# What is the distance between the following polar coordinates?:  (7,(-2pi)/3), (5,(-pi)/6)

Aug 19, 2017

$\sqrt{74}$ units.

#### Explanation:

It's probably easiest to do this by first converting your polar coordinates into Cartesian coordinates—of course, polar coordinates are non-unique so theoretically they could translate to different Cartesian coordinates. We shall take the obvious set of points, however.

To convert, remember that for a set of polar coordinates $\left[r , \theta\right]$, $r = \sqrt{{x}^{2} + {y}^{2}}$. Secondly, $x = r \cos \theta$ and $y = r \sin \theta$.

For the point $\left[7 , - \frac{2 \pi}{3}\right]$, we get:
$x = 7 \cos \left(- \frac{2 \pi}{3}\right)$ and $y = 7 \sin \left(- \frac{2 \pi}{3}\right)$.

$\cos \left(- \frac{2 \pi}{3}\right) = \cos \left(\frac{2 \pi}{3}\right) = - \cos \left(\frac{\pi}{3}\right) = - \frac{1}{2}$
$\sin \left(- \frac{2 \pi}{3}\right) = - \sin \left(\frac{2 \pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$

With a bit of further calculation:

$x = - \frac{7}{2}$ and $y = \frac{- 7 \sqrt{3}}{2}$

For the point $\left[5 , - \frac{\pi}{6}\right]$, we get:
$x = 5 \cos \left(- \frac{\pi}{6}\right)$ and $y = 5 \sin \left(- \frac{\pi}{6}\right)$.

$\cos \left(- \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.
$\sin \left(- \frac{\pi}{6}\right) = - \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

With a further bit of calculation:

$x = \frac{5 \sqrt{3}}{2}$ and $y = - \frac{5}{2}$.

The next bit is the easiest; simply apply the formula for the distance between two points ${D}_{\text{xy}} = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$ with the two points $\left(- \frac{7}{2} , \frac{- 7 \sqrt{3}}{2}\right)$ and $\left(\frac{5 \sqrt{3}}{2} , - \frac{5}{2}\right)$.

The final answer after inserting those numbers is:

${D}_{\text{xy}} = \sqrt{74}$