How do you convert #(-sqrt3,-1)# into polar form?

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Answer 1 · 2017-02-12T07:26:47

The polar coordinates are #(2,7/6pi)#

To convert from cartesian coordinates #(x,y)# into polar coordinates, #(r, theta)# we use the following equations

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

#tan theta=y/x#

#r=sqrt((sqrt3)^2+1^2)#

#=sqrt4=2#

#tan theta=-1/-sqrt3=1/sqrt3#

We are in the 3rd quadrant

#theta=pi+pi/6=7/6pi#

The polar coordinates are #(2,7/6pi)#