What is the distance between #(2 ,(5 pi)/6 )# and #(-1 , (19 pi )/12 )#?

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Answer 1 · 2018-03-18T00:34:51

#d = sqrt{5-2\sqrt(2)} \approx 1.47#

Distance for cartesian coordinate systems in two-dimensions would be given as:

#d = sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}#

In polar coordinates, we have to change into using #r# and #\theta#. We do this by using:

#x = rcos(\theta)#
#y = rsin(\theta)#

I'll spare the steps of deriving and give the result:

#d = \sqrt{r_{1}^{2} + r_{2}^{2} - 2r_{1}r_{2} cos (\theta_{1} -\theta_{2})}#

We are given:
#r_1 = 2#, #\theta_1 = (5\pi)/6#
#r_2 = -1#, #\theta_2 = (19\pi)/12#

The distance is then:

#d = \sqrt{(2)^2+(-1)^2-2(2)(-1) cos((5\pi)/6-(19\pi)/12)}#

#= \sqrt{4+1+4cos((-9\pi)/12)}#

# = \sqrt{5+4 cos((-3\pi)/(4))}#

# => color(green)(d = sqrt{5-2\sqrt(2)} \approx 1.47)#