How to solve for the x-coordinate of the top of an isosceles triangle

I think the following picture helps explain it easier than words:

I know the coordinates of point a, which is at a corner of the triangle, as well as the coordinates of point b, which is somewhere along the edge of the opposite side. I also know the y-coordinate of point c, and I'm trying to find its x-coordinate.

What I've got so far is that as the y-coordinate of b moves up to c, the x coordinate of c approaches the x coordinate of b. And as the y-coordinate of b moves down to a, the x coordinate of c moves to half-way between the x coordinates of a and b. But I don't know the direct relationship.

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Seems like there might be some additional information that needs to be given. Do we know any of the angles? distances? Is the bottom side parallel to the x axis? –  Aryabhata Nov 10 '10 at 23:28

Assuming the base is horizontal as you have drawn it, we have enough information. Let us call the coordinates $(xa,ya), (xb,yb),$ and $(xc,yc)$. Then $$\frac{bx-cx}{by-cy}=-\frac{ax-cx}{ay-cy}$$ because the slopes have to be negatives of each other. You have five of the six variables.

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Looks like a reasonable assumption to me. Especially given the last paragraph of the question. –  Aryabhata Nov 11 '10 at 0:04
Thanks, looks good. And yes, the assumption is correct that the base is parallel to the x-axis. –  Ed Marty Nov 11 '10 at 15:16