Question

The equation of the line through the origin making an angle of ${60}^o$ with the line $x+y\sqrt{3}+3\sqrt{3}=0$ are

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

Answer 1

Answer: (B)

$x-y\sqrt{3}=0$

Let the equation of the line through the origin is $y=mx$.....(1) where $m$ is the slope.

Given the equation of the line is $x+y\sqrt{3}+3\sqrt{3}=0$.....(2) this line has the slope $-\frac{1}{\sqrt{3}}$.

Given the angle between (1) and (2) is ${60}^o$.

Then

$\left|\frac{m+\frac{1}{\sqrt{3}}}{1-\frac{m}{\sqrt{3}}}\right|=\tan {60}^o$

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

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

or, $\sqrt{3}m-1=\pm \sqrt{3}(\sqrt{3}-m)$

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

So the equation of the line is $y=\frac{x}{\sqrt{3}}$ or, $x-y\sqrt{3}$.