# What is the polar form of ( -1,121 )?

Jan 23, 2016

Rectangular form: $\left(- 1 , 121\right) \Leftrightarrow$ Polar form: $\left(\sqrt{14642} , \text{arccos} \left(- \frac{1}{\sqrt{14642}}\right)\right)$

#### Explanation:

Radius is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{x}^{2} + {y}^{2}}$

For the given values $\left(x , y\right) = \left(- 1 , 121\right)$
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(- 1\right)}^{1} + {121}^{2}} = \sqrt{14642}$

The given point is in Quadrant II, so it is convenient to use the cos/arccos" functions.
$\textcolor{w h i t e}{\text{XXX}} \cos \left(\theta\right) = \frac{x}{r}$

color(white)("XXX")rarr theta = "arccos"(x/r)

color(white)("XXXXXX")="arccos" ((-1)/sqrt(14642))

Note: If this had any practical applications we would probably convert the above values into (approximate) values (using a calculator) as
$\textcolor{w h i t e}{\text{XXX")(r,theta)~=(121,1.58 " (radians)}}$