# How do you find the possible values for a if the points (a,6), (-6,2) has a distance of 4sqrt10?

Jul 9, 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 for the distance and from the points in the problem gives:

$4 \sqrt{10} = \sqrt{{\left(\textcolor{red}{- 6} - \textcolor{b l u e}{a}\right)}^{2} + {\left(\textcolor{red}{2} - \textcolor{b l u e}{6}\right)}^{2}}$

We can now solve for $a$:

First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

${\left(4 \sqrt{10}\right)}^{2} = {\left(\sqrt{{\left(\textcolor{red}{- 6} - \textcolor{b l u e}{a}\right)}^{2} + {\left(\textcolor{red}{2} - \textcolor{b l u e}{6}\right)}^{2}}\right)}^{2}$

$16 \cdot 10 = {\left(\textcolor{red}{- 6} - \textcolor{b l u e}{a}\right)}^{2} + {\left(\textcolor{red}{2} - \textcolor{b l u e}{6}\right)}^{2}$

$160 = {\left(\textcolor{red}{- 6} - \textcolor{b l u e}{a}\right)}^{2} + {\left(- 4\right)}^{2}$

$160 = {\left(\textcolor{red}{- 6} - \textcolor{b l u e}{a}\right)}^{2} + 16$

$160 = {\left(- 6\right)}^{2} + 6 a + 6 a + {a}^{2} + 16$

$160 = 36 + 12 a + {a}^{2} + 16$

$160 = {a}^{2} + 12 a + 52$

$160 - \textcolor{red}{160} = {a}^{2} + 12 a + 52 - \textcolor{red}{160}$

$0 = {a}^{2} + 12 a - 108$

We can factor this as:

0 = (a + 18)(a - 6)

Now, equate each term on the right side of the equation to $0$ and solve for $a$:

Solution 1)

$a + 18 = 0$

$a + 18 - \textcolor{red}{18} = 0 - \textcolor{red}{18}$

$a + 0 = - 18$

$a = - 18$

Solution 3)

$a - 6 = 0$

$a - 6 + \textcolor{red}{6} = 0 + \textcolor{red}{6}$

$a - 0 = 6$

$a = 6$

The solution is, a could be $- 18$ or $6$

Jul 9, 2017

$a = - 18$
or
$a = 6$

#### Explanation:

Distance between two points, $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$

= sqrt((x_1-x_2)^2+(y_1-y_2)^2#

In this case
$\left(a , 6\right)$ and $\left(- 6 , 2\right)$

$4 \sqrt{10} = \sqrt{{\left[a - \left(- 6\right)\right]}^{2} + {\left(6 - 2\right)}^{2}}$
${\left(4 \sqrt{10}\right)}^{2} = {\left\{\sqrt{{\left(a + 6\right)}^{2} + {\left(4\right)}^{2}}\right\}}^{2}$
$16 \cdot 10 = {\left(a + 6\right)}^{2} + 16$
$160 = {a}^{2} + 6 x + 6 x + 36 + 16$
${a}^{2} + 12 x - 108 = 0$

Carry out factorisation
${a}^{2} + 12 x - 108 = 0$
$\left(a + 18\right) \left(a - 6\right) = 0$

Divide RHS $\left(0\right)$ with $\left(a + 18\right)$ and $\left(a - 6\right)$
You will have
$\left(a + 18\right) = 0$ and $\left(a - 6\right) = 0$ which give you
$a = - 18$ or $a = 6$ respectively