Linear Interpolation to find coord in triangle

Question

Suppose you have the following three points A, B, and C as shown in the following picture:

The points are always sorted according to their vertical offset, so the top most point is always A. Sometimes B and C could have the same y coordinate.

I'm trying to find the x coordinate for point D. I can find the Y coordinate for D by interpolating points A.y and C.y at (B.y / (C.y - A.y)). I'm doing my interpolation using the following formula (in C++)

float linearInterpolation(float a, float b, float t)
{
    return a + (t * (b - a));
}

So in other words, D.y = linearInterpolation(A.y, C.y, (B.y - A.y) / (C.y - A.y))

So to summarize, my question is: how do I find D.x?

Thanks

--

Answer:

Just to clarify, here's the solution that was suggested and worked:

D.x = A.x + (B.y - A.y) * (C.x - A.x) / (C.y - A.y);
D.y = B.y;

As illustrated in the image below:

Answer 1

It it is the x coordinate that requires interpolation. The y coordinates of B and D are equal on your diagram.

D.x = A.x + (B.y - A.y) * (C.x - A.x) / (C.y - A.y);
D.y = B.y;

You should also make a provision for the case of C.y == A.y, where D.x could be anywhere between A.x and C.x. One way to do this is to not draw triangles, for which abs(C.y - A.y) < delta, with the delta being on the order of magnitude of 1 pixel.

Answer 2

D.y = B.y

delta_x = C.x - A.x 
delta_y = C.y - A.y 

dist_y = B.y - A.y

percent = dist_y / delta_y

D.x = A.x + percent * delta_x

Answer 3

The function for the line AC is y = mx + b.

m = (A.y - C.y)/(A.x - C.x)

You can then substitute A in: A.y = A.x * m + b

b = A.y - A.x *m

You need to calculate x from y, so swap the function around.

mx = y -b

x = (y -b)/m

Those are three steps to find the x from the y along that side of a triangle. Note that you don't need to do any interpolation to find Dy. Simply, D.y = B.y.

Note that you can probably optimize what I have just written into a smaller series of steps. I think its better to write easier to read code though.