First slide

Equation of line in Straight lines

Question

The polar coordinates equivalent to $(- 3, \sqrt{3})$ are

Moderate

A

$(2 \sqrt{3}, \frac{π}{6})$
B

$(- 2 \sqrt{3}, \frac{5 π}{6})$
C

$(2 \sqrt{3}, \frac{7 π}{6})$
D

$(2 \sqrt{3}, \frac{5 π}{6})$

Solution

$r = \sqrt{x^{2} + y^{2}} = \sqrt{9 + 3} = 2 \sqrt{3}$
$α = \tan^{- 1} \frac{| y |}{| x |}$
$α = \tan^{- 1} (\frac{\sqrt{3}}{3})$
$α = \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$
Point $(- 3, \sqrt{3})$ lies in the second quadrant.
$∴ θ = π - α = π - \frac{π}{6} = \frac{5 π}{6}$
So, polar coordinates are $(2 \sqrt{3}, \frac{5 π}{6})$

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