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

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Answer 1 · 2018-01-27T11:59:58

#sqrt(43)#

This problem is more easily done in rectangular coordinates:

#(-6,pi/2) to (0,-6)#
#(1,pi/6) to (sqrt(3)/2, 1/2)#

Now we can just use the distance formula:

#sqrt((sqrt(3)/2 - 0)^2+(1/2-(-6))^2)#

#=sqrt(3/4+(13/2)^2)#

#=sqrt(3/4+169/4)#

#=sqrt(172/4)#

#sqrt(43)#

More generally, if we rewrite the first point as #(6, -pi/2)#, then we can use the Law of Cosines, as shown in this video.

Answer 2 · 2018-01-27T12:37:18

Distance between points is #6.56 # unit .

Polar coordinates of point A is #r_1= -6 ,theta_1=pi/2#

Polar coordinates of point B is #r_2=1, theta_2=pi/6#

Distance between points #A and B# is

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

# D = sqrt(36+1+12*cos (pi/2-pi/6)) ~~ 6.56(2dp) # unit

Distance between points is #6.56 # unit [Ans]