What is the distance between #(-5 ,( 5 pi)/6 )# and #(3 , ( pi )/2 )#?

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Answer 1 · 2016-01-08T12:21:25

7

by using the distance formula in Polar form :

# d^2= r_1^2 + r_2^2 - (2 r_1 r_2 cos (theta_2 - theta_1 )) #

where # (r_1 , theta_1) , ( r_2, theta_2 ) # are 2 points.

in this case the 2 points are # (r_1 , theta_1 ) = (3 , pi/3 ) #

and # (r_2 , theta_2 ) = ( - 5 , (5pi) /6 ) #

substituting these points into the formula :

# d^2 = (- 5 )^2 + 3^2 - (2 xx (- 5 ) xx3 xx cos( (5pi)/6 - (pi/2) ) #

evaluating : # d^2 = 25 + 9 - ( - 30 cos(pi/3 )) #

# d^2 = 34 - (- 30 xx 1/2 ) # = 34 + 15 = 49

#rArr d^2 = 49 rArr d = sqrt49 = 7#