# What is the polar form of ( -5,-1 )?

$\left(\sqrt{26} , \arctan \left(\frac{1}{5}\right) - \pi\right)$
Let $A \left(- 5 , - 1\right)$.The polar form will be something like $\left(r , \theta\right)$ with r non-negative and $\theta \in \left[0 , 2 \pi\right]$.
The module will be given by the norm of the vector $O A$ which is $\sqrt{{\left(- 5\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{26}$.
The angle between the $\left(O x\right)$ axis and the vector $O A$ will be given by $\arctan \left(\frac{y}{x}\right) - \pi = \arctan \left(\frac{- 1}{- 5}\right) - \pi = \arctan \left(\frac{1}{5}\right) - \pi$ (we substract $\pi$ because $x < 0$ and $y < 0$, and it will give us the principal measure of the angle ie the angle in ]-pi,pi]).