[tex]\bf \begin{array}{llll}
5&,&-5\\
x&&y
\end{array}\qquad
\begin{cases}
r=\sqrt{x^2+y^2}\\\\
\theta=tan^{-1}\left( \frac{y}{x} \right)
\end{cases}\\\\
-----------------------------\\\\
r=\sqrt{(5)^2+(-5)^2}\implies r\sqrt{50}\implies \boxed{r=5\sqrt{2}}
\\\\\\
\theta=tan^{-1}\left( \frac{-5}{5} \right)\implies 0=tan^{-1}(-1)[/tex]
now.... the tangent is -1 when the "y" and "x" are of different signs
now, that happens in the 2nd and 4th quadrant, however, let's take a look at our terminal point, 5 , -5
the "x" is positive, the "y" is negative, that means, is the one on the 4th quadrant [tex]\bf \theta=\frac{7\pi }{4}[/tex]
thus [tex]\bf \left( 5\sqrt{2}\ ,\ \frac{7\pi }{4} \right)[/tex]
