How do you find the polar coordinate for (-12, -12)?

Feb 10, 2016

The point (-12,-12) forms a triangle with (0,0) and (-12,0). Now you can use Pythagoras' theorem to find the magnitude and trigonometry to find the angle. You'll find the polar coordinates are: $\left(12 \sqrt{2} , - \frac{3}{4} \pi\right)$

Explanation:

The point (-12,-12) forms a triangle with (0,0) and (-12,0). Now you can use Pythagoras' theorem to find the magnitude (length) and trigonometry to find the angle. You'll find the polar coordinates are: $\left(12 \sqrt{2} , - \frac{3}{4} \pi\right)$

Pythagoras' theorem to find the magnitude:
${a}^{2} + {b}^{2} = {c}^{2}$
${\left(- 12\right)}^{2} + {\left(- 12\right)}^{2} = m a g n i t u {\mathrm{de}}^{2}$
$m a g n i t u \mathrm{de} = \sqrt{288} = 12 \sqrt{2}$

To use trigonometry to find the angle, it's best to use the arctan() function. This is because you can do this using the original values given in the question. For arcsin() or arccos(), you'd have to use the magnitude, which you calculated yourself and could have made a mistake with.

$\tan \left(\theta\right) = \frac{\textrm{o p p o s i t e}}{\textrm{a \mathrm{dj} a c e n t}} = \frac{- 12}{-} 12 = 1$
$\theta = \arctan \left(1\right) = \frac{\pi}{4} \textrm{O R} \frac{- 3 \pi}{4}$

Looking at the geometry of the problem, we can see that the latter of these two possible angles is the correct one. Therefore the polar coordinates are:

$\left(12 \sqrt{2} , \frac{- 3 \pi}{4}\right)$

Best regards,

Rory.