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 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 $\alpha ={60}^{\circ }$, is given by
$\frac{y-y_1}{x-x_1}=tan(\theta \pm \alpha )$


$\Rightarrow \sqrt{3}x+y=1$
$\Rightarrow y=-\sqrt{3}x+1$, then $tan\theta =-\sqrt{3}$
$\Rightarrow \frac{y+2}{x-3}=\frac{tan\theta \pm tan\alpha }{1\mp tan\theta tan\alpha }\frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})\left(\sqrt{3}\right)}and\frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})\left(\sqrt{3}\right)}\Rightarrow y+2=0and\frac{y+2}{x-3}=\frac{-2\sqrt{3}}{1-3}=\sqrt{3}y+2=\sqrt{3}x-3\sqrt{3}....whichistherequiredequation$
$(Neglecting,y+2=0,asitdoesnotintersectthex-axis.)$