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) = √(4−2)2+(y−3)2
The problem tells us that the distance between (4, y) and (2, 3) is √40 , so...
√(4−2)2+(y−3)2=√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=±√36
y−3=±6
Add 3 to both sides.
y=3±6
y=3+6ory=3−6 y=9ory=−3
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.