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

Friday, 6 December 2013

$\displaystyle{\left(-\sqrt{{{3}}},-{1}\right)},{\left({0},-{2}\right)}$ Find the inclination 'theta' (in radians and degrees) of the line with slope m.

Given are the coordinates $\displaystyle{\left(-\sqrt{{{3}}},-{1}\right)},{\left({0},-{2}\right)}$

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

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

m= $\displaystyle\frac{{-{2}-{\left(-{1}\right)}}}{{{0}-{\left(-\sqrt{{{3}}}\right)}}}$

m= $\displaystyle\frac{{-{2}+{1}}}{\sqrt{{{3}}}}$

m= $\displaystyle\frac{{-{1}}}{\sqrt{{{3}}}}$

Inclination $\displaystyle{\left(\theta\right)}={\arctan{{\left(\frac{{-{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle\theta=-{{30}}^{\circ}$

$\displaystyle\Rightarrow\theta={360}-{30}={{330}}^{\circ}$

Now convert this into radians,

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

$\displaystyle\therefore{{330}}^{\circ}=\frac{{{2}\pi}}{{360}}\cdot{330}$

= $\displaystyle\frac{{{33}\pi}}{{18}}$ radians

Given are the coordinates $\displaystyle{\left(-\sqrt{{{3}}},-{1}\right)},{\left({0},-{2}\right)}$

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

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

m= $\displaystyle\frac{{-{2}-{\left(-{1}\right)}}}{{{0}-{\left(-\sqrt{{{3}}}\right)}}}$

m= $\displaystyle\frac{{-{2}+{1}}}{\sqrt{{{3}}}}$

m= $\displaystyle\frac{{-{1}}}{\sqrt{{{3}}}}$

Inclination $\displaystyle{\left(\theta\right)}={\arctan{{\left(\frac{{-{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle\theta=-{{30}}^{\circ}$

$\displaystyle\Rightarrow\theta={360}-{30}={{330}}^{\circ}$

Now convert this into radians,

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

$\displaystyle\therefore{{330}}^{\circ}=\frac{{{2}\pi}}{{360}}\cdot{330}$

= $\displaystyle\frac{{{33}\pi}}{{18}}$ radians

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