Question

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

Answer: (B)

$\sqrt{3}x+y=1$ has slope angle $120^{\circ }$
$\Rightarrow$ Any line with inclination of $60^{\circ }$ with above line has either slope angle $=180\%$ '(Parallel to $x$ -axis, not passing through origin, does not intersect $x$ -axis) or has slope angle $60^{\circ }$ which is required.
$\therefore$ Its equation must be $(y+2)=\tan 60^{\circ }(x-3)$
$\Rightarrow y-\sqrt{3}x+3\sqrt{3}+2=0$