Question

Find the equation of the straight line which passes through the origin and making angle ${60}^{\circ }$ with the line $x+\sqrt{3}y+3\sqrt{3}=0$

Answer 1

Given line is $x+\sqrt{3}y+3\sqrt{3}=0$
$\Rightarrow y=(-\frac{1}{\sqrt{3}})x-3$
$\therefore$ Slope of (1) = $-\frac{1}{\sqrt{3}}$
Let slope of the required line be m Also between there lines is given to be ${60}^{\circ }$
$\Rightarrow \tan {60}^{\circ }=\left|\frac{m-(-1/\sqrt{3})}{1+m(-1/\sqrt{3})}\right|\Rightarrow \sqrt{3}=\left|\frac{\sqrt{3}m+1}{\sqrt{3}-m}\right|\Rightarrow \frac{\sqrt{3}m+1}{\sqrt{3}-m}=\pm \sqrt{3}$
$\frac{\sqrt{3}m+1}{\sqrt{3}-m}=\sqrt{3}$
$\Rightarrow \sqrt{3}m+1=3-\sqrt{3}m$
$\Rightarrow m=\frac{1}{\sqrt{3}}$
Using y = mx + c the equation of the required line is $y=\frac{1}{\sqrt{3}}x+0$
i.e. $x-\sqrt{3}y=0$ ($\because$ This passes through origin so c = 0)
$\frac{\sqrt{3}m+1}{\sqrt{3}-m}=-\sqrt{3}$
$\Rightarrow \sqrt{3}m+1=-3+\sqrt{3}m$
$\Rightarrow$ m is not defined
$\therefore$ The slope of the required line is not defined Thus the required line is a vertical line This line is to pass through the origin
$\therefore$ The equation of the required line is x = 0