find the inclination angle

Question

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

(3,√3)(6,-2√3)

Answer 1

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

The given points are (3,√3)(6,-2√3)

Slope formula is (y₂-y₁)⁄(x₂-x₁)

Substitute the values in slope formula

m = (-2√3-√3)⁄(6-3)                    ((x₁,y₁)=(3,√3)and(x₂,y₂) = (6,-2√3))

m = -3√3⁄3                                   (Divide:-3√3⁄3 = -√3)

m = -√3

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

Slope = -√3

m = -√3

Re call m = tanθ, where θ is inclination angle

m = tanθ = -√3  

θ = 150⁰ or -600 and 150 × ∏/180

θ = -60° or  150° and 150 × 3.14/180

θ = -60° or  150° and 2.61 (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 150⁰.

 

Comments

m = tan θ = - √3

θ = tan^{- 1}(-√3) = - 60^o = tan^{- 1}(tan(180^o - 60^o)) = 120^o.

Now in θ = radians

θ = (120 × 3.14)/180

θ = 2.093radians..

Therefore, the inclination angle is 120⁰ (or) 2.093radians.