Question

Find the equation of the straight line, if the perpendicular from the origin makes an angle 120 ° with x axis and length of the perpendicular from the origin is 6 units. ​

Answer 1

If the normal drawn to a straight line from the origin has magnitude $p,$ and it makes an angle $\omega$ with positive x axis, then the equation of the straight line is,

$\longrightarrow x\cos\omega+y\sin\omega=p$

This is called normal form of the equation of the line.

In the question,

• $\omega=120^o$

• $p=6$

Then, the equation of the line is,

$\longrightarrow x\cos\omega+y\sin\omega=p$

$\longrightarrow x\cos120^o+y\sin120^o=6$

$\longrightarrow x\cdot-\dfrac{1}{2}+y\cdot\dfrac{\sqrt3}{2}=6$

$\longrightarrow -\dfrac{x}{2}+\dfrac{y\sqrt3}{2}=6$

$\longrightarrow-x+y\sqrt3=12$

$\longrightarrow\underline{\underline{x-y\sqrt3+12=0}}$