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 $(2,7/6pi)$

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