Find the equation of a straight line on which the perpendicular from the origin makes an angle of `30^0` with x-axis and which forms a triangle of area `50sqrt(3)` with the axes.

Explanation

Let consider `AB` as given line and `OL=p` as the perpendicular drawn

And,`alpha=30^0` and
equation of line AB is given by

`xcosalpha+ysinalpha=p`

`=>xcos30+ysin30=p`

`=>x/2+sqrt(3)/(2y)
=2p`

`=>x+3y=2p............(1)`
...