I tryed to write code for calculating point-line distance. I found a lot of calculations on the internet but I am not sure if I understand right. I found some equations for lines and it seems like equations for calculating point-plane distance in 3D. I think it is the same but in 2 dimensional.
I have some experiences with 3D point-plane distance calculations. I calculated parameters A,B,C,D from 3 points vec3 (plane definition) and to get distance just apply this equation with vec3 point (x,y,z).
Plane equation: Ax + By + Cz = D
Line equation should be: Ax + By = C
Line equation should work similary by applying equation with vec2 point (x,y).
My problem is how to calculate parameters A,B,C from this equation with 2 points vec2 (line definition)?
Any simply programming/mathematic explanation?
Plane with 3 points:
Plane(vec3 p0, vec3 p1, vec3 p2) {
vec3 v = p1 - p0;
vec3 u = p2 - p0;
vec3 n = cross(v, u);
normalize(n);
//Result A,B,C,D
A = n.x;
B = n.y;
C = n.z;
D = dot(vec3(-n.x, -n.y, -n.z), p0);
}
Plane with 2 points:
Line(vec2 p0, vec2 p1) {
//...
//Result A,B,C
A = (?);
B = (?);
C = (?);
}
Thanks.
PS: Sorry for my english. :/
Update:
Solved!
I found answer after hour of calculations and it's similar to plane equation.
Programming:
Line(vec2 p0, vec2 p1) {
vec2 l = p1 - p0;
vec2 n = l.cross();
n.normalize();
//Result a,b,c
a = n.x;
b = n.y;
c = vec2::dot(vec2(-n.x, -n.y), p0);
}
Difference is in cross product of vec2. It's someting like this:
vec2 cross(vec2 p) { //Only one parameter
return vec2(y, -x);
}
Mathematic:
P1 - start point of line
P2 - end point of line
1) N = x(P2 - P1)
Where "x" is cross product of vector (swap elemets and negate element x) ---> x(V) = [Vy, -Vx]
(I'm not sure is this operation is official but result of that should be vector which is perpendicular to that parameter)
2) N = N / |N| (normalize vector N)
Where |N| is length of vector N
3) Result A: a = Nx
4) Result B: b = Ny
5) N' = -N
6) Result C: c = (N').(P1)
Where "(N').(P1)" is dot product of vectors N' and P1
PS:
This formula d = (|(x_2-x_1)x(x_1-x_0)|)/(|x_2-x_1|) is right. It works. But I need to use equation ax + by + c = 0 because I need to know if it's on left or right side from line (positive of negaive distance) and it's better for programming. Thanks for answers.
Still sorry for my english. :D
