# What is the distance between (-6 , pi/2 ) and (1 , pi/6 )?

Jan 27, 2018

$\sqrt{43}$

#### Explanation:

This problem is more easily done in rectangular coordinates:

$\left(- 6 , \frac{\pi}{2}\right) \to \left(0 , - 6\right)$
$\left(1 , \frac{\pi}{6}\right) \to \left(\frac{\sqrt{3}}{2} , \frac{1}{2}\right)$

Now we can just use the distance formula:

$\sqrt{{\left(\frac{\sqrt{3}}{2} - 0\right)}^{2} + {\left(\frac{1}{2} - \left(- 6\right)\right)}^{2}}$

$= \sqrt{\frac{3}{4} + {\left(\frac{13}{2}\right)}^{2}}$

$= \sqrt{\frac{3}{4} + \frac{169}{4}}$

$= \sqrt{\frac{172}{4}}$

$\sqrt{43}$

More generally, if we rewrite the first point as $\left(6 , - \frac{\pi}{2}\right)$, then we can use the Law of Cosines, as shown in this video.

Jan 27, 2018

Distance between points is $6.56$ unit .

#### Explanation:

Polar coordinates of point A is ${r}_{1} = - 6 , {\theta}_{1} = \frac{\pi}{2}$

Polar coordinates of point B is ${r}_{2} = 1 , {\theta}_{2} = \frac{\pi}{6}$

Distance between points $A \mathmr{and} B$ is

D= sqrt(r_1^2+r_2^2-2r_1r_2cos(theta1-theta2) or

$D = \sqrt{36 + 1 + 12 \cdot \cos \left(\frac{\pi}{2} - \frac{\pi}{6}\right)} \approx 6.56 \left(2 \mathrm{dp}\right)$ unit

Distance between points is $6.56$ unit [Ans]