# find the inclination angle

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

(3,√3)(6,-2√3)

asked Oct 30, 2013 in GEOMETRY

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.

answered Oct 30, 2013

m = tan θ = - √3

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

Now in θ = radians

θ = (120 × 3.14)/180

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