Alice is one the line x = 4 , so the x-coordinate of whatever point Alice is on is 4 .
Let's call the point that Alice is located (4, y) .
By the distance forumula...
the distance between (4, y) and (2, 3) = \(\sqrt{(4-2)^2+(y-3)^2}\)
The problem tells us that the distance between (4, y) and (2, 3) is √40 , so...
\(\sqrt{(4-2)^2+(y-3)^2}=\sqrt{40}\)
Now let's solve this equation for y . First square both sides.
\((4-2)^2+(y-3)^2=40\)
Simplify \((4-2)^2\) to \(4\) .
\(4+(y-3)^2=40\)
Subtract 4 from both sides of the equation.
\((y-3)^2=36\)
Take the ± square root of both sides.
\(y-3=\pm\sqrt{36}\)
\(y-3=\pm6\)
Add 3 to both sides.
\(y=3\pm6\)
\(\begin{array}{ccc} y=3+6&\qquad\text{or}\qquad &y=3-6\\~\\ y=9&\text{or}&y=-3 \end{array}\)
The possible y-coordinates of the point Alice at are 9 and -3 .
The sum of the possible y-coordinates of the point Alice is at = 9 + -3 = 6
To help check the answer, we can see on this graph that (4, 9) and (4, -3) , the two possible points Alice is at, are the same distance away from (2, 3) ...over 2 units and up or down 6 units.