Saturday, March 14, 2009

`(-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

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