Question

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

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

A straight line passing through $P$ and making an angle of $\alpha ={60}^o$ is given by
$\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha )$
$\Rightarrow \sqrt{3}x+y=1$ $\Rightarrow y=-\sqrt{3}x+1$
$\tan \theta =-\sqrt{3}$
$\Rightarrow \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha }{1\mp \tan \theta \tan \alpha }$
$\Rightarrow \frac{y+2}{x-3}=\frac{-\sqrt{3}\pm \sqrt{3}}{1-(-\sqrt{3})\left(\sqrt{3}\right)}$
$\Rightarrow y+2=0;\frac{y+2}{x-3}=\frac{-2\sqrt{3}}{1-3}=\sqrt{3}$
$\Rightarrow y+2=\sqrt{3}x-3\sqrt{3}$

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