How do you find the polar coordinates given #(-2, -2sqrt3)#?

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Answer 1 · 2016-11-17T22:30:41

Please see the explanation.

Polar coordinates are an order pair of #(r, theta)#

The conversion from #(x, y)# to r is:

#r = sqrt(x^2 + y^2)#

The conversion from #(x, y)# to #theta# is more complicated:

If #x > 0 and y >= 0#, then #theta = tan^-1(y/x)" [1]"#

If #x = 0 and y > 0#, then #theta = pi/2" [2]"#

If #x = 0 and y < 0#, then #theta = (3pi)/2" [3]"#

If #x < 0#, then #theta = pi + tan^-1(y/x)" [4]"#

If #x > 0 and y < 0#, then #theta = 2pi + tan^-1(y/x)" [5]"#

For the given point #(-2, -2sqrt(3))#

#r = sqrt((-2)^2 + (-2sqrt(3))^2)#

#r = sqrt(4 + 12)#

#r = sqrt(16)#

#r = 4#

Because the x coordinate is less than zero, we use equation [4]:

#theta = pi + tan^-1((-2sqrt(3))/(-2))#

#theta = pi + pi/3#

#theta = (4pi)/3#

The polar point is #(4, (4pi)/3)#