# What is the distance between (-4,-11) and (13,-41)?

##### 3 Answers
Mar 26, 2018

Distance$= 34.482 \ldots$

#### Explanation:

Apply Pythagorean theorem, where $d$ is the distance between the two points.

$d = \sqrt{{\left(13 - - 4\right)}^{2} + {\left(- 41 - - 11\right)}^{2}}$
$\textcolor{w h i t e}{d} = \sqrt{{\left(17\right)}^{2} + {\left(- 30\right)}^{2}}$
$\textcolor{w h i t e}{d} = \sqrt{1189}$
$\textcolor{w h i t e}{d} = 34.482 \ldots$

Mar 26, 2018

#### Explanation:

D=sqrt((y_2-y_1)^2+(x_2-x_1)^2

subbing in

$D = \sqrt{{\left(- 41 - \left(- 11\right)\right)}^{2} + {\left(13 - \left(- 4\right)\right)}^{2}}$
$D = \sqrt{900 + 289}$
$D = \sqrt{1189} u n i t s$

Mar 26, 2018

$d = \sqrt{1189}$

#### Explanation:

distance between $A \left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$:

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

in this case : $d = \sqrt{{\left(- 4 - 13\right)}^{2} + {\left(- 11 - \left(- 41\right)\right)}^{2}}$

$d = \sqrt{289 + 900} = \sqrt{1189}$