Question

Determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°

Answer 1

Answer:

$(3 \sqrt{2} ,135 \degree) \: \: and \: \: (3 \sqrt{2} ,315 \degree)$

Step-by-step explanation:

Polar coordinates are of the form:

$(r,\theta)$

where

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

and

$\theta = \tan^{ - 1} ( \frac{y}{x} )$

The given rectangular coordinates are: (3,-3).

$r = \sqrt{ {3}^{2} + {( - 3)}^{2} }$

$r = \sqrt{9 + 9} = \sqrt{18}$

$r = 3 \sqrt{2}$

$\theta = \tan^{ - 1} ( \frac{ - 3}{3} )$

$\theta = \tan^{ - 1} ( - 1 )$

$\theta =135 \degree \: \: or \: \: 315 \degree$

The two polar coordinates are:

$(3 \sqrt{2} ,135 \degree) \: \: and \: \: (3 \sqrt{2} ,315 \degree)$

Answer 2

Answer:

$\left ( 3\sqrt{2},135^{\circ} \right )\,,\,\left ( 3\sqrt{2},315^{\circ} \right )$

Step-by-step explanation:

Let (x,y) be the rectangular coordinates of the point.

Here, $(x,y)=(3,-3)$

Let polar coordinates be $(r,\theta )$ such that $r=\sqrt{x^2+y^2}\,,\,\theta =\arctan \left ( \frac{y}{x} \right )$

$r=\sqrt{3^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

$\theta =\arctan \left ( \frac{-3}{3} \right )= \arctan (-1)$

We know that tan is negative in first and fourth quadrant, we get

$\theta =\pi-\frac{\pi}{4}=\frac{3\pi}{4}=135^{\circ}\\\theta =2\pi-\frac{\pi}{4}=\frac{7\pi}{4}=315^{\circ}$

So, polar coordinates are $\left ( 3\sqrt{2},135^{\circ} \right )\,,\,\left ( 3\sqrt{2},315^{\circ} \right )$