How to check if a point lies on a line between 2 other points

Assuming that point1 and point2 are different, first you check whether the point lies on the line. For that you simply need a “cross-product” of vectors point1 -> currPoint and point1 -> point2.

dxc = currPoint.x - point1.x;
dyc = currPoint.y - point1.y;

dxl = point2.x - point1.x;
dyl = point2.y - point1.y;

cross = dxc * dyl - dyc * dxl;

Your point lies on the line if and only if cross is equal to zero.

if (cross != 0)
  return false;

Now, as you know that the point does lie on the line, it is time to check whether it lies between the original points. This can be easily done by comparing the x coordinates, if the line is “more horizontal than vertical”, or y coordinates otherwise

if (abs(dxl) >= abs(dyl))
  return dxl > 0 ? 
    point1.x <= currPoint.x && currPoint.x <= point2.x :
    point2.x <= currPoint.x && currPoint.x <= point1.x;
else
  return dyl > 0 ? 
    point1.y <= currPoint.y && currPoint.y <= point2.y :
    point2.y <= currPoint.y && currPoint.y <= point1.y;

Note that the above algorithm if entirely integral if the input data is integral, i.e. it requires no floating-point calculations for integer input. Beware of potential overflow when calculating cross though.

P.S. This algorithm is absolutely precise, meaning that it will reject points that lie very close to the line but not precisely on the line. Sometimes this is not what’s needed. But that’s a different story.