What are the polar coordinates of #(6, -6)#?

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Answer 1 · 2016-01-09T00:20:15

#(6sqrt2,-pi/4)#

Polar coordinates are in the form #(r,theta)#.

#r# is the radius and #theta# is the angle.

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Since #r# is the hypotenuse of a right triangle, use the Pythagorean theorem to say that #r=sqrt(x^2+y^2)#.

To determine #theta#, you can say that #tantheta=y/x#, so #theta=tan^-1(y/x)#.

Thus,

#r=sqrt(6^2+(-6)^2)=sqrt72=6sqrt2#

#theta=tan^-1(6/(-6))=tan^-1(-1)=-pi/4#

Note that the radian angle #-pi/4# will take us to Quadrant #"IV"#, which is where the original point #(6,-6)# is.

The polar point is #(6sqrt2,-pi/4)#.