What is the distance between #(2 ,(5 pi)/6 )# and #(-4 , (7 pi )/4 )#?

1 Answer
Mar 26, 2017

#d = 2 sqrt(5 - sqrt(2) - sqrt(6)) ~~ 2.132#

Explanation:

Polar coordinates: #(r, theta)#

Find location of #(-4, (7pi)/4)#:
Start at the x-axis and go #(7pi)/4 = 315^@#, Since #r = -4#, rotate #180^@#: #(-4, (7pi)/4) = (4, (3pi)/4)#

Convert both to rectangular coordinates #(r cos theta, r sin theta)#

#(2, (5pi)/6) = (2 sin ((5pi)/6), 2 cos ((5pi)/6)) = (-2sqrt(3)/2, 2*1/2) = (-sqrt(3), 1)#

#(4, (3pi)/4) = (4 cos ((3pi)/4), 4 sin ((3pi)/4)) = (4*-sqrt(2)/2, 4*sqrt(2)/2) = (-2sqrt(2), 2sqrt(2))#

Find the distance between the points #d = sqrt((y_2-y_1)^2 + (x_2-x_1)^2)#:

#d = sqrt((2sqrt(2) - 1)^2 + (-2sqrt(2)--sqrt(3))^2)#
#d = sqrt((2sqrt(2) - 1)(2sqrt(2) - 1) + (-2sqrt(2)+sqrt(3)) (-2sqrt(2)+sqrt(3)))#
#d = sqrt(9-4sqrt(2) + 11 - 4sqrt(6) )#
#d = sqrt(20 - 4 sqrt(2) - 4 sqrt(6))#
# d = sqrt(4(5 - sqrt(2) - sqrt(6))#
#d = 2 sqrt(5 - sqrt(2) - sqrt(6)) ~~ 2.132#