# What is the distance between (2,-14) and (-1,21)?

Jan 19, 2016

Distance between $= \sqrt{1234} \approx 35.128$ to 3 decimal places

#### Explanation:

This is treated like a triangle where the line between the points is the hypotenuse.

The distance we are after is that of AC

Given :
$\left({x}_{1} , {y}_{1}\right) \to \left(2 , - 14\right)$
$\left({x}_{2} , {y}_{2}\right) \to \left(- 1 , 21\right)$

So by Pythagoras

${\left(A C\right)}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$

${\left(A C\right)}^{2} = {\left(\textcolor{w h i t e}{.} \left(- 1\right) - 2\right)}^{2} + {\left(21 - \left(- 14\right) \textcolor{w h i t e}{.}\right)}^{2}$

${\left(A C\right)}^{2} = {\left(- 3\right)}^{2} + {\left(35\right)}^{2}$

$A C = \sqrt{1234} \approx 35.128$ to 3 decimal places