A line forms a triangle of area `54sqrt(3)s q u a r e` units with the coordinate axes. Find the equation of the line if the perpendicular drawn from the origin to the line makes an angle of `60^0` with the X-axis.

Options

A. `sqrt(3)x+y=18`
B. `x+sqrt(3)y=0`
C. `x+sqrt(3)y=18`
D. none of these

Answer

C

Explanation

Let p be the length of the perpendicular drawn from the origin to the required line. The perpendicular makes an angle of `60^(@)` with the a-axis . Putting `alpha=60^(@)` in `xcos alpha+y sinalpha=p` the equation of the line is `xcos60^(@) +ysin60^(@)=p`
or , `x+sqrt(3)y=2p`
This cuts the coordinate axes at `A(2p,0) and B(0,(2p)/(sqrt(3)))`
It is given that
Area of `DeltaOAB=54sqrt(3)` sq. units.
`implies (1)/(2)(OA)(OB)=54sqrt(3)`
`implies (1)/(2)xx2pxx(2p)/(sqrt(3))=54sqrt(3) implies p^(2)=81 implies p=9`
Hence, the equation of the line is `x+sqrt(3)y=18`