If the polar co-ordinates of a point are $\left(\sqrt{2}, \frac{\pi}{4}\right)$, then its Cartesian co-ordinates are

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Aicharya · Accepted Answer

Polar coordinates $\mathrm{z}=\mathrm{a}+\mathrm{ib}$ are $\left(\sqrt{2}, \frac{\pi}{4}ight)$ $	herefore \sqrt{2}=\sqrt{\mathrm{a}^2+\mathrm{b}^2} \Rightarrow \mathrm{a}^2+\mathrm{b}^2=2$ and $	an \left(\frac{\pi}{4}ight)=\frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=1$ $\Rightarrow \mathrm{a}=\mathrm{b}$ $	herefore \mathrm{a}^2+\mathrm{b}^2=2 \Rightarrow 2 \mathrm{a}^2=2 \Rightarrow \mathrm{a}^2=1 \Rightarrow \mathrm{a}= \pm 1$ Since point lies in $1^{	ext {st }}$ quadrant, $a=1 \Rightarrow b=1$ $	herefore$ Cartesian coordinates are $(1,1)$

If the polar co-ordinates of a point are $\left(\sqrt{2}, \frac{\pi}{4}\right)$, then its Cartesian…

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