# What is the distance between (4 ,( 7 pi)/6 ) and (3 , ( - pi )/2 )?

Feb 27, 2016

sqrt13

#### Explanation:

(x, y) = (r cos $\theta$, r sin $\theta$ ) are components of the position vector ( r, $\theta$ ).
The radial vectors are ( $- 2$ sqrt3, $- 2$ ) and ( 0, $- 3$ ). The vector between the given points is the difference ($- 2$ sqrt3, 1 ). .
The distance is the length of this vector = sqrt13.

Feb 27, 2016

$3.606$

#### Explanation:

Point $\left(4 , \frac{7 \pi}{6}\right)$ in Cartesian coordinates represents (4cos(7pi)/6), 4sin(7pi)/6))

i.e. $\left(4 \times \left(- \frac{\sqrt{3}}{2}\right) , 4 \times \left(\frac{- 1}{2}\right)\right)$ or $\left(- 2 \sqrt{3} , - 2\right)$

Point $\left(3 , - \frac{\pi}{2}\right)$ in Cartesian coordinates represents $\left(3 \cos \left(- \frac{\pi}{2}\right) , 3 \sin \left(- \frac{\pi}{2}\right)\right)$

i.e. $\left(3 \times 0 , 3 \times \left(- 1\right)\right)$ or $\left(0 , - 3\right)$

Hence distance between $\left(- 2 \sqrt{3} , - 2\right)$ and $\left(0 , - 3\right)$

is sqrt((0-(-2sqrt3))^2+(-3-(-2))^2

= sqrt((2sqrt3)^2+(-1)^2

= $\sqrt{12 + 1}$

= $\sqrt{13}$ = $3.606$