If (-sqrt{2}, sqrt{2}) are the Cartesian coor | vedclass

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(B) Given Cartesian coordinates $(x, y) = (-\sqrt{2}, \sqrt{2})$.To find polar coordinates $(r, 	heta)$:$r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{2})^2 +

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$\left(2, \frac{3 \pi}{4}\right)$

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(B) Given Cartesian coordinates $(x, y) = (-\sqrt{2}, \sqrt{2})$.To find polar coordinates $(r, 	heta)$:$r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.Since $x = r \cos 	heta$ and $y = r \sin 	heta$,we have:$\cos 	heta = \frac{-\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$ and $\sin 	heta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Since $\cos 	heta  0$,the point lies in the second quadrant.$	heta = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.Thus,the polar coordinates are $\left(2, \frac{3 \pi}{4}ight)$.

If $(-\sqrt{2}, \sqrt{2})$ are the Cartesian coordinates of a point,then its polar coordinates are $.....$

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