How do you convert (-5, 6) into polar coordinates?

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Answer 1 · 2015-11-27T02:19:28

Find $r and theta$ using the following ...

$r=sqrt(-5^2+6^2)=sqrt61$

Now, the coordinate is in Quadrant II , so
$90^o < theta < 180^o$

$theta=arctan(6/5)=50.19^o$ , this is called the reference angle

Since we know it lies in Quadrant II ...

$theta = 180-50.19=129.81^o$

Polar Coordinate $=(sqrt61, 129.81^o)$ or $(sqrt61, 0.72pi)$

hope that helped