The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line is

Given equation is √3x + y = 1

and θ = 60°

Firstly, we find the slope of the given equation

√3x + y = 1

⇒ y = 1 - √3x

or y = (-√3)x + 1

Since, the above equation is in y = mx + b form.

So, the slope of this equation m₁ = -√3

Let m₂ be the slope of the required line.

Now, we find the value of m₂, by using the formula

Putting the values of m₁ and m₂ in above eq., we get

[∵ tan 60° = √3]

Taking (+) sign, we get

⇒ -√3 – m₂ = √3(1 - √3m₂)

⇒ -√3 – m₂ = √3 - 3m₂

⇒ 3m₂ – m₂ = √3 + √3

⇒ 2m₂ = 2√3

⇒ m₂ = √3

Taking (-) sign, we get

⇒ √3 + m₂ = √3(1 - √3m₂)

⇒ √3 + m₂ = √3 - 3m₂

⇒ 3m₂ + m₂ = 0

⇒ 4m₂ = 0

⇒ m₂ = 0

∴, Equation of line passing through (3, -2) with slope √3 is

y – y₁ = m(x – x₁)

⇒ y + 2 = √3x – 3√3

⇒ √3x – y – 3√3 – 2 = 0

⇒ √3x – y – (3√3 + 2) = 0

and Equation of line passing through (3, -2) with slope 0 is

y – y₁ = m(x – x₁)

⇒y – (-2) = 0 (x – 3)

⇒ y + 2 = 0

Hence, the required equations are √3x – y – (3√3 + 2) = 0 and y + 2 = 0

Hence, the correct option is (a)