Question

A straight line L through the poit $(3,-2)$ is inclined at an angle ${60}^0$ 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 $y-\sqrt{3}x+2+3\sqrt{3}=0$

Let the slope of line L be $m$.
Then $\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|=\sqrt{3}$
$\Rightarrow m+\sqrt{3}=\pm (\sqrt{3}-3m)$


$\Rightarrow 4m=0\text{or}2m=2\sqrt{3}\Rightarrow m=0\text{or}m=\sqrt{3}\because \text{L intersects x - axis,}\therefore m=\sqrt{3}\therefore \text{Equation of L is}y+2=\sqrt{3}(x-3)\text{or}\sqrt{3}x-y-(2+3\sqrt{3})=0$