Check is a point (x,y) is between two points drawn on a straight line

I have drawn a line between two points A(x,y)---B(x,y) Now I have a third point C(x,y). I want to know that if C lies on the line which is drawn between A and B. I want to do it in java language. I have found couple of answers similar to this. But, all have some problems and no one is perfect.

if (distance(A, C) + distance(B, C) == distance(A, B))
    return true; // C is on the line.
return false;    // C is not on the line.

or just:

return distance(A, C) + distance(B, C) == distance(A, B);

The way this works is rather simple. If C lies on the AB line, you'll get the following scenario:

A-C------B

and, regardless of where it lies on that line, dist(AC) + dist(CB) == dist(AB). For any other case, you have a triangle of some description and 'dist(AC) + dist(CB) > dist(AB)':

A-----B
 \   /
  \ /
   C

In fact, this even works if C lies on the extrapolated line:

C---A-------B

provided that the distances are kept unsigned. The distance dist(AB) can be calculated as:

  ___________________________
 /           2              2
V (A.x - B.x)  + (A.y - B.y)

Keep in mind the inherent limitations (limited precision) of floating point operations. It's possible that you may need to opt for a "close enough" test (say, less than one part per million error) to ensure correct functioning of the equality.

ATTENTION! Math-only!

You can try this formula. Put your A(x1, y1) and B(x2, y2) coordinates to formula, then you'll get something like

y = k*x + b; // k and b - numbers

Then, any point which will satisfy this equation, will lie on your line. To check that C(x, y) is between A(x1, y1) and B(x2, y2), check this: (x1<x<x2 && y1<y<y2) || (x1>x>x2 && y1>y>y2).

Example

A(2,3) B(6,5)

The equation of line:

(y - 3)/(5 - 3) = (x - 2)/(6 - 2)
(y - 3)/2 = (x - 2)/4
4*(y - 3) = 2*(x - 2)
4y - 12 = 2x - 4
4y = 2x + 8
y = 1/2 * x + 2; // equation of line. k = 1/2, b = 2;

Let's check if C(4,4) lies on this line.

2<4<6 & 3<4<5 // C between A and B

Now put C coordinates to equation:

4 = 1/2 * 4 + 2
4 = 2 + 2 // equal, C is on line AB

PS: as @paxdiablo wrote, you need to check if line is horizontal or vertical before calculating. Just check

y1 == y2 || x1 == x2