MHT CET · Maths · Straight Lines
If the polar co-ordinates of a point are \(\left(\sqrt{2}, \frac{\pi}{4}\right)\), then its Cartesian co-ordinates are
- A \((\sqrt{2}, 2)\)
- B \((1,-1)\)
- C \((2, \sqrt{2})\)
- D \((1,1)\)
Answer & Solution
Correct Answer
(D) \((1,1)\)
Step-by-step Solution
Detailed explanation
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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