find the inclination angle

Question

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

(-√3,-1)(0,-2)

Answer 1

The slope - intercept form is y = mx + b, m is slope and b is y-intercept

The given points are(-√3,-1)(0,-2)

Slope formula is (y₂-y₁)/(x₂-x₁)

Substitute the values in slope formula

= (-2-(-1))⁄(0-(- √3)))                         ((x₁,y₁)= (-√3,-1)and(x₂,y₂) = (0,-2))

m = -1⁄  √3                                    (Simplify )

m = -1⁄ √3 

The slope intercept form is y = -1⁄ √3 x + b

Slope = -1⁄ √3

m =  -1⁄ √3

Re call m = tanθ, where θ is inclination angle

m = tanθ =-1⁄ √3                    

θ = 120⁰ or -30⁰and 120 × ∏/180

θ = 120° or  -30° and 120 × 3.14/180

θ = 120° or  -30° and 1.04 (radians)              (Simplify)

Since the slope is positive Tangent sign should be positive, so the line passes through the I and III quadrant.

So, inclination angle is 120⁰.

Comments

m = tan θ = -1⁄ √3.

θ = 150⁰ (or) 150 × ∏/180 radians

θ = 150° (or) 150 × 3.14/180 radians

θ = 150° (or) 2.616radians

Therefore, the inclination angle (θ) is 150° (or) 2.616radians.