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 `50/sqrt(3)` with the axes.

Options

A. `sqrt(3)x + y-10=0`
B. `sqrt(3)x + y+10=0`
C. `x+sqrt(3)y-10=0`
D. `x-sqrt(3)y-10=0`

Answer

A, B

Explanation

Let p be the length of the perpendicular from the origin on the given line. Then its equation in normal form is
`x "cos " 30^(@) + y "sin " 30^(@) = p`
`"or " sqrt(3) x+y = 2p`
This meets the coordintates axes at `A(2p//sqrt(3), 0)` and B(0, 2p).
Therefore, the area of `DeltaAOB` is
`(1)/(2)((2p)/(sqrt(3)))2p = (2p^(2))/(sqrt(3))`
By hypothesis,
`(2p^(2))/(sqrt(3)) = (50)/(sqrt(3)) "or "p = +-5`
Hence, the line are `sqrt(3)x +y+-10=0.`