Answer
$\Rightarrow y=\dfrac {\sqrt {3}}{3}x$
Work Step by Step
Lets assume that Equation is
$y=kx+b$
If the line passes origin then $b=0$ and $k=\tan \alpha =\tan 30=\dfrac {\sin 30}{\cos 30}=\dfrac {\dfrac {1}{2}}{\dfrac {\sqrt {3}}{2}}=\dfrac {1}{\sqrt {3}}=\dfrac {\sqrt {3}}{3}$
$\Rightarrow y=\dfrac {\sqrt {3}}{3}x$
