How to determine if a point is in a 2D triangle? [closed]

Is there an easy way to determine if a point is inside a triangle? It's 2D, not 3D.

In general, the simplest (and quite optimal) algorithm is checking on which side of the half-plane created by the edges the point is.

Here's some high quality info in this topic on GameDev, including performance issues.

And here's some code to get you started:

float sign (fPoint p1, fPoint p2, fPoint p3)
{
    return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}

bool PointInTriangle (fPoint pt, fPoint v1, fPoint v2, fPoint v3)
{
    float d1, d2, d3;
    bool has_neg, has_pos;

    d1 = sign(pt, v1, v2);
    d2 = sign(pt, v2, v3);
    d3 = sign(pt, v3, v1);

    has_neg = (d1 < 0) || (d2 < 0) || (d3 < 0);
    has_pos = (d1 > 0) || (d2 > 0) || (d3 > 0);

    return !(has_neg && has_pos);
}

Solve the following equation system:

p = p0 + (p1 - p0) * s + (p2 - p0) * t

The point p is inside the triangle if 0 <= s <= 1 and 0 <= t <= 1 and s + t <= 1.

s,t and 1 - s - t are called the barycentric coordinates of the point p.

I agree with Andreas Brinck, barycentric coordinates are very convenient for this task. Note that there is no need to solve an equation system every time: just evaluate the analytical solution. Using Andreas' notation, the solution is:

s = 1/(2*Area)*(p0y*p2x - p0x*p2y + (p2y - p0y)*px + (p0x - p2x)*py);
t = 1/(2*Area)*(p0x*p1y - p0y*p1x + (p0y - p1y)*px + (p1x - p0x)*py);

where Area is the (signed) area of the triangle:

Area = 0.5 *(-p1y*p2x + p0y*(-p1x + p2x) + p0x*(p1y - p2y) + p1x*p2y);

Just evaluate s, t and 1-s-t. The point p is inside the triangle if and only if they are all positive.

EDIT: Note that the above expression for the area assumes that the triangle node numbering is counter-clockwise. If the numbering is clockwise, this expression will return a negative area (but with correct magnitude). The test itself (s>0 && t>0 && 1-s-t>0) doesn't depend on the direction of the numbering, however, since the expressions above that are multiplied by 1/(2*Area) also change sign if the triangle node orientation changes.

EDIT 2: For an even better computational efficiency, see coproc's comment below (which makes the point that if the orientation of the triangle nodes (clockwise or counter-clockwise) is known beforehand, the division by 2*Area in the expressions for s and t can be avoided). See also Perro Azul's jsfiddle-code in the comments under Andreas Brinck's answer.