# Question #14ca3

Oct 15, 2016

${d}^{2} = \frac{5}{9} {x}^{2} + \frac{52}{9} x + \frac{169}{9}$

#### Explanation:

For any $x$, the Cartesian coordinates of the point on the equation $y = \frac{2}{3} x + \frac{13}{3}$
would be $\left(x , \frac{2}{3} x + \frac{13}{3}\right)$

The square of the distance from the origin, $\left(0 , 0\right)$,
using the Pythagorean Theorem, would be
$\textcolor{w h i t e}{\text{XXX}} {d}^{2} = {\left(x - 0\right)}^{2} + {\left(\frac{2}{3} x + \frac{13}{3} - 0\right)}^{2}$

$\textcolor{w h i t e}{\text{XXX}} = {x}^{2} - \left(\frac{4}{9}\right) {x}^{2} + \frac{52}{9} x + \frac{169}{9}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{5}{9} {x}^{2} + \frac{52}{9} x + \frac{169}{9}$