Mathematics - Straight Lines Question with Solution | TestHub

A straight line passing through $P$ and making an angle of $\alpha =60^{\circ }$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha )$
where, $\sqrt{3}x+y=1$ $$\Rightarrow y=-\sqrt{3}x+1$$ Then, $\tan \theta =-\sqrt{3}$ $$\begin{array}{ll}\Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha }{1\mp \tan \theta \tan \alpha }\\ \Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})}\\ \text{ and }& \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}\end{array}$$ $$\begin{array}{ll}\Rightarrow & y+2=0\\ \text{ and }& \frac{y+2}{x-3}=\frac{-2\sqrt{3}}{1-3}=\sqrt{3}\\ \Rightarrow & y+2=\sqrt{3}x-3\sqrt{3}\end{array}$$ Neglecting, $y+2=0$ as it does not intersect $Y$-axis.