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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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Verified Answer
The correct answer is:
$(1,1)$
Polar coordinates $\mathrm{z}=\mathrm{a}+\mathrm{ib}$ are $\left(\sqrt{2}, \frac{\pi}{4}\right)$
$\therefore \sqrt{2}=\sqrt{\mathrm{a}^2+\mathrm{b}^2} \Rightarrow \mathrm{a}^2+\mathrm{b}^2=2$ and $\tan \left(\frac{\pi}{4}\right)=\frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=1$
$\Rightarrow \mathrm{a}=\mathrm{b}$
$\therefore \mathrm{a}^2+\mathrm{b}^2=2 \Rightarrow 2 \mathrm{a}^2=2 \Rightarrow \mathrm{a}^2=1 \Rightarrow \mathrm{a}= \pm 1$
Since point lies in $1^{\text {st }}$ quadrant, $a=1 \Rightarrow b=1$
$\therefore$ Cartesian coordinates are $(1,1)$
$\therefore \sqrt{2}=\sqrt{\mathrm{a}^2+\mathrm{b}^2} \Rightarrow \mathrm{a}^2+\mathrm{b}^2=2$ and $\tan \left(\frac{\pi}{4}\right)=\frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=1$
$\Rightarrow \mathrm{a}=\mathrm{b}$
$\therefore \mathrm{a}^2+\mathrm{b}^2=2 \Rightarrow 2 \mathrm{a}^2=2 \Rightarrow \mathrm{a}^2=1 \Rightarrow \mathrm{a}= \pm 1$
Since point lies in $1^{\text {st }}$ quadrant, $a=1 \Rightarrow b=1$
$\therefore$ Cartesian coordinates are $(1,1)$
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