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

Answer 1

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

The question is too simple from the diagram, the given line

$\sqrt{3}+y=1$ makes an angle ${120}^{\circ }$ with x-axis and intersects at $(\frac{1}{\sqrt{3}},0)$. A line making an angle ${60}^{\circ }$ with the given line is either x-axis or different from x-axis. By observation it is clear that the straight line $y-\sqrt{3}x+2+3\sqrt{3}$ is the required line.

$\text{Second Solution}$

The equation of the line through $(–3,–2)$ may be written as $y+2=m(x–3)$
which will make ${60}^{\circ }$ with $\sqrt{3}x+y=1$ if
$\Rightarrow \tan {60}^{\circ }=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|$
$\Rightarrow \sqrt{3}=\pm \frac{m+\sqrt{3}}{1-\sqrt{3}m}$
$\Rightarrow m=\sqrt{3}$ or $m=0$

Since the line intersects x-axis also, hence $m\ne 0$ consequently $m=\sqrt{3}$ and the required line is

$y\ne 2\sqrt{3}(x-3)$
$\Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0$