# How would I find a point on the y-axis equidistant from two other points?

The points are$$(5,-5) and (1,1)$$

I tried doing this visually and came up with (0,-5). This wasn't correct once I applied the distance formula to check the distance between that point and the two others.

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Let the point be $(0,y)$. Then $5^2+(y+5)^2=1^2+(y-1)^2$. Expand and solve. (There are other ways.) – André Nicolas Aug 26 '14 at 11:49
As for what you did wrong, I'm not sure, but $(-5,0)$ is not on the $y$-axis... – fixedp Aug 26 '14 at 11:51
that was a typo – Cherry_Developer Aug 26 '14 at 11:52

A point on the $y-$axis is of the form $(0,y)$.

The distance between $(0,y)$ and $(5,-5)$ is: $$\sqrt{(5-0)^2+(-5-y)^2}=\sqrt{25+(5+y)^2}$$

The distance between $(0,y)$ and $(1,1)$ is: $$\sqrt{(1-0)^2+(1-y)^2}=\sqrt{1+(1-y)^2}$$

The two distances are equal, so also their squares:

$$25+(5+y)^2=1+(1-y)^2$$

Now you have to solve for $y$.

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It's probably more convenient to equate the squares of the distances, rather than the distances themselves. – fixedp Aug 26 '14 at 11:50
I have doubts but I think it's (0,-4) – Cherry_Developer Aug 26 '14 at 11:59
It is correct!! – Mary Star Aug 26 '14 at 12:04

You want the point where the perpendicular bisector of the two points cuts the $y$-axis.

The slope of the line between $(5,-5)$ and $(1,1)$ is $-\frac{3}{2}$, so the slope of the normal is $\frac23$. Hence the normal through the mid-point $(3,-2)$ cuts the $y$-axis at $(0,-4)$.

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+1 because on seeing the question I thought to myself, "well personally I'd just solve the equation, but what would that teach you about the set of points equidistant between A and B?" :-) – Steve Jessop Aug 26 '14 at 13:19
@Steve: Yes! I was disturbed by all those quadratic terms in the other answers, which always cancelled each other out. – TonyK Aug 26 '14 at 13:22
That's an insight in itself, mind you. They cancel out so that the set of equidistant points can be a straight line, as opposed to something curved if there were quadratic terms left in there! – Steve Jessop Aug 26 '14 at 13:23
This also illustrates why it's always possible to find such a point, unless the given points lie on a horizontal line (which would make the perpendicular bisector parallel to the $y$-axis. – cjm Aug 26 '14 at 14:31