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    <p>This is working well for me. Taken from <a href="http://workshop.evolutionzone.com/2007/09/10/code-2d-line-intersection/" rel="noreferrer">here</a>.</p> <pre><code> // calculates intersection and checks for parallel lines. // also checks that the intersection point is actually on // the line segment p1-p2 Point findIntersection(Point p1,Point p2, Point p3,Point p4) { float xD1,yD1,xD2,yD2,xD3,yD3; float dot,deg,len1,len2; float segmentLen1,segmentLen2; float ua,ub,div; // calculate differences xD1=p2.x-p1.x; xD2=p4.x-p3.x; yD1=p2.y-p1.y; yD2=p4.y-p3.y; xD3=p1.x-p3.x; yD3=p1.y-p3.y; // calculate the lengths of the two lines len1=sqrt(xD1*xD1+yD1*yD1); len2=sqrt(xD2*xD2+yD2*yD2); // calculate angle between the two lines. dot=(xD1*xD2+yD1*yD2); // dot product deg=dot/(len1*len2); // if abs(angle)==1 then the lines are parallell, // so no intersection is possible if(abs(deg)==1) return null; // find intersection Pt between two lines Point pt=new Point(0,0); div=yD2*xD1-xD2*yD1; ua=(xD2*yD3-yD2*xD3)/div; ub=(xD1*yD3-yD1*xD3)/div; pt.x=p1.x+ua*xD1; pt.y=p1.y+ua*yD1; // calculate the combined length of the two segments // between Pt-p1 and Pt-p2 xD1=pt.x-p1.x; xD2=pt.x-p2.x; yD1=pt.y-p1.y; yD2=pt.y-p2.y; segmentLen1=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2); // calculate the combined length of the two segments // between Pt-p3 and Pt-p4 xD1=pt.x-p3.x; xD2=pt.x-p4.x; yD1=pt.y-p3.y; yD2=pt.y-p4.y; segmentLen2=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2); // if the lengths of both sets of segments are the same as // the lenghts of the two lines the point is actually // on the line segment. // if the point isn’t on the line, return null if(abs(len1-segmentLen1)&gt;0.01 || abs(len2-segmentLen2)&gt;0.01) return null; // return the valid intersection return pt; } class Point{ float x,y; Point(float x, float y){ this.x = x; this.y = y; } void set(float x, float y){ this.x = x; this.y = y; } } </code></pre>
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    1. POHow do you detect where two line segments intersect?
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    1. PTAnswer
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    1. USKingNestor
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      1. VTUpMod
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    1. COThere are several problems with this code. It can raise an exception due to division by zero; it's slow because it takes square roots; and it sometimes returns false positives because it uses a fudge factor. You can do better than this!
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      1. USGareth Rees
    2. COOkay as a solution but that given by Jason is definitely computationally quicker and avoids a lot of the problems with this solution
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      1. USElemental
 

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