Try this code which makes use of a cross product:
public bool isLeft(Point a, Point b, Point c){
return ((b.X - a.X)*(c.Y - a.Y) - (b.Y - a.Y)*(c.X - a.X)) > 0;
}
Where a = line point 1; b = line point 2; c = point to check against.
If the formula is equal to 0, the points are colinear.
If the line is horizontal, then this returns true if the point is above the line.
Use the sign of the determinant of vectors (AB,AM)
, where M(X,Y)
is the query point:
position = sign((Bx - Ax) * (Y - Ay) - (By - Ay) * (X - Ax))
It is 0
on the line, and +1
on one side, -1
on the other side.
You look at the sign of the determinant of
| x2-x1 x3-x1 |
| y2-y1 y3-y1 |
It will be positive for points on one side, and negative on the other (and zero for points on the line itself).
The vector (y1 - y2, x2 - x1)
is perpendicular to the line, and always pointing right (or always pointing left, if you plane orientation is different from mine).
You can then compute the dot product of that vector and (x3 - x1, y3 - y1)
to determine if the point lies on the same side of the line as the perpendicular vector (dot product > 0
) or not.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With