How do you convert $(11,-9)$ into polar coordinates?

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Answer 1 · 2018-05-06T07:33:54

$(sqrt202,tan^-1(-9/11)+2pi) or (14.2,5.60^c)$

$(x,y)->(r,theta);(r,theta)=(sqrt(x^2+y^2),tan^-1(y/x))$

$r=sqrt(x^2+y^2)=sqrt(11^2+(-9)^2)=sqrt(121+81)=sqrt202~~14.2$

$theta=tan^-1(-9/11)$

However, $(11,-9)$ is in quadrant 4, and so we must add $2pi$ to our answer.

$theta=tan^-1(-9/11)+2pi ~~5.60^c$

$(sqrt202,tan^-1(-9/11)+2pi) or (14.2,5.60^c)$

How do you convert (11,-9)(11,-9) into polar coordinates?