# How do you find the value of a given the points (-4,1), (a,8) with a distance of sqrt50?

Mar 24, 2018

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 information from the problem and solving for $a$ gives:

$\sqrt{50} = \sqrt{{\left(\textcolor{red}{a} - \textcolor{b l u e}{- 4}\right)}^{2} + {\left(\textcolor{red}{8} - \textcolor{b l u e}{1}\right)}^{2}}$

$\sqrt{50} = \sqrt{{\left(\textcolor{red}{a} + \textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{8} - \textcolor{b l u e}{1}\right)}^{2}}$

$\sqrt{50} = \sqrt{{\left(\textcolor{red}{a} + \textcolor{b l u e}{4}\right)}^{2} + {7}^{2}}$

$\sqrt{50} = \sqrt{{a}^{2} + 8 a + 16 + 49}$

$\sqrt{50} = \sqrt{{a}^{2} + 8 a + 65}$

${\left(\sqrt{50}\right)}^{2} = {\left(\sqrt{{a}^{2} + 8 a + 65}\right)}^{2}$

$50 = {a}^{2} + 8 a + 65$

$50 - \textcolor{red}{50} = {a}^{2} + 8 a + 65 - \textcolor{red}{50}$

$0 = {a}^{2} + 8 a + 15$

$0 = \left(a + 3\right) \left(a + 5\right)$

Solution 1:

$a + 3 = 0$

$a + 3 - \textcolor{red}{3} = 0 - \textcolor{red}{3}$

$a + 0 = - 3$

$a = - 3$

Solution 2:

$a + 5 = 0$

$a + 5 - \textcolor{red}{5} = 0 - \textcolor{red}{5}$

$a + 0 = - 5$

$a = - 5$

The Solution Is:

$\textcolor{red}{a}$ can be either $- 3$ or $- 5$ for the two points to have a distance of $\sqrt{50}$