Question

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 line $L$ is

• A. $y+\sqrt{3}x+2-3\sqrt{3}=0$
• B. $y-\sqrt{3}x+2+3\sqrt{3}=0$
• C. $\sqrt{3}y-x+3+2\sqrt{3}=0$
• D. $\sqrt{3}y+x-3+2\sqrt{3}=0$

Answer 1

The correct option is B $y-\sqrt{3}x+2+3\sqrt{3}=0$

Let the slope of line $L$ be $m.$
Slope of the line $\sqrt{3}x+y=1$ is $-\sqrt{3}.$
Then, $\tan {60}^{\circ }=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|$
$\Rightarrow \left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|=\sqrt{3}$
$\Rightarrow m=0,\sqrt{3}$
Since, $L$ intersects the $x$-axis,
$m=\sqrt{3}$
Hence, equation of line $L$ is
$y-(-2)=\sqrt{3}(x-3)$
$\Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0$