Question

The polar co-ordinates of a point (–$\sqrt{2}$,$\sqrt{2}$) is-

• A. $\left( 2\sqrt{2},\frac{–\pi}{4} \right)$
• B. $\left( 2\sqrt{2},\frac{3\pi}{4} \right)$
• C. $\left( –2\sqrt{2},\frac{\pi}{4} \right)$
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

(–$\sqrt{2}$,$\sqrt{2}$)

r = $\sqrt{x^{2} + y^{2}}$ = $\sqrt{4 + 4}$ = 2$\sqrt{2}$

Now a = tan^–1 $\frac{|y|}{|x|}$

a = tan^–1 $\left( \frac{\sqrt{2}}{\sqrt{2}} \right)$ = $\frac{\pi}{4}$

Point lies in II quadrant

\ q = p – a = p – $\frac{\pi}{4}$ = $\frac{3\pi}{4}$

Ž$\left( 2\sqrt{2},\frac{3\pi}{4} \right)$ or $\left( –2\sqrt{2},\frac{–\pi}{4} \right)$