# How do you find the distance between (-2,-4) and (-5,-3)?

Aug 10, 2017

$d = \sqrt{10}$ units

#### Explanation:

Use the distance formula in coordinate geometry.

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

Label your points with $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$.

Let's let $\left(- 2 , - 4\right)$ be $\left({x}_{2} , {y}_{2}\right)$, and $\left(- 5 , - 3\right)$ be $\left({x}_{1} , {y}_{1}\right)$. Then we substitute in the values into the equation and solve.

$d = \sqrt{{\left(\left(- 4\right) - \left(- 3\right)\right)}^{2} + {\left(\left(- 2\right) - \left(- 5\right)\right)}^{2}}$

$d = \sqrt{1 + 9}$

$d = \sqrt{10}$

Therefore, the distance is $\sqrt{10}$ units.

Aug 10, 2017

color(magenta)(sqrt 10 or  color(magenta)(+-3.162 units to the nearest 3 decimal places

#### Explanation:

$\therefore {y}_{2} - {y}_{1}$

$\therefore \left(- 4\right) - \left(- 3\right) = - 1 = o p p o s i t e$

$\therefore {x}_{2} - {x}_{1}$

$\therefore \left(- 2\right) - \left(5\right) = 3 = a \mathrm{dj} a c e n t$

Pythagoras:

$\therefore {d}^{2} = {\left(- 1\right)}^{2} + {\left(3\right)}^{2}$

$\therefore {d}^{2} = 1 + 9$

$\therefore {d}^{2} = 10$

:.color(magenta)(d=sqrt 10=+-3.162 to 3 decimal places