# How do you find the distance between (3,2) and (1,-1)?

Sep 11, 2015

Distance between $\left(3 , 2\right)$ and $\left(1 , - 1\right)$ is$\sqrt{13}$

#### Explanation:

(see image below)

The horizontal distance between $\left(1 , - 1\right)$ and $\left(3 , 2\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \Delta x = \left\mid 3 - 1 \right\mid = 2$

The vertical distance between $\left(- 1 , - 1\right)$ and $\left(3 , 2\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \Delta y = \left\mid \left(- 1\right) - 2 \right\mid = 3$

By the Pythagorean Theorem the distance between $\left(- 1 , 1\right)$ and $\left(3 , 2\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{2}^{2} + {3}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \sqrt{13}$