Q.
Let us consider the situation when the axes are inclined at an angle $\omega$.
If the coordinates of a point $P$ are $\left( x _{1}, y _{1}\right)$ then $PN = x _{1}, PM = y _{1}$, where $PM$ is parallel to the $y$-axis and PN is parallel to the $x$-axis. The straight line through $P$ that makes an angle $\theta$ with the $x$-axis is
$RQ = y - y _{1}, PQ = x - x _{1}$
From $\triangle PQR$, we have
$\frac{ PQ }{\sin (\omega-\theta)}=\frac{ RQ }{\sin \theta}$
or $y-y_{1}=\frac{\sin \theta}{\sin (\omega-\theta)}\left(x-x_{1}\right)$
written in the form of $y-y_{1}=m\left(x-x_{1}\right)$ where
$m =\frac{\sin \theta}{\sin (\omega-\theta)}$
Therefore, if the slope of the line is $m$,
then the angle of inclination of the line with the $x$-axis is given by
$\tan \theta=\left(\frac{ m \sin \omega}{1+ m \cos \omega}\right)$
The axes being inclined at an angle of $30^{\circ}$, the equation of the straight line which makes an angle of $60^{\circ}$ with the positive direction of the $x$-axis and $x$-intercept 2 is
Straight Lines
Solution: