What is the distance between (2 ,(5 pi)/6 ) and (-4 , (7 pi )/4 )?

Mar 26, 2017

$d = 2 \sqrt{5 - \sqrt{2} - \sqrt{6}} \approx 2.132$

Explanation:

Polar coordinates: $\left(r , \theta\right)$

Find location of $\left(- 4 , \frac{7 \pi}{4}\right)$:
Start at the x-axis and go $\frac{7 \pi}{4} = {315}^{\circ}$, Since $r = - 4$, rotate ${180}^{\circ}$: $\left(- 4 , \frac{7 \pi}{4}\right) = \left(4 , \frac{3 \pi}{4}\right)$

Convert both to rectangular coordinates $\left(r \cos \theta , r \sin \theta\right)$

$\left(2 , \frac{5 \pi}{6}\right) = \left(2 \sin \left(\frac{5 \pi}{6}\right) , 2 \cos \left(\frac{5 \pi}{6}\right)\right) = \left(- 2 \frac{\sqrt{3}}{2} , 2 \cdot \frac{1}{2}\right) = \left(- \sqrt{3} , 1\right)$

$\left(4 , \frac{3 \pi}{4}\right) = \left(4 \cos \left(\frac{3 \pi}{4}\right) , 4 \sin \left(\frac{3 \pi}{4}\right)\right) = \left(4 \cdot - \frac{\sqrt{2}}{2} , 4 \cdot \frac{\sqrt{2}}{2}\right) = \left(- 2 \sqrt{2} , 2 \sqrt{2}\right)$

Find the distance between the points $d = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$:

$d = \sqrt{{\left(2 \sqrt{2} - 1\right)}^{2} + {\left(- 2 \sqrt{2} - - \sqrt{3}\right)}^{2}}$
$d = \sqrt{\left(2 \sqrt{2} - 1\right) \left(2 \sqrt{2} - 1\right) + \left(- 2 \sqrt{2} + \sqrt{3}\right) \left(- 2 \sqrt{2} + \sqrt{3}\right)}$
$d = \sqrt{9 - 4 \sqrt{2} + 11 - 4 \sqrt{6}}$
$d = \sqrt{20 - 4 \sqrt{2} - 4 \sqrt{6}}$
 d = sqrt(4(5 - sqrt(2) - sqrt(6))
$d = 2 \sqrt{5 - \sqrt{2} - \sqrt{6}} \approx 2.132$