Convert the point (-sqrt3, -sqrt 3) in rectangular coordinates to polar coordinates that satisfies r > 0, -pi < theta leq pi

Question

Convert the point

(- \sqrt{3}, - \sqrt{3})

in rectangular coordinates to polar coordinates that satisfies

r > 0, - π < θ \leq π

Answer 1

We have to convert the rectangular coordinates to polar coordinates. $(- \sqrt{3}, - \sqrt{3})$ as we know that the rectangular coordinates are (x, y) and the polar coordinates are $(r, θ)$ where $r = \sqrt{x^{2} + y^{2}}$ here $r = {\sqrt{(- \sqrt{3})^{2} + (- \sqrt{3})}}^{2}$
$r = \sqrt{3 + 3}$
$r = \sqrt{6}$ also $θ = t a n^{- 1} \frac{x}{y}$ $θ = \tan^{- 1} (\frac{- \sqrt{3}}{- \sqrt{3}})$
$θ = \tan^{- 1} 1$
$θ = \frac{π}{4}$ Therefore The polar coordinates are $(\sqrt{6}, \frac{π}{4})$

Convert the point (-sqrt3, -sqrt 3) in rectangular coordinates to polar coordinates that satisfies r > 0, -pi < theta leq pi

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