# What is the distance between the following polar coordinates?:  (3,(15pi)/8), (9,(-2pi)/8)

Nov 20, 2017

Distance is $6.33$unit

#### Explanation:

Polar coordinates of point A is ${r}_{1} = 3.0 , {\theta}_{1} = \frac{15 \pi}{8} = {337.5}^{0}$

Polar coordinates of point B is ${r}_{2} = 9 , {\theta}_{2} = - \frac{2 \pi}{8} , {\theta}_{2} = - {45}^{0} = {315}^{0}$

Cartesian coordinates of point A is ${x}_{1} = {r}_{1} \cos {\theta}_{1}$

or ${x}_{1} = 3.0 \cos 337.5 \approx 2.77 , {y}_{1} = {r}_{1} \sin \theta$ or

${y}_{1} = 3.0 \sin 337.5 \approx - 1.15 \therefore$ Cartesian coordinates of

point A is $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left(2.77 , - 1.15\right)$

Cartesian coordinates of point B is ${x}_{2} = {r}_{2} \cos {\theta}_{2}$

or ${x}_{2} = 9.0 \cos 315 = 6.36 , {y}_{2} = {r}_{2} \sin {\theta}_{2}$ or

${y}_{2} = 9.0 \sin 315 = - 6.36$. Cartesian coordinates of

point B is $\left({x}_{2} , {y}_{2}\right) \mathmr{and} \left(6.36 , - 6.36\right)$

Distance between them D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2

$D = \sqrt{{\left(2.77 - 6.36\right)}^{2} + {\left(- 1.15 + 6.36\right)}^{2}} \approx 6.33$unit [Ans]