# How do you find the perimeter of a triangle whose coordinates are (10,2) and (7,-3)?

Jun 4, 2015

Assuming the third coordinate is the origin $\left(0 , 0\right)$

Using the Pythagorean Theorem:

$\textcolor{w h i t e}{\text{XXXX}}$Distance $\left(0 , 0\right)$ to $\left(10 , 2\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt{{10}^{2} + {2}^{2}} = \sqrt{104}$

$\textcolor{w h i t e}{\text{XXXX}}$Distance $\left(0 , 0\right)$ to $\left(7 , - 3\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt{{7}^{2} + {\left(- 3\right)}^{2}} = \sqrt{58}$

$\textcolor{w h i t e}{\text{XXXX}}$Distance $\left(10 , 2\right)$ to $\left(7 , - 3\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\sqrt{{3}^{2} + {5}^{2}} = \sqrt{34}$

Perimeter = $\sqrt{104} + \sqrt{58} + \sqrt{34}$

(perhaps this would have been a prettier answer with a third point other than the origin)