Find equation of the line through the point $(0,2)$ making an angle $\frac{2 \pi}{3}$ with the positive $x$-axis. Also, find the equation of line parallel to it and crossing the $y$-axis at a distance of 2 units below the origin.
The slope of the line making an angle $\frac{2 \pi}{3}$ with the positive $x$-axis is $m=\tan \left(\frac{2 \pi}{3}\right)=-\sqrt{3}$
Now, the equation of the line passing through point $(0,2)$ and having a slope $-\sqrt{3}$ is $(y-2)=-\sqrt{3}(x-0)$.
$y-2=-\sqrt{3} x$
i.e., $\sqrt{3} x+y-2=0$
The slope of line parallel to line $\sqrt{3} x+y-2=0$ is $-\sqrt{3}$.
It is given that the line parallel to line $\sqrt{3} x+y-2=0$ crosses the $y$-axis 2 units below the origin i.e., it passes through point ( $0,-2$ ).
Hence, the equation of the line passing through point $(0,-2)$ and having a slope $-\sqrt{3}$ is
$y-(-2)=-\sqrt{3}(x-0)$
$y+2=-\sqrt{3} x$
$\sqrt{3} x+y+2=0$
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