How to find the intersection point between a line

Question

I have a line that goes from points A to B; I have (x,y) of both points. I also have a rectangle that's centered at B and the width and height of the rectangle.

I need to find the point in the line that intersects the rectangle. Is there a formula that gives me the (x,y) of that point?

Answer 1

Here is a solution in Java that returns true if a line segment (the first 4 parameters) intersects an axis aligned rectangle (the last 4 parameters). It would be trivial to return the intersection point instead of a boolean. It works by first checking if completely outside, else using the line equation y=m*x+b. We know the lines that make up the rectangle are axis aligned, so the checks are easy.

public boolean aabbContainsSegment (float x1, float y1, float x2, float y2, float minX, float minY, float maxX, float maxY) {  
    // Completely outside.
    if ((x1 <= minX && x2 <= minX) || (y1 <= minY && y2 <= minY) || (x1 >= maxX && x2 >= maxX) || (y1 >= maxY && y2 >= maxY))
        return false;

    float m = (y2 - y1) / (x2 - x1);

    float y = m * (minX - x1) + y1;
    if (y > minY && y < maxY) return true;

    y = m * (maxX - x1) + y1;
    if (y > minY && y < maxY) return true;

    float x = (minY - y1) / m + x1;
    if (x > minX && x < maxX) return true;

    x = (maxY - y1) / m + x1;
    if (x > minX && x < maxX) return true;

    return false;
}

It is possible to shortcut if the start or end of the segment is inside the rectangle, but probably it is better to just do the math, which will always return true if either or both segment ends are inside. If you want the shortcut anyway, insert the code below after the "completely outside" check.

// Start or end inside.
if ((x1 > minX && x1 < maxX && y1 > minY && y1 < maxY) || (x2 > minX && x2 < maxX && y2 > minY && y2 < maxY)) return true;

Answer 2

The point A is always outside of the rectangle and the point B is always at the center of the rectangle

Assuming the rectangle is axis-aligned, this makes things pretty simple:

The slope of the line is s = (Ay - By)/(Ax - Bx).

• If -h/2 <= s * w/2 <= h/2 then the line intersects:
  • The right edge if Ax > Bx
  • The left edge if Ax < Bx.
• If -w/2 <= (h/2)/s <= w/2 then the line intersects:
  • The top edge if Ay > By
  • The bottom edge if Ay < By.

Once you know the edge it intersects you know one coordinate: x = Bx ± w/2 or y = By ± h/2 depending on which edge you hit. The other coordinate is given by y = By + s * w/2 or x = Bx + (h/2)/s.

Answer 3

Another option that you can consider especially if you are planning on testing many lines with the same rectangle is to transform your coordinate system to have the axes align with diagonals of the rectangle. Then since your line or ray starts at the center of the rectangle you can determine the angle then you can tell which segment it will intersect by the angle (i.e. <90deg seg 1, 90deg< <180deg seg 2 etc...). Then of course you have to transform back to the original coordinate system

Although this seems like more work the transformation matrix and its inverse can be calculated once and then reused. This also extends to higher dimensional rectangles more easily where you would have to consider quadrants and intersections with faces in 3D and so on.

Answer 4

Here is a solution that works for me. I assume that the rect is aligned to the axes.

Data:

// Center of the Rectangle
let Cx: number
let Cy: number
// Width
let w: number
// Height
let h: number

// Other Point
let Ax: number
let Ay: number

Now translate point A by the center of the rectangle so the rect is centered in O(0,0) and consider the problem in the first quarter (i.e. x > 0 and y > 0).

// Coordinates Translated
let Px = Math.abs(Ax - Cx)
let Py = Math.abs(Ay - Cy)

// Slope of line from Point P to Center
let Pm = Py / Px

// Slope of rectangle Diagonal
let Rm = h / w

// If the point is inside the rectangle, return the center
let res: [number, number] = [0, 0]

// Check if the point is inside and if so do not calculate
if (!(Px < w / 2 && Py < h / 2)) {

    // Calculate point in first quarter: Px >= 0 && Py >= 0
    if (Pm <= Rm) {
        res[0] = w / 2
        res[1] = (w * Pm) / 2
    } else {
        res[0] = h / (Pm * 2)
        res[1] = h / 2
    }

    // Set original sign 
    if (Ax - Cx < 0) res[0] *= -1
    if (Ay - Cy < 0) res[1] *= -1
}

// Translate back
return [res[0] + Cx, res[1] + Cy]

Answer 5

I'll not give you a program to do that, but here is how you can do it:

• calculate the angle of the line
• calculate the angle of a line from the center of the rectangle to one of it's corners
• based on the angles determine on which side does the line intersect the rectangle
• calculate intersection between the side of the rectangle and the line

Answer 6

I don't know if this is the best way, but what you could do is to figure out the proportion of the line that is inside the rectangle. You can get that from the width of the rectangle and the difference between the x coordinates of A and B (or height and y coordinates; based on the width and height you can check which case applies, and the other case will be on the extension of a side of the rectangle). When you have this, just take that proportion of the vector from B to A and you have your intersection point's coordinates.