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Question Number 44891 by peter frank last updated on 06/Oct/18

Answered by MrW3 last updated on 07/Oct/18

$$LineL:ax+by+c=0$$$$Point{P}_1(x_1,y_1)$$$$PointP(u,v)with{PP}_1\mathrm{\perp }L.$$$$\frac{v-y_1}{u-x_1}=\frac{b}{a}$$$$(u-x_1)b=(v-y_1)a=k$$$$let\lambda =distancebetweenPand{P}_1$$$$\lambda =\sqrt{{(u-x_1)}^2+{(v-y_1)}^2}=\sqrt{\frac{k^2}{b^2}+\frac{k^2}{a^2}}=\frac{k\sqrt{a^2+b^2}}{ab}$$$$\Rightarrow k=\lambda \frac{ab}{\sqrt{a^2+b^2}}$$$$u=x_1+\frac{k}{b}=x_1+\frac{\lambda a}{\sqrt{a^2+b^2}}$$$$v=y_1+\frac{k}{a}=y_1+\frac{\lambda b}{\sqrt{a^2+b^2}}dd$$$$ifPisonthelineL,$$$$au+bv+c=0$$$$\Rightarrow {ax}_1+\frac{\lambda a^2}{\sqrt{a^2+b^2}}+{by}_1+\frac{\lambda b^2}{\sqrt{a^2+b^2}}+c=0$$$$\Rightarrow {ax}_1+{by}_1+c+\lambda \sqrt{a^2+b^2}=0$$$$\Rightarrow \lambda =-\frac{{ax}_1+{by}_1+c}{\sqrt{a^2+b^2}}$$$$distancefrom{P}_{1}tolineL=\mid P_{1}P\mid =\mid \lambda \mid$$$$=\frac{\mid {ax}_1+{by}_1+c\mid }{\sqrt{a^2+b^2}}$$

Commented by peter frank last updated on 07/Oct/18

$$\mathrm{perfect}!\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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