Let the points be A(x1,y1), B(x2,y2), C(x3,y3). We need y3.
The slope of AB is (y2-y1)/(x2-x1), which is the same slope as AC and BC. We don't know how far away C is from A or B, but we do know that C lies on the line AB or AB extended.
So y-y1=[(y2-y1)/(x2-x1)](x-x1) is the equation of AB, that is, y=(y2-y1)(x-x1)/(x2-x1)+y1.
This can be written y=[(y2-y1)(x-x1)+y1(x2-x1)]/(x-x1),
y=(xy2-x1y2-xy1+x2y1)/(x-x1).
C can be any point on this line. If x=a, we have the point C(a,y3) where:
y3=(ay2-x1y2-ay1+x2y1)/(a-x1),
y3=(a(y2-y1)-x1y2+x2y1)/(a-x1).
By plugging in the value of a and the coordinates of A and B, y3 can be calculated.