We have to convert the rectangular coordinates to polar coordinates.\((-\sqrt{3}, -\sqrt{3})\)
as we know that the rectangular coordinates are (x, y)
and the polar coordinates are \((r, \theta)\)
where
\(r = \sqrt{x^2 + y^2}\)
here
\(r = \sqrt{(-sqrt3)^2 + (-sqrt3)}^2\)
\(r = \sqrt{3 + 3}\)
\(r = \sqrt6\) also \(\theta = tan^{-1} \frac{x}{y}\) \(\theta = tan^{-1} \frac{(-sqrt3}{-sqrt3})\)
\(\theta = tan^{-1} 1\)
\(\theta = \frac{\pi}{4}\) Therefore The polar coordinates are (\sqrt6, \frac{\pi}{4})
\(r = \sqrt{3 + 3}\)
\(r = \sqrt6\) also \(\theta = tan^{-1} \frac{x}{y}\) \(\theta = tan^{-1} \frac{(-sqrt3}{-sqrt3})\)
\(\theta = tan^{-1} 1\)
\(\theta = \frac{\pi}{4}\) Therefore The polar coordinates are (\sqrt6, \frac{\pi}{4})
