Question

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 $△PQR$, 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

• A. $y-\sqrt{3}x=0$
• B. $\sqrt{3}y=x$
• C. $y+\sqrt{3}x=2\sqrt{3}$
• D. $y+2x=0$

Answer 1

Answer: (C)

$m=\frac{\sin 60^{\circ }}{\sin \left(30^{\circ }-60^{\circ }\right)}=-\sqrt{3}$
Therefore, the equation of the line is
$y-0=-\sqrt{3}(x-2)$
i.e., $\sqrt{3}x+y=2\sqrt{3}$