coordinates question?

Question

hey i have a math problem i'm having issues with:

The point (5, y) is equidistant from (1, 4) and (10, -3). Find y.

My first instinct was, of course, to plug the points into the midpoint formula, but the x coordinate comes out to be 5.5 rather than 5.

Help would be greatly appreciated :)

To John:
First of all, I do not post questions simply to get the answers. The answers themselves aren't worth crap, it's about the process to obtain them.

Second, your answer is totally illogical. You can't just assume that the points (1,4) and (5, y) will form a right triangle, which of course is a prerequisite for using the pythagorean theorum in the first place.

I'm looking for help from people who actually know what they're doing.

Alright, thanks for your help Ratlover and Huneebee, I've solved it.

But Huneebee, you made a computational error. -8y+32=6y+34 doesn't equal -2y=2. You probably subtracted 6y from -8y and accidentally perceived it as being -2, while in reality it is -14.

Answer 1

If it's equidistant, that means the distance from (5, y) to (1, 4) and (5, y) to (10, -3) are the same. You can compare these with the distance formula.
The distance formula has a square root in it, so comparing square distances is easier... and just as correct!

Squared distance from (5, y) to (1, 4):
(5-1)^2 + (y - 4)^2

Squared distance from (5, y) to (10, -3):
(5-10)^2 + (y + 3)^2

Since the point is equidistant, these distances are the same:
16 + (y - 4)^2 = 25 + (y + 3)^2

This is the equation you want to solve.
Multiply out (y - 4)^2 and (y + 3)^2 (use FOIL) and the y^2 terms will cancel out.
You can then solve for y. Good luck!

Answer 2

The midpoint formula didn't work because the point does not have to lie on a straight between (1,4) and (10,-3) to be equidistant from them. It could make a V with the 2 points.

The best thing to use is the distance formula. Since the distances are equal, you can set the 2 equal to eachother.

D^2 = (y2-y1)^2 + (x2-x1)^2 So use (5,y) and one of the coordinates on each side

(4-y)^2 + (1-5)^2 = (-3-y)^2 + (10-5)^2

(4-y)^2 + 16 = (-3-y)^2 + 25

16 -8y + y^2 + 16 = 9 + 6y + y^2 + 25

y^2 - 8y + 32 = y^2 + 6y + 34

-2y = 2
y = -1

~~~~~~~~
Edit: You're right, I did make a boo-boo. But did I at least help you understand how to do it?

Answer 3

Instead of dooing your homework for you, let me encourage you to use the pythagorean theorem twice. Use Pythagorean's theorem to calculate the distance between (1,4) and (5,y) and set that equal to the distance between (10,-3) and (5,y)