Question Number 44891 by peter frank last updated on 06/Oct/18
Answered by MrW3 last updated on 07/Oct/18
$${LineL}:\:{ax}+{by}+{c}=\mathrm{0} \\ $$$${Point}\:{P}_{\mathrm{1}} \:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right) \\ $$$${Point}\:{P}\left({u},{v}\right)\:{with}\:{PP}_{\mathrm{1}} \bot{L}. \\ $$$$\frac{{v}−{y}_{\mathrm{1}} }{{u}−{x}_{\mathrm{1}} }=\frac{{b}}{{a}} \\ $$$$\left({u}−{x}_{\mathrm{1}} \right){b}=\left({v}−{y}_{\mathrm{1}} \right){a}={k} \\ $$$${let}\:\lambda={distance}\:{between}\:{P}\:{and}\:{P}_{\mathrm{1}} \\ $$$$\lambda=\sqrt{\left({u}−{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({v}−{y}_{\mathrm{1}} \right)^{\mathrm{2}} }=\sqrt{\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}=\frac{{k}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}} \\ $$$$\Rightarrow{k}=\lambda\frac{{ab}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${u}={x}_{\mathrm{1}} +\frac{{k}}{{b}}={x}_{\mathrm{1}} +\frac{\lambda{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${v}={y}_{\mathrm{1}} +\frac{{k}}{{a}}={y}_{\mathrm{1}} +\frac{\lambda{b}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{dd} \\ $$$${if}\:{P}\:{is}\:{on}\:{the}\:{line}\:{L}, \\ $$$${au}+{bv}+{c}=\mathrm{0} \\ $$$$\Rightarrow{ax}_{\mathrm{1}} +\frac{\lambda{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+{by}_{\mathrm{1}} +\frac{\lambda{b}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+{c}=\mathrm{0} \\ $$$$\Rightarrow{ax}_{\mathrm{1}} +{by}_{\mathrm{1}} +{c}+\lambda\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\lambda=−\frac{{ax}_{\mathrm{1}} +{by}_{\mathrm{1}} +{c}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${distance}\:{from}\:{P}_{\mathrm{1}} \:{to}\:{line}\:{L}=\mid{P}_{\mathrm{1}} {P}\mid=\mid\lambda\mid \\ $$$$=\frac{\mid{ax}_{\mathrm{1}} +{by}_{\mathrm{1}} +{c}\mid}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$
Commented by peter frank last updated on 07/Oct/18
$$\mathrm{perfect}!\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$