Question Number 55071 by peter frank last updated on 17/Feb/19
Answered by mr W last updated on 17/Feb/19
$${the}\:{line}\:{through}\:{origin}\:{and}\:{perpendicular}\:{to} \\ $$$${line}\:{lx}+{my}+{n}=\mathrm{0}\:{is}: \\ $$$${mx}−{ly}=\mathrm{0}\:{or}\:{y}=\frac{{m}}{{l}}{x} \\ $$$${let}\:\mathrm{tan}\:\theta_{\mathrm{0}} =\frac{{m}}{{l}} \\ $$$${line}\:\mathrm{1}\:{is}\:{y}=\mathrm{tan}\:\theta_{\mathrm{1}} {x}\:{with}\:\theta_{\mathrm{1}} =\theta_{\mathrm{0}} −\frac{\pi}{\mathrm{4}} \\ $$$${line}\:\mathrm{2}\:{is}\:{y}=\mathrm{tan}\:\theta_{\mathrm{2}} {x}\:{with}\:\theta_{\mathrm{2}} =\theta_{\mathrm{0}} +\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\frac{{m}}{{l}}−\mathrm{1}}{\mathrm{1}+\frac{{m}}{{l}}}=\frac{{m}−{l}}{{m}+{l}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\frac{{m}}{{l}}+\mathrm{1}}{\mathrm{1}−\frac{{m}}{{l}}}=\frac{{m}+{l}}{{l}−{m}} \\ $$$${line}\:\mathrm{1}: \\ $$$${y}=\frac{{m}−{l}}{{m}+{l}}{x}\:{or}\:\left(\boldsymbol{{l}}−\boldsymbol{{m}}\right)\boldsymbol{{x}}+\left(\boldsymbol{{l}}+\boldsymbol{{m}}\right)\boldsymbol{{y}}=\mathrm{0} \\ $$$${line}\:\mathrm{2}: \\ $$$${y}=\frac{{m}+{l}}{{l}−{m}}{x}\:{or}\:\left(\boldsymbol{{l}}+\boldsymbol{{m}}\right)\boldsymbol{{x}}+\left(\boldsymbol{{m}}−\boldsymbol{{l}}\right)\boldsymbol{{y}}=\mathrm{0} \\ $$