# How do you find the possible values for a if the points (5,8), (a,2) has a distance of 3sqrt5?

Apr 16, 2017

Use the distance formula and solve for $a$:
$\left\{a : a = 2 , 8\right\}$

#### Explanation:

The distance between the two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given as $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$.

For this problem, it is given that $\left({x}_{1} , {y}_{1}\right) = \left(5 , 8\right)$, $\left({x}_{2} , {y}_{2}\right) = \left(a , 2\right)$, and $d = 3 \sqrt{5}$. We need to solve for $a$ by plugging these values into the distance formula.
$3 \sqrt{5} = \sqrt{{\left(a - 5\right)}^{2} + {\left(2 - 8\right)}^{2}}$.

To solve for $a$, begin by squaring both sides of the equation and simplify a bit.
${\left(3 \sqrt{5}\right)}^{\textcolor{g r e e n}{2}} = {\sqrt{{\left(a - 5\right)}^{2} + {\left(2 - 8\right)}^{2}}}^{\textcolor{g r e e n}{2}}$
${3}^{2} {\sqrt{5}}^{2} = {\sqrt{{\left(a - 5\right)}^{2} + {\left(2 - 8\right)}^{2}}}^{2}$
$9 \cdot 5 = {\left(a - 5\right)}^{2} + {\left(2 - 8\right)}^{2}$
$45 = {\left(a - 5\right)}^{2} + {\left(\text{-} 6\right)}^{2}$
$45 = {a}^{2} - 10 a + 25 + 36$
$45 = {a}^{2} - 10 a + 61$

Now subtract 45 from both sides to get an equation in quadratic form.
$45 \textcolor{g r e e n}{- 45} = {a}^{2} - 10 a + 61 \textcolor{g r e e n}{- 45}$
$0 = {a}^{2} - 10 a + 16$

Finally, solve for $a$ using either the quadratic formula or the factoring method. I'll be factoring.
$0 = {a}^{2} - 2 a - 8 a + 16$
$0 = a \left(a - 2\right) - 8 \left(a - 2\right)$
$0 = \left(a - 8\right) \left(a - 2\right)$
$a = 8 \mathmr{and} 2$

Therefore the possible values for $a$ are defined by the set $\left\{a : a = 2 , 8\right\}$