The given equation of a line is y = mx + c ... (1) Slope of (1) is m Let n be the slope of the required line, then tan θ = $\left|\frac{n-m}{1+nm}\right|$ ⇒ tan θ = ± $\left(\frac{n-m}{1+nm}\right)$ Case (i) : When tan θ = $\frac{n-m}{1+nm}$ Then tanθ + nm tanθ = n – m ⇒ nm tanθ – n = – m – tanθ ⇒ n = $\frac{m+\tan\theta}{1-m\tan\theta}$ ∴ Equation of the required line through the origin is y – 0 = n (x – 0) i.e., y = $\left(\frac{m+\tan\theta}{1-m\tan\theta}\right)x$ ⇒ $\frac{y}{x}$ = $\frac{m+\tan\theta}{1-m\tan\theta}$ ... (2) Case (ii) : When tan θ = − $\left(\frac{n-m}{1+mn}\right)$ Then tanθ + mn tanθ = – n + m ⇒ n(1 + m tanθ) = m – tanθ ⇒ n = $\frac{m-\tan\theta}{1+m\tan\theta}$ Hence, from (2) and (3), equation of line is $\frac{y}{x}$ = $\frac{m\pm\tan\theta}{1\mp m\tan\theta}$. Hence proved.