A straight line L passes through (3,-2) also inclined at an angle `60^@` to the line `sqrt(3x)+y=1`.If L also intersects the x-axis, then the equation of L is: (A) `y+sqrt(3x)+2-3sqrt3=0` (B) `y-sqrt(3x)+2+3sqrt3=0` (C) `sqrt(3y)-x+3+2sqrt3=0` (D) `sqrt(3y)+x-3+2sqrt3=0`

Options

A. `y+sqrt(3)x +2-3sqrt(3) = 0`
B. `y-sqrt(3)x +2+3sqrt(3) = 0`
C. `sqrt(3)y -x +3 +2sqrt(3) = 0`
D. `sqrt(3)y +x -3 +2sqrt(3) = 0`

Explanation

Let the slope of the required line be m. Then
`|(m+sqrt(3))/(1-sqrt(3)m)| = sqrt(3)`
`therefore m+sqrt(3) = +-(sqrt(3)-3m)`
`therefore m=0 " or " m =sqrt(3)`
Therefore, the equation is
`y+2 = sqrt(3)(x-3)(m ne 0 " as given that line cuts the x-axis")`
`"or " sqrt(3)x-y-(2+3sqrt(3)) = 0`