How do you convert $(2, - 1)$ $(2,-1)$ into polar form?

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Answer 1 · 2017-09-12T20:24:22

see below

Use the formulas $r^{2} = x^{2} + y^{2}, tan θ = \frac{y}{x}$ $r^2=x^2+y^2, tan theta=y/x$

$r^{=} \sqrt{2^{2} + {(- 1)}^{2}} = \sqrt{5}$ $r^=sqrt(2^2+(-1)^2)=sqrt5$

$tan θ = - \frac{1}{2}$ $tan theta =-1/2$

$θ = {tan}^{- 1} (- \frac{1}{2}) \approx - 0.46 + π n$ $theta = tan^-1 (-1/2) ~~-0.46+pin$

So one possible solution is $(\sqrt{5}, - 0.46) or (2.24, - 0.46)$ $(sqrt 5, -0.46) or (2.24,-0.46)$

How do you convert (2,-1)(2,−1)(2,-1) into polar form?