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Question:

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

√3y - x + 3 + 2√3 = 0

y - √3x + 2 + 3√3 = 0

y + √3x + 2 - √3 = 0

√3y + x - 2√3 = 0

Solution:

Given line is √3x + y = 1
Slope of this line is m2 = -√3
Let the slope of line L be m1.
Angle between the lines is 60°.
Then, tan 60° = |(m1 - m2) / (1 + m1m2)|
√3 = |(m1 + √3) / (1 - √3m1)|
√3(1 - √3m1) = m1 + √3
√3 - 3m1 = m1 + √3
4m1 = 0
m1 = 0
This is not possible since the angle is 60°.
√3(1 - √3m1) = -(m1 + √3)
√3 - 3m1 = -m1 - √3
2m1 = 2√3
m1 = √3
Equation of line L passing through (3, 2) and having slope √3 is
y - 2 = √3(x - 3)
y - 2 = √3x - 3√3
y - √3x + 3√3 - 2 = 0
√3y - 3x + 9 - 2√3 = 0
Comparing with options, none matches. Let's check the other case.
Let θ be the angle between the lines.
Then tan θ = |(m1 + √3) / (1 - m1√3)|
If θ = 60°, then √3 = |(m1 + √3) / (1 - m1√3)|
√3 - 3m1 = m1 + √3 or √3 - 3m1 = -m1 - √3
4m1 = 0 or 2m1 = 2√3
m1 = 0 or m1 = √3
m1 = √3
The equation of line is y - 2 = √3(x - 3)
y - 2 = √3x - 3√3
y - √3x + 3√3 - 2 = 0
This matches none of the options.