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 $\frac{y}{x}=\frac{m±tanθ}{1-+m\space tanθ}.$