Q66 / 150 · MHT CET 2019 (Shift 2) · Mathematics

If $(- \sqrt{2}, \sqrt{2})$ are cartesian co-ordinates of the point, then its polar co-ordinates are…..

A
$(1, \frac{4 π}{3})$
Correct answer
B
$(2, \frac{3 π}{4})$
C
$(3, \frac{7 π}{4})$
D
$(4, \frac{5 π}{4})$

Solution

We have,

r c o s θ = - \sqrt{2}

and

r s i n θ = \sqrt{2}

∴ r^{2} {c o s}^{2} θ + r^{2} {s i n}^{2} θ = 2 + 2 = 4

\Rightarrow r^{2} ({c o s}^{2} θ + {s i n}^{2} θ) = 4

\Rightarrow r = \pm 2 a n d θ = {t a n}^{- 1} (\frac{\sqrt{2}}{- \sqrt{2}})

θ = {t a n}^{- 1} (- 1) \Rightarrow θ = \frac{3 π}{4}

∴

Required polar co-ordinate is

(r, θ)

= (2, \frac{3 π}{4})

Hence, the correct answer is option: 'B'.

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