Alice is on the line x=4, and she is √40 away from (2,3). What is the sum of the possible y-coordinates of the point Alice is at?

Please help! I think the answer is 0 but I'm not sure.

Guest Aug 19, 2018

edited by
Guest
Aug 19, 2018

#1**+1 **

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.

hectictar Aug 19, 2018

#1**+1 **

Best Answer

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.

hectictar Aug 19, 2018