Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com