how to determine whether a point lies inside a rectangle? [duplicate]

Answer 1

Pick a point that's definitely outside the rectangle. Then create a segment from that point to the point in question. Solve the linear equations for intersections between that segment and the segments that make up the rectangle. If you get exactly one intersection, the point is inside the rectangle. Otherwise (0 or 2 intersections), it's outside.

This is trivial to extend to essentially any polygon -- an odd number of intersections means the point is inside the polygon, and an even number means it's outside.

Edit: It may not be immediately obvious, so I'll emphasize that the point we pick outside the rectangle (polygon) is entirely arbitrary. We can pick whatever point we want as long as we're sure it's outside the polygon. To keep our computations easy, what we'll typically do is pick (P_x, infinity) (where P_x is the x coordinate of the point P that we're testing) -- that is, what we're creating is essentially a vertical ray. That simplifies testing a bit, because we only have to test against one end-point to find an intersection. It also simplifies solving the linear equations to the point that it's barely recognizable as solving linear equations anymore. We really just need to compute the Y coordinate for the line at the P_x, and see if it's greater than P_y. As such, solving the linear equation breaks down to:

1. checking whether that X value is within the range of X values for the segment
2. if it is, plugging the X value into the equation of the line
3. testing whether the resulting Y value is greater than P_y

If those pass, we have an intersection. Also note that the tests can be carried out in parallel (handy if we're doing this on parallel hardware like a GPU).

Answer 2

Finding whether a point lies within a bounded region like rectangle is part of the classic clipping algorithms. Refer to the wikipedia articles on Clipping and Line Clipping to know more about it.

Answer 3

This is far from the best solution... But if you have the points in consecutive order, call them a, b, c, and d with an x and a y field, you can use the cross product of the vectors between your point p and each of the consecutive pairs.

If you always get the same sign for the result (i.e., all are positive or all are negative) then you're inside the rectangle; otherwise, you're outside.

Answer 4

Define a new coordinate system with two rectangle sides as unit vectors and transform the coordinate of the point into the new coordinate system. If both coordinates are between 0 and 1, it's inside.

In equations (assuming A,B,C,D are corners of the rectangle, P is the point, _x and _y are the x and y components):

P_x = A_x + x * (B_x - A_x) + y * (D_x - A_x)
P_y = A_y + x * (B_y - A_y) + y * (D_y - A_y)

Solve for x and y and check if they are between 0 and 1

Written as linear equation system (A,B,C,D,P are vectors of length 2):

[    |    ]   [x]   [   ]
[B-A | D-A] * [ ] = [P-A]
[    |    ]   [y]   [   ]

Solving is easy as it has only two dimensions and you can be sure that you are not singular.

Answer 5

Following the spirit of @Jerry Coffin: create segments from rectangle corners to the point in question. Solve the linear equations. Slope is tan(a). Sum up all seq arctangents diff, if it is 2*PI and each diff < PI - point is inside the rectangle.

Edit Probably enough just check for each sequential difference < Pi...

Answer 6

Since the rectangle could be rotated, you might want to consider an algorithm that is used to determine whether a point is interior to a convex polygon.

You could also compute the rotation angle of the rectangle, then transform both the rectangle and the point to axially align the rectangle. Then check to see if the transformed point is inside the axially aligned rectangle.