A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3}…

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

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

Asked in: JEE Advanced 2011 (Paper 1)

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