Question

A straight line $L$ through the point $(3,-2)$ is inclined at an angle ${60}^o$ to the line $\sqrt{3}x+y=1$. lf $L$ 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 $y-\sqrt{3}x+2+3\sqrt{3}=0$

Equation of given line is

$\sqrt{3}x+y=1$
or, $y=-\sqrt{3}x+1$
So, slope of this line $=-\sqrt{3}$
Let the slope of the required line be m.

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan {60}^0=\left|\frac{m-(-\sqrt{3})}{1-\sqrt{3}m}\right|$

$\sqrt{3}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|$

or, $\frac{m+\sqrt{3}}{1-\sqrt{3}m}=\pm \sqrt{3}$

Taking (+) sign , $\frac{m+\sqrt{3}}{1-\sqrt{3}m}=\sqrt{3}$

$m+\sqrt{3}=\sqrt{3}-3m$

$m=0$

Taking (-) sign , $\frac{m+\sqrt{3}}{1-\sqrt{3}m}=-\sqrt{3}$

$m+\sqrt{3}=-\sqrt{3}+3m$

$m=\sqrt{3}$

Since, the required line intersects $x-$ axis , so $m$ will be $\sqrt{3}$.

Required line passes through the point $(3,-2)$, its equation is given by
$y-(-2)=\sqrt{3}(x-3)$

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

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