Question

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

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

$\sqrt{3}x+y=1⟹m=\tan \theta =-\sqrt{3}⟹$ this line makes an angle of ${120}^{\circ }$ with x-axis.

Since $L$ makes an angle ${60}^{\circ }$ with this line.

$⟹L$ must be either x- axis $($ which is not correct because given $L$ intersect the x-axis$)$ or a line making an angle ${60}^{\circ }$ with the x-axis.

$⟹m=\tan {60}^{\circ }=\sqrt{3}$

$L:y=mx+c$

$L:y=\sqrt{3}x+c$

$L$ passes through $(3,-2)⟹-2=3\sqrt{3}+c⟹c=-2-3\sqrt{3}$

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