Perpendicular on a line segment from a given point

Question

I want to calculate a point on a given line that is perpendicular from a given point.

I have a line segment AB and have a point C outside line segment. I want to ca

Answer 1

I didn't see this answer offered, but Ron Warholic had a great suggestion with the Vector Projection. ACD is merely a right triangle.

1. Create the vector AC i.e (Cx - Ax, Cy - Ay)
2. Create the Vector AB i.e (Bx - Ax, By - Ay)
3. Dot product of AC and AB is equal to the cosine of the angle between the vectors. i.e cos(theta) = ACx*ABx + ACy*ABy.
4. Length of a vector is sqrt(x*x + y*y)
5. Length of AD = cos(theta)*length(AC)
6. Normalize AB i.e (ABx/length(AB), ABy/length(AB))
7. D = A + NAB*length(AD)

Answer 2

A point on line AB can be parametrized by:

M(x)=A+x*(B-A), for x real.

You want D=M(x) such that DC and AB are orthogonal:

dot(B-A,C-M(x))=0.

That is: dot(B-A,C-A-x*(B-A))=0, or dot(B-A,C-A)=x*dot(B-A,B-A), giving:

x=dot(B-A,C-A)/dot(B-A,B-A) which is defined unless A=B.

Answer 3

function getSpPoint(A,B,C){
    var x1=A.x, y1=A.y, x2=B.x, y2=B.y, x3=C.x, y3=C.y;
    var px = x2-x1, py = y2-y1, dAB = px*px + py*py;
    var u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
    var x = x1 + u * px, y = y1 + u * py;
    return {x:x, y:y}; //this is D
}