# How to solve for 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 vertex 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.