Simple Point-Line question

you could do a cross product of vP2P3 x vP2P1 and see if the normal goes up or down
normal points up -> P1 is on the left of L1
normal points down -> P1 is on the right of L1

there might be other, better ways, but this one looks good enough

You could find the cartesian equation of the line that is ax + by + c = 0. Then you define the function f(x,y) = ax+by+c and if f(x,y) > 0 then the point lies "above" the line, if f(x,y) < 0 then the point lies "below" the line (and ofcourse if f(x,y) = 0 then the point lies on the line).

E = P3 - P2;
D = P1 - P2;
N = Vector(-E.y, E.x);
side = D.DotProduct(N);

Thanks for all the replies! I think I'm getting it, but I might just be getting myself more confused. I'm a math-tard. It looks like oliii's math may be the easiest for me to understand, but there are parts I don't quite get...

Here's what I've got so far (I think)...

E = P3 - P2;
or
Ex = P3x - P2x;
Ey = P3y - P2y;

D = P1 - P2;
or
Dx = P1x - P2x;
Dy = P1y - P2y;

N = Vector(-E.y, E.x);
or
Nx = -Ey;
Ny = Ex;

side = D.DotProduct(N);
or
side = D.N;
or
side = (Dx * Nx) + (Dy * Ny);

Is that correct? And if so, do I need to normalize 'Vector' or would the raw (-Ey, Ex) result work?

Thanks for the clarification!

Quote:

Original post by Dookie Thanks for all the replies! I think I'm getting it, but I might just be getting myself more confused. I'm a math-tard. It looks like oliii's math may be the easiest for me to understand, but there are parts I don't quite get... Here's what I've got so far (I think)... E = P3 - P2; or Ex = P3x - P2x; Ey = P3y - P2y; D = P1 - P2; or Dx = P1x - P2x; Dy = P1y - P2y; N = Vector(-E.y, E.x); or Nx = -Ey; Ny = Ex; side = D.DotProduct(N); or side = D.N; or side = (Dx * Nx) + (Dy * Ny); Is that correct? And if so, do I need to normalize 'Vector' or would the raw (-Ey, Ex) result work? Thanks for the clarification!

I think thats pretty much correct, you dont really need to normalize N since what gives you the side is the sign of the 'side' variable (negative is on one side, positive the other) rather than its value.