# How do you find a possible value for a if the points (7,5), (-9,a) has a distance of d=2sqrt65?

Mar 19, 2017

$a = \left\{3 , 7\right\}$

#### Explanation:

To solve for the value of $a$, we will use the distance formula:

"distance" = sqrt(("change in x")^2 + ("change in y")^2)

The change in $x$ is the difference between x-coordinates:

$7 - \left(- 9\right) = 7 + 9 = 16$

The change in $y$ is the difference between y-coordinates:

$5 - a$

Now, plug these values into the distance formula:

"distance" = sqrt(("change in x")^2 + ("change in y")^2)

$2 \sqrt{65} = \sqrt{{\left(16\right)}^{2} + {\left(5 - a\right)}^{2}}$

Squaring both sides gives:
$260 = {\left(16\right)}^{2} + {\left(5 - a\right)}^{2}$
$260 = 256 + {\left(5 - a\right)}^{2}$

$4 = {\left(5 - a\right)}^{2}$
$\pm 2 = 5 - a$
$a = 5 \pm 2$

Therefore, $a$ could be either 3 or 7.