Projection of a point on line defined by 2 points

Question

I have been trying to figure this out for sometime now..

The problem to solve..

Say I have 3 Points..

P1 ---------- P2, and P3 can be anywhe         


        

Answer 1

The problem is that you Point has integer values for X and Y and therefore you are doing integer division. Try to cast your values into float or double, do the calculations and then return them back to the integers.

Note that when you are doing this: (P1.Y - P0.Y) * ((P.X - P0.X) / (P1.X - P0.X)) you are actualy loosing the precision since the result of 5/2 is 2, not 2.5 but when your values are real numbers then 5.0/2.0 is indeed 2.5.

You should try this:

double y1 = P0.Y + (double)(P1.Y - P0.Y) * ((double)(P.X - P0.X) / (double)(P1.X - P0.X));
double x1 = P.X; //these two are implicit casts

double y2 = P.Y;
double x2 = P0.X + (double)(P1.X - P0.X) * ((double)(P.Y - P0.Y) / (double)(P1.Y - P0.Y));

return new Point((x1 + x2) / 2.0, (y1 + y2) / 2.0); //put 2.0 for any case even though x1+x2 is `double`

Also, then you are converting from double to int, decimal part of the number is automatically cut off so for instance 3.87 will become 3. Than your last line should be more precise if you could use this:

   return new Point((x1 + x2) / 2.0 + 0.5, (y1 + y2) / 2.0 + 0.5);

which will effectively round double values to the closer integer value.

EDIT:

But if you just want to find the point p3 on the line between the two points, than it is easier to use this approach:

public Point lerp(Point P0, Point P1) 
{
      double x = ((double)P0.X + P1.X)/2.0;

      double y = (double)P0.Y + (double)(P1.Y - P0.Y) * ((double)(x - P0.X) / (double)(P1.X - P0.X));
      return new Point(x + 0.5, y + 0.5);
}