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

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Answer 1 · 2017-03-19T04:06:20

$a = {3, 7}$

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:

$7 - (-9) = 7 + 9 = 16$

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

$5 - a$

Now, plug these values into the distance formula:

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

$2sqrt(65) = sqrt((16)^2+(5-a)^2)$

Squaring both sides gives:
$260 = (16)^2+(5-a)^2$
$260 = 256 + (5-a)^2$

$4 = (5-a)^2$
$+-2 = 5-a$
$a = 5 +-2$

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

How do you find a possible value for a if the points (7,5), (-9,a) has a distance of d=2sqrt65d=2sqrt65?