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Question-55071




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=0 is:  mx−ly=0 or y=(m/l)x  let tan θ_0 =(m/l)  line 1 is y=tan θ_1 x with θ_1 =θ_0 −(π/4)  line 2 is y=tan θ_2 x with θ_2 =θ_0 +(π/4)  tan θ_1 =(((m/l)−1)/(1+(m/l)))=((m−l)/(m+l))  tan θ_2 =(((m/l)+1)/(1−(m/l)))=((m+l)/(l−m))  line 1:  y=((m−l)/(m+l))x or (l−m)x+(l+m)y=0  line 2:  y=((m+l)/(l−m))x or (l+m)x+(m−l)y=0
$${the}\:{line}\:{through}\:{origin}\:{and}\:{perpendicular}\:{to} \\ $$$${line}\:{lx}+{my}+{n}=\mathrm{0}\:{is}: \\ $$$${mx}−{ly}=\mathrm{0}\:{or}\:{y}=\frac{{m}}{{l}}{x} \\ $$$${let}\:\mathrm{tan}\:\theta_{\mathrm{0}} =\frac{{m}}{{l}} \\ $$$${line}\:\mathrm{1}\:{is}\:{y}=\mathrm{tan}\:\theta_{\mathrm{1}} {x}\:{with}\:\theta_{\mathrm{1}} =\theta_{\mathrm{0}} −\frac{\pi}{\mathrm{4}} \\ $$$${line}\:\mathrm{2}\:{is}\:{y}=\mathrm{tan}\:\theta_{\mathrm{2}} {x}\:{with}\:\theta_{\mathrm{2}} =\theta_{\mathrm{0}} +\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\frac{{m}}{{l}}−\mathrm{1}}{\mathrm{1}+\frac{{m}}{{l}}}=\frac{{m}−{l}}{{m}+{l}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\frac{{m}}{{l}}+\mathrm{1}}{\mathrm{1}−\frac{{m}}{{l}}}=\frac{{m}+{l}}{{l}−{m}} \\ $$$${line}\:\mathrm{1}: \\ $$$${y}=\frac{{m}−{l}}{{m}+{l}}{x}\:{or}\:\left(\boldsymbol{{l}}−\boldsymbol{{m}}\right)\boldsymbol{{x}}+\left(\boldsymbol{{l}}+\boldsymbol{{m}}\right)\boldsymbol{{y}}=\mathrm{0} \\ $$$${line}\:\mathrm{2}: \\ $$$${y}=\frac{{m}+{l}}{{l}−{m}}{x}\:{or}\:\left(\boldsymbol{{l}}+\boldsymbol{{m}}\right)\boldsymbol{{x}}+\left(\boldsymbol{{m}}−\boldsymbol{{l}}\right)\boldsymbol{{y}}=\mathrm{0} \\ $$

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