How do you convert $(0, - 5)$ $(0, -5)$ into polar form?

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Answer 1 · 2017-09-12T20:18:26

see below

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

$r^{=} \sqrt{0^{2} + {(- 5)}^{2}} = \sqrt{25} = 5$ $r^=sqrt(0^2+(-5)^2)=sqrt25 =5$

$tan θ = - \frac{5}{0} = u n d e f$ $tan theta =-5/0=undef$

$θ = - \frac{π}{2} + 2 π n or \frac{3 π}{2} + 2 π n$ $theta=-pi/2+2pin or (3pi)/2+2pin$

So one possible solution is $(5, - \frac{π}{2})$ $(5, -pi/2)$

How do you convert (0, -5)(0,−5)(0, -5) into polar form?