# Question #a99ba

Oct 11, 2017

The distance is $\setminus \sqrt{72}$ units on the coordinate plane.

#### Explanation:

You can solve this using the Distance Formula, which is derived from the Pythagorean Theorem.

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

where $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are two points.

$\left({x}_{1} , {y}_{1}\right) = \left(- 4 , 8\right)$

and

$\left({x}_{2} , {y}_{2}\right) = \left(2 , 2\right)$

So, we can write the distance formula as:

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

$d = \setminus \sqrt{{\left(6\right)}^{2} + {\left(- 6\right)}^{2}}$

$d = \setminus \sqrt{36 + 36}$

$d = \setminus \sqrt{72}$

$\setminus \therefore$ the distance between $\left(- 4 , 8\right)$ and $\left(2 , 2\right)$ is $\setminus \sqrt{72}$

Oct 11, 2017

See a solution process below:

#### Explanation:

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2}}$

Substituting the values from the points in the problem gives:

$d = \sqrt{{\left(\textcolor{red}{2} - \textcolor{b l u e}{- 4}\right)}^{2} + {\left(\textcolor{red}{2} - \textcolor{b l u e}{8}\right)}^{2}}$

$d = \sqrt{{\left(\textcolor{red}{2} + \textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{2} - \textcolor{b l u e}{8}\right)}^{2}}$

$d = \sqrt{{6}^{2} + {\left(- 6\right)}^{2}}$

$d = \sqrt{36 + 36}$

$d = \sqrt{36 \cdot 2}$

$d = \sqrt{36} \sqrt{2}$

$d = 6 \sqrt{2}$