`(-sqrt(3), -1), (0, -2)` Find the inclination 'theta' (in radians and degrees) of the line with slope m.

Given are the coordinates `(-sqrt(3),-1),(0,-2)`

Slope (m)=`(y_2-y_1)/(x_2-x_1)`

Plug in the coordinates in the above formula to get the slope(m),

m=`(-2-(-1))/(0-(-sqrt(3)))`

m=`(-2+1)/sqrt(3)`

m=`(-1)/sqrt(3)` 

Inclination `(theta)=arctan((-1)/sqrt(3))`

`theta=-30^@`

`=>theta=360-30=330^@`

Now convert this into radians,

`360^@`  =`2pi` radians

`:.330^@=(2pi)/360*330`

= `(33pi)/18` radians

Given are the coordinates `(-sqrt(3),-1),(0,-2)`

Slope (m)=`(y_2-y_1)/(x_2-x_1)`

Plug in the coordinates in the above formula to get the slope(m),

m=`(-2-(-1))/(0-(-sqrt(3)))`

m=`(-2+1)/sqrt(3)`

m=`(-1)/sqrt(3)` 

Inclination `(theta)=arctan((-1)/sqrt(3))`

`theta=-30^@`

`=>theta=360-30=330^@`

Now convert this into radians,

`360^@`  =`2pi` radians

`:.330^@=(2pi)/360*330`

= `(33pi)/18` radians