A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is

• A

  $y+\sqrt{3} x+2-3 \sqrt{3}=0$
• Correct answer
  B

  $y-\sqrt{3} x+2+3 \sqrt{3}=0$
• C

  $\sqrt{3} y-x+3+2 \sqrt{3}=0$
• D

  $\sqrt{3} y+x-3+2 \sqrt{3}=0$

Solution

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, $\quad \sqrt{3} x+y=1$ $$ \Rightarrow \quad y=-\sqrt{3} x+1 $$ Then, $\quad \tan \theta=-\sqrt{3}$ $$ \begin{array}{ll} \Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\ \Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\ \text { and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})} \end{array} $$ $$ \begin{array}{ll} \Rightarrow & y+2=0 \\ \text { and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\ \Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3} \end{array} $$ Neglecting, $y+2=0$ as it does not intersect $Y$-axis.

Hence, the correct answer is option: 'B'.