Distance Problem (Between 2 Ordered Pairs)

[Danforth]

Find a point on the y-axis that is equidistant from the points (5, -5) and (1, 1).


Answer: (0, -4)



I thought that the mid-point formula would work, but that obviously isn't how to solve this problem. Thanks in advance to anyone who is able to help.

 

[pka]

First note that any point on the y-axis looks like (0,y).

So that \(\displaystyle \L
\sqrt {\left( {5 - 0} \right)^2 + \left( { - 5 - y} \right)^2 } = \sqrt {\left( {1 - 0} \right)^2 + \left( {1 - y} \right)^2 }\)

Now solve for y!

 

[Unco]

First plot the points.

The point lies on the y-axis so has a x-ordinate of 0.

Let its coordinates be (0, p).

The distance between (0, p) and (5, -5) is equal to the distance between (1, 1) and (0, p).

The distance formula may be useful.

 

[Danforth]

Thank you both very much. It didn't even occur to me that the x coordinate would be 0. I tried setting up the distance formula, but with 2 variables, was unable to solve.

Anyway, thank you both very much once again.

 

[Danforth]

Still Need Help!

I still need help. I apologize for bothering you again. I'll show you my steps.

We'll start with what you put:
\(\displaystyle \L
\sqrt {\left( {5 - 0} \right)^2 + \left( { - 5 - y} \right)^2 } = \sqrt {\left( {1 - 0} \right)^2 + \left( {1 - y} \right)^2 }\)

(Sorry..I just copy and pasted your TeX above, but now it's confusing so I'm just typing it out. I hope that you can understand)

So I end up with this:
[square root of (5)^2 + (-5-y)(-5-y)] = [square root of (1)^2 + (1-y)(1-y)]

Next:
(square root of 25 + y^2 + 10y + 25) = (square root of 1 + y^2 - 2y + 1)

Next:
(square root of y^2 + 10y + 50) = (square root of y^2 - 2y + 2)

I have no idea where to go from here. I assume that I've messed up before this point.

 

[Unco]

Your work is excellent.

Now square both sides.

Check your solution works in the original equation (with the square roots).

 

[Danforth]

Your work is excellent.

Now square both sides.

Check your solution works in the original equation (with the square roots).

Thank you so much. I didn't realize how close I was to finishing the problem. I completely forgot that I could square both sides to get rid of the square root from both sides. I spent so long trying to work out the quadratic formula on both sides instead of just squaring them.

Anyway, thank you both so much for your help!