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

Question

5283 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-02T09:44:10

$(- 2, 2)$ $(-2,2)$ in polar form is $(2 \sqrt{2}, \frac{3 π}{4})$ $(2sqrt2,(3pi)/4)$

$(a, b)$ $(a,b)$ in polar form is $(r, θ)$ $(r,theta)$ where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $tan θ = \frac{b}{a}$ $tantheta=b/a$

Hence $(- 2, 2)$ $(-2,2)$ in polar form is $(r, θ)$ $(r,theta)$ where

$r = \sqrt{{(- 2)}^{2} + 2^{2}} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$ $r=sqrt((-2)^2+2^2)=sqrt(4+4)=sqrt8=2sqrt2$ and

$tan θ = \frac{2}{- 2} = - 1 = \frac{3 π}{4}$ $tantheta=2/(-2)=-1=(3pi)/4$

Hence $(- 2, 2)$ $(-2,2)$ in polar form is $(2 \sqrt{2}, \frac{3 π}{4})$ $(2sqrt2,(3pi)/4)$

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