The parallelogram contains a point

Question

What is the fastest way to decide if a point is inside a parallelogram / rhomboid?

Answer 1

Hi and thanks for all your answers. In the meantime, I myself came up with what I think would be pretty quick:

Suppose we have a parallelogram spanned by PQ and PR, where PQ and PR are vectors (P, Q and R are angles). In addition, we have a point that we want to check by the name A.

We know that the vector PA can be divided into two vectors parallel to PQ and PR:

PA=n*PQ+m*PR

Now we know that n and m MUST be in the interval [0; 1], we solve n and m:

n = -det(PA, PQ)/det(PQ, PR)
m = det(PA, PR)/det(PQ, PR)

Where det (PA, PQ) is the determinant of the vectors PA and PQ:

det(PA, PQ) = PA.x*PQ.y-PQ.x*PA.y

A , 0 <= n <= 1 0 <= m <= 1, :

var d:Number = det(PQ, PR);
if (0 <= -det(PA, PQ)/d <= 1 && 0 <= det(PA, PR)/d <= 1)
{
    //inside
}
else
{
    //outside
}

Answer 2

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Answer 3

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Answer 4

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Answer 5

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Answer 6

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I.e., a, b, c ,

if ( ((a1*x+b1*y+c1)>0) && ((a2*x+b2*y+c2)<0) && 
        ((a3*x+b3*y+c3)<0) && ((a4*x+b4*y+c4)>0) {
    // it in!
}

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Answer 7

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Answer 8

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Edit:

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if (y >= y1 && y <= y3) {
   var k = (x4 - x1) / (y4 - y1);
   var m = x1 - k * y1;
   if (x >= k * y + m) {
     k = (x3 - x2) / (y3 - y2);
     m = x2 - k * y2;
     if (x <= k * y + m) {
       // inside
     }
   }
}