Find the inclination (in radians and degrees) of the line passing through the points.

Question

points are (-√3,-1)(0,-2)

Answer 1

Given points are (-√3,-1)(0,-2)

Slope(m) of (x1,y1) and (x2,y2) is = y2 - y1/x2 - x1

Slope(m) = -2 - (-1)/0 - (-√3)

Product of two same sign is positive

Slope(m) = -2 + 1/0 + √3

Slope(m) = -1/√3

Here Tanθ = m

Tanθ = -1/√3

Tanθ = Tan(150)

θ = 150° (or) 150*π/180

θ = 150°(or) 0.73π

So, here the inclination angle is θ = 150°  or 0.73π radians.

Comments

The function tan (θ) is negative in second and fourth quadrant.

tan (θ) = - 1/√3.

In second Quadrant, π/2 ≤ θ ≤ π.

- 1/√3 = tan (- π/6) = tan ( π - π/6) = tan (5π/6).

In fourth Quadrant, 3π/2 ≤ θ ≤ 2π.

- 1/√3 = tan (- π/6) = tan ( 2π - π/6) = tan (11π/6).

The solutions are θ = 5π/6 and θ = 11π/6.

So, the inclination angles are θ = 5π/6 radians or θ = 150°and θ = 11π/6 radians or θ = 330°.