How do you convert $(-2, 5)$ to polar form?

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Answer 1 · 2017-12-10T13:52:14

$(sqrt(29), pi-tan^(-1)(5/2))$ or $(-sqrt(29), tan^(-1)(-5/2))$

To convert rectangular $(a,b)$ to polar we use the two formulas:

$r=sqrt(a^2+b^2)$ and $hat theta=tan^(1)(|b/a|)$

then we have to "fix" the quadrant for $theta$ .

$r=sqrt((-2)^2+5^2)=sqrt(29)$

$hat theta = tan^(-1)(5/2)$

the point $(-2,5)$ is in QII, so $theta = pi-hat theta$

therefore $theta = pi-tan^(-1)(5/2)$ .

An alternate, but equivalent representation is $(-sqrt(29), tan^(-1)(-5/2))$ , which uses a QIV angle but a negative $r$ to end up in QII.

How do you convert (-2, 5)(-2, 5) to polar form?