find the inclination angle

Question

Find the inclination (in radians and degrees) of the line

x+√3y+1=0

Answer 1

Write the line equation in slope-intercept form y = mx + b, m is slope and b is y-intercept.

x +√3y + 1 = 0

Add -1 to each side

x +√3y + 1-1= -1

x +√3y = -1

Add -1 to each side

x +√3y - x = -1-x

√3y = -1 - x

Divide each side by √3

√3y/√3 = -1 - x√3

y =-x-1/√3

Slope = -1/√3

m = -1/√3

Re call m = tanθ, where θ is inclination angle

m = tanθ = -1/√3

θ = -30° or  -0.785radians

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

So, inclination angle is -30°.

Comments

m = tan θ = - 1/√3

θ = tan^{- 1}(- 1/√3) = - 30^o = tan^{- 1}(tan(180^o - 30^o)) = 150^o.

Now in θ = radians

θ = (150 × 3.14)/180

θ = 2.616radians..

Therefore, the inclination angle is 150⁰ (or) 2.616radians.