# How do you convert the Cartesian coordinates (-7sqrt3/2,7/2) to polar coordinates?

Oct 18, 2015

$\left(- 7 \frac{\sqrt{3}}{2} , \frac{7}{2}\right)$ [Cartesian]
$\textcolor{w h i t e}{\text{XXXXXXXX}} = \left(- \frac{\pi}{3} , 7\right)$ [Polar]

#### Explanation:

If $\left(\textcolor{red}{x} , \textcolor{b l u e}{y}\right) = \left(\textcolor{red}{- 7 \frac{\sqrt{3}}{2}} , \textcolor{b l u e}{\frac{7}{2}}\right)$

Then for an angle $\theta$ between the positive X-axis and the point $\left(\textcolor{red}{x} , c o 9 l \mathmr{and} \left(b l u e\right) \left(y\right)\right)$ with vertex at the origin,
by definition of $\tan$
$\textcolor{w h i t e}{\text{XXXX}} \tan \left(\theta\right) = \frac{\textcolor{red}{- 7 \frac{\sqrt{3}}{2}}}{\textcolor{b l u e}{\frac{7}{2}}}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = - \sqrt{3}$

This value for $\tan \left(\theta\right)$ is one found in the standard $\frac{\pi}{3} = 6 {O}^{\circ}$ triangle
and tells us that the angle is either $- \frac{\pi}{3}$ or $2 \frac{\pi}{3}$

$\left(\textcolor{red}{- 7 \frac{\sqrt{3}}{2}} , \textcolor{b l u e}{\frac{7}{2}}\right)$ is in Quadrant IV, so $\theta = - \frac{\pi}{3}$

The radius for the polar coordinate is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXXX}} r = \sqrt{{\left(\textcolor{red}{- 7 \frac{\sqrt{3}}{2}}\right)}^{2} + {\left(\textcolor{b l u e}{\frac{7}{2}}\right)}^{2}}$

color(white)("XXXXX")=sqrt((color(red)(3*7^2)+color(blue)(7^2))/4

$\textcolor{w h i t e}{\text{XXXXX}} = \sqrt{\frac{\cancel{4} \cdot {7}^{2}}{\cancel{4}}}$

$\textcolor{w h i t e}{\text{XXXXX}} = 7$

The Polar coordinate $\left(\theta , r\right) = \left(- \frac{\pi}{3} , 7\right)$