Question

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

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

The equations of lines passing through $(3,-2)$ and inclined at ${60}^{\circ }$ to the line $\sqrt{3}x+y=1$ are given by
$y+2=m(x-3)\dots \left(1\right)$
and
Slope of given line $m_1=-\sqrt{3}$
$\tan {60}^{\circ }=\left|\frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})}\right|$ $[\because \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|]$
$\Rightarrow \sqrt{3}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|$
$\Rightarrow \pm \sqrt{3}(1-\sqrt{3}m)=m+\sqrt{3}$
$\Rightarrow \sqrt{3}-3m=m+\sqrt{3}$ or $-\sqrt{3}+3m=m+\sqrt{3}$
$\Rightarrow m=0$ or $m=\sqrt{3}$
given that line intersects the $x-$axis, So $m=\sqrt{3}$
From equation $\left(1\right)$
$y+2=\sqrt{3}(x-3)$
$\Rightarrow y-\sqrt{3}x+3\sqrt{3}+2=0$

Alternative Solution:
Inclination of the given line $\sqrt{3}x+y=1$ is ${120}^{\circ }$


From the diagram, line inclined at ${60}^{\circ }$ to $L_1$ can have inclination ${120}^{\circ }\pm {60}^{\circ }={180}^{\circ }$ (or) ${60}^{\circ }$
$\because$ Line $L$ also intersects $x$ -axis, its inclination $={60}^{\circ }$
$\therefore$ Required line equation is $y+2=\sqrt{3}(x-3)$
$\Rightarrow \sqrt{3}x-y-3\sqrt{3}-2=0$ or
$y-\sqrt{3}x+3\sqrt{3}+2=0$