What is the polar form of #(-2,3)#?

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Answer 1 · 2016-03-19T04:27:36

#(sqrt13,pi-tan^{-1}(3/2))#

To write in polar form, you need to know

the distance from the point to the origin
the angle the line passing through it and the origin makes with the positive #x# axis.

To solve 1. we use Pythagoras Theorem

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

#= sqrt13#

To solve 2. we first find the quadrant that the point lies in.

#y# is positive while #x# is negative #=># quadrant II

Then we find the basic angle by taking inverse tangent of #|y/x|#.

#alpha = tan^{-1}(|3/{-2}|)#

#= tan^{-1}(3/2)#

The angle that we are looking for would be

#theta = pi-alpha#

#= pi-tan^{-1}(3/2)#

#~~ 2.16#

Therefore, the polar coordinate is #(sqrt13,pi-tan^{-1}(3/2))#.

Note that the answer above is not unique. You can add any integer multiples of #2pi# to #theta# to get other representations of the same point.