Question

A straight line through the point $(3,-2)$ is inclined at an angle ${60}^{\circ }$ to the line $\sqrt{3}x+y=1$. If it intersects the
$x$-axis, then the equation will be

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

Answer 1

The correct option is B $y-x\sqrt{3}+2+3\sqrt{3}=0$

Slope of line $\sqrt{3}x+y=1$ is $-\sqrt{3}$.
So, if it makes an angle $\theta$ with positive $x$-axis.
Then $\tan \theta =-\sqrt{3}$.
$\Rightarrow \theta ={120}^{\circ }$.
Now, the required line makes either ${180}^{\circ }$ or ${60}^{\circ }$ angle with $+x$-axis.
But, required line is not parallel to $x$-axis because it intersects the $x$-axis.

So, slope of required line is, $\tan {60}^{\circ }=\sqrt{3}$
$\Rightarrow$Required line is $(y+2)=\sqrt{3}(x-3)$ $\Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0$