# What is the polar form of (-6,12)?

Feb 12, 2016

$\left(6 \sqrt{5} , 2.04\right)$

#### Explanation:

using formulae that links Cartesian to Polar coordinates.

• r^2 = x^2 + y^2

•theta = tan^-1 (y/x)

hence ${r}^{2} = {\left(- 6\right)}^{2} + {12}^{2} = 36 + 144 = 180$

${r}^{2} = 180 \Rightarrow r = \sqrt{180} = 6 \sqrt{5}$

since (-6 , 12 ) is a point in the 2nd quadrant , care must be taken to

ensure that $\theta \textcolor{b l a c k}{\text{ is in the 2nd quadrant }}$

$\theta = {\tan}^{-} 1 \left(\frac{12}{-} 6\right) = {\tan}^{-} 1 \left(- 2\right) = - 1.1 \textcolor{b l a c k}{\text{ radians }}$

rArr theta = (pi-1.1) ≈ 2.04color(black)(" radians ")