Notes: `(3, sqrt(3)), (6, -2sqrt(3))` Find the inclination 'theta' (in radians and degrees) of the line with slope m.

Tuesday, October 21, 2008

Given coordinates are $\displaystyle{\left({3},\sqrt{{{3}}}\right)},{\left({6},-{2}\sqrt{{{3}}}\right)}$

Slope (m)= $\displaystyle\frac{{{y}_{{2}}-{y}_{{1}}}}{{{x}_{{2}}-{x}_{{1}}}}$

Plug in the given coordinates in the above formula,

$\displaystyle{m}=\frac{{-{2}\sqrt{{{3}}}-\sqrt{{{3}}}}}{{{6}-{3}}}$

$\displaystyle{m}=\frac{{-{3}\sqrt{{{3}}}}}{{{3}}}$

$\displaystyle{m}=-\sqrt{{{3}}}$

So the angle of inclination( $\displaystyle\theta$ )= $\displaystyle{\arctan{{\left(-\sqrt{{{3}}}\right)}}}$

$\displaystyle\theta=-{{60}}^{\circ}$

or $\displaystyle\theta={360}-{60}={{300}}^{\circ}$

Use the conversion $\displaystyle{{360}}^{\circ}$ = $\displaystyle{2}\pi$ radians

$\displaystyle\therefore{{300}}^{\circ}={\left(\frac{{{2}\pi}}{{360}}\right)}\cdot{300}$

$\displaystyle=\frac{{{5}\pi}}{{3}}$

Answer: $\displaystyle{{300}}^{\circ}$

and $\displaystyle\frac{{{5}\pi}}{{3}}$ radians

Notes