That should work, but I'd bet that my method is faster.
Don
On Jul 29, 6:02 am, Udit Gupta <uditgupta...@gmail.com> wrote:
> Join the given point with all the vertices of the triangle and calculate the
> area of each of the three sub-triangles thus formed
> now compare the area of original triangle with the sum of the area of those
> 3 triangles
> if that comes out to be equal, then the point lies inside
> otherwise not.
>
> On Fri, Jul 29, 2011 at 1:34 AM, Don <dondod...@gmail.com> wrote:
> > // True if point (x,y) is above line defined by two points
> > // If line is vertical, "above" is defined as to the right
> > bool above(double lx1, double ly1, double lx2, double ly2, double x,
> > double y)
> > {
> > return (lx1 != lx2) ? (y > (ly1+(x-lx1)*(ly2-ly1)/(lx2-lx1))) : (x
> > > lx1);
> > }
>
> > // True if point 1 and point 2 are on the same side of the line
> > defined by l1 and l2
> > bool sameSide(double lx1, double ly1, double lx2, double ly2, double
> > px1, double py1, double px2, double py2)
> > {
> > return above(lx1, ly1, lx2, ly2, px1, py1) == above(lx1, ly1, lx2,
> > ly2, px2, py2);
> > }
>
> > // True if point (x,y) is inside triangle
> > bool pointInTriangle(double x1, double y1, double x2, double y2,
> > double x3, double y3, double x, double y)
> > {
> > return sameSide(x1,y1,x2,y2,x,y,x3,y3) &&
> > sameSide(x1,y1,x3,y3,x,y,x2,y2) &&
> > sameSide(x2,y2,x3,y3,x,y,x1,y1);
> > }
>
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