# What is the distance between (8, -3) and (4, -7)?

Mar 25, 2018

$4 \sqrt{2} \approx 5.66 \text{ to 2 dec. places}$

#### Explanation:

$\text{calculate the distance using the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(8,-3)" and } \left({x}_{2} , {y}_{2}\right) = \left(4 , - 7\right)$

$d = \sqrt{{\left(4 - 8\right)}^{2} + {\left(- 7 + 3\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2} \approx 5.66$

Mar 25, 2018

$\implies d = 4 \sqrt{2}$

#### Explanation:

In 2D cartesian coordinates, distance between two points is given as:

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

So for our problem:

$\implies d = \sqrt{{\left(4 - 8\right)}^{2} + {\left(- 7 - \left(- 3\right)\right)}^{2}}$

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

$\implies d = \sqrt{16 + 16}$

$\implies d = \sqrt{32}$

Now we can simplify by trying to find the largest square number we can pull out of the square root.

$\implies d = \sqrt{2 \times 16}$

$\implies d = \sqrt{2 \times {4}^{2}}$

$\implies d = 4 \sqrt{2}$