How do you convert (6, -6) into polar coordinates?

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How do you convert (6, -6) into polar coordinates?

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Answer 1 · 2016-01-31T12:13:32

The point that has coordinates $(6, - 6)$ $(6, -6)$ in rectangular coordinates has the polar coordinates $(\sqrt{72}, - \frac{π}{4})$ $(sqrt72, -pi/4)$ or $(8.5, - 0.79)$ $(8.5, -0.79)$ or (to give a positive value to $θ$ $theta$ ) as $(\sqrt{72}, \frac{7 π}{4})$ $(sqrt72, (7pi)/4)$ .

Explanation:

Polar coordinates are in the form $(r, θ)$ $(r,theta)$ where $r$ $r$ is the distance from the origin $(0, 0)$ $(0, 0)$ to the point and $θ$ $theta$ is the angle in radians from the positive x-axis.

To find the radius, use:

$r = \sqrt{6^{2} + {(- 6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 8.5$ $r=sqrt(6^2+(-6)^2) = sqrt(36+36) = sqrt72 = 8.5$

(some may prefer to leave it in the form $\sqrt{72}$ $sqrt72$ )

To find the value of $θ$ $theta$ , know that $6$ $6$ is the opposite and $- 6$ $-6$ is the adjacent side of a right-angled triangle, so:

$tan θ = \frac{6}{- 6} = - 1$ $tan theta = 6/-6 = -1$

Therefore $θ = {tan}^{- 1} (- 1) = - \frac{π}{4}$ $theta=tan^-1(-1) = -pi/4$ $r a d$ $rad$ .

This means the polar coordinates can be expressed as $(\sqrt{72}, - \frac{π}{4})$ $(sqrt72, -pi/4)$ or $(8.5, - 0.79)$ $(8.5, -0.79)$ or (to give a positive value to $θ$ $theta$ ) as $(\sqrt{72}, \frac{7 π}{4})$ $(sqrt72, (7pi)/4)$ .

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How do you convert (6, -6) into polar coordinates?

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