Question

Show that the equation of the line passing through the origin and making an angle θ with the line \(y=mx+c\) is \(\frac{y}{x}=\frac{m±tanθ}{1-+mtanθ}\)

Answer 1

Let the equation of the line passing through the origin be \(y = m_1x\). 
If this line makes an angle of θ with line \(y = mx + c\), then angle θ is given by
\(tanθ=\left|\frac{m_1-m}{1+m_1m}\right|\)

\(⇒tanθ=\left|\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right|\)

\(⇒tanθ=±\left(\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right)\)

\(⇒tanθ=\left(\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right) \space or\space tanθ=-\left(\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right)\)
 

Case I:
\(tanθ=\left(\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right)\)

\(⇒ tanθ+\frac{y}{x}m\space tanθ=\frac{y}{x}-m\)

\(⇒ m+tanθ=\frac{y}{x}(1-m \space tanθ)\)

\(⇒ \frac{y}{x}=\frac{m+tanθ}{1-m\space tanθ}\)

Case II:
\(tanθ=-\left(\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right)\)

\(⇒ tanθ+\frac{y}{x}m\space tanθ=-\frac{y}{x}+m\)

\(⇒\frac{ y}{x}(1+m\space tanθ)=m-tanθ\)

\(⇒ \frac{y}{x}=\frac{m-tanθ}{1+m\space tanθ}\)

 Therefore, the required line is given by \(\frac{y}{x}=\frac{m±tanθ}{1-+m\space tanθ}.\)