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 $(6sqrt2,-pi/4)$ .

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