x}_p+b{y}_p+c)\\ \\ {(y_q-y_p)}^2& =& {\left\{\frac{-b}{a^2+b^2}(a{x}_p+b{y}_p+c)\right\}}^2\\ & =& {\left(\frac{-b}{a^2+b^2}\right)}^2{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{{(-b)}^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{b^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2\\ \\ {(x_q-x_p)}^2+{(y_q-y_p)}^2& =& \frac{a^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2+\frac{b^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2\\ & =& \{\frac{a^2}{{(a^2+b^2)}^2}+\frac{b^2}{{(a^2+b^2)}^2}\}{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{a^2+b^2}{{(a^2+b^2)}^2}{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{1}{a^2+b^2}{(a{x}_p+b{y}_p+c)}^2\\ & =& \frac{{(a{x}_p+b{y}_p+c)}^2}{a^2+b^2}\\ \\ \sqrt{{(x_q-x_p)}^2+{(y_q-y_p)}^2}& =& \sqrt{\frac{{(a{x}_p+b{y}_p+c)}^2}{a^2+b^2}}\\ & =& \frac{|a{x}_p+b{y}_p+c|}{\sqrt{a^2+b^2}}\end{array}$$ 与えられる$\mathrm{直}\mathrm{線}$が$\mathrm{直}\mathrm{線}y=ax+b$の形の場合，上記公式の$b=1$で$c$を$b$と表すことになる．よって公式は以下のように変わることになる． $$\begin{array}{rcl}\sqrt{{(x_q-x_p)}^2+{(y_q-y_p)}^2}& =& \frac{|a{x}_p+y_p+b|}{\sqrt{a^2+1^2}}\end{array}$$