A straight line L through the point (3, 2) is inclined at an angle 60° to the line √3x + y = 1. If L also intersects the x-axis, then the equation of L is

Let the equation of line L be y - 2 = m(x - 3).
The slope of the line √3x + y = 1 is -√3.
Since line L is inclined at an angle of 60° to the line √3x + y = 1, the angle between the lines is 60°.
Let m1 be the slope of the line √3x + y = 1 and m2 be the slope of line L. Then m1 = -√3.
The angle θ between two lines with slopes m1 and m2 is given by:
tan θ = |(m2 - m1) / (1 + m1m2)|
Here, θ = 60°, m1 = -√3. Therefore,
tan 60° = |(m2 + √3) / (1 - m2√3)|
√3 = |(m2 + √3) / (1 - m2√3)|
√3(1 - m2√3) = ±(m2 + √3)
√3 - 3m2 = m2 + √3 or √3 - 3m2 = -(m2 + √3)
4m2 = 0 or 2m2 = 2√3
m2 = 0 or m2 = √3
If m2 = 0, then the equation of L is y - 2 = 0, which is a horizontal line and does not intersect the x-axis.
If m2 = √3, then the equation of L is y - 2 = √3(x - 3)
y - 2 = √3x - 3√3
y - √3x + 3√3 - 2 = 0
This line intersects the x-axis when y = 0.
-√3x + 3√3 - 2 = 0
√3x = 3√3 - 2
x = 3 - 2/√3
Therefore, the equation of line L is y - √3x + 3√3 - 2 = 0 or y - √3x + 2 + 3√3 = 0.