Find the equation of the straight line which passes through the origin and makes angle `60^0` with the line `x+sqrt(3)y+sqrt(3)=0` .

Answer

x=0, x`-sqrt(3)` y =0

Explanation

The given line is `x+sqrt(3)y+3sqrt(3) = 0.`
`"or " y=(-(1)/(sqrt(3)))x-3`
Therefore, the slope of (1) is `-1//sqrt(3).`
Let the slope of the required line be m.
Also, the angle between these line is `60^(@)`. Therefore,
`" tan " 60^(@) = |(m-(-1//sqrt(3)))/(1+m(1//sqrt(3)))|`
`"or " sqrt(3) = |(sqrt(3)m+1)/(sqrt(3)-m)|`
`"If "(sqrt(3)m+1)/(sqrt(3)-m) =sqrt(3)`
`"or "m=(1)/(sqrt(3))`
Using y=mx +c, the equation of the required line is
`y=(1)/(sqrt(3))x +0`
`"i.e.," x-sqrt(3) y = 0` (as the line passes through the origin, c=0)
`(sqrt(3)m-1)/(sqrt(3)-m) = -sqrt(3)`
`"or " sqrt(3)m+1 =-3+sqrt(3)m`
Therefore, finite m does not exist.
Therefore, the slope of the required line is infinity. Thus, the required line is a vertical line. This line passes through the origin.
Therefore, the equation of the required line is x=0.