Check whether a point (x,y) is on the line between two other points

Question

I have drawn a line between two points: A(x,y) and B(x,y). I have a third point C(x,y). How can I know whether C lies on the straight line drawn between A and B?

I want to do this in the Java language. I have found a couple of answers similar to this, but all have some problems and none is perfect.

Answer 1

One good library is available for this by turfjs.

var pt = turf.point([0, 0]);
var line = turf.lineString([[-1, -1],[1, 1],[1.5, 2.2]]);
var isPointOnLine = turf.booleanPointOnLine(pt, line);

If you want to reduce the accuracy use epsilon option

turf.booleanPointOnLine([c1, c2], turf.lineString([[x1, y1], [x2, y2]]), {epsilon: 10}))

Hope this is helpful

https://turfjs.org/docs/#booleanPointOnLine

Answer 2

If you just want to check whether the point C is on the infinite line passing through the points A and B (rather than check whether C is on the line segment from A to B, i.e. C is also between them), the simplest implementation is:

// Are a, b and c on the same line?
public static boolean inLine(Point a, Point b, Point c) {
   return (a.x - c.x)*(c.y - b.y) == (c.x - b.x)*(a.y - c.y);
}

This is morally equivalent to checking that gradient(A, C) == gradient(C, B), but rearranged to use multiplication instead of division to avoid divide-by-zero when one of the gradients is vertical (also, it gives exact results if using integers).

It is equivalent to checking that the cross product of (A - C) and (C - B) is equal to 0. The property of three points being on the same line is known as collinearity.

Note: there is a different test to see if C appears on the line between A and B if you draw it on a screen. Maths assumes that A, B, C are infinitely small points.

Answer 3

if ( (ymid - y1) * (x2-x1) == (xmid - x1) * (y2-y1) )  **is true, Z lies on line AB**

Start Point : A (x1, y1),
End Point : B (x2, y2),
Point That is on Line AB or Not : Z (xmid, ymid)

I just condensed everyone's answers and this formula works the best for me.

1. It avoids division by zero
2. No distance calculation required
3. Simple to implement

Edit: In case you are dealing with floats, which you most probably are, use this:

   if( (ymid - y1) * (x2-x1) - (xmid - x1) * (y2-y1) < DELTA )

where the tolerance DELTA is a value close to zero. I usually set it to 0.05

Answer 4

Here is a JavaScript function I made. You pass it three points (three objects with an x and y property). Points 1 and 2 define your line, and point 3 is the point you are testing.

You will receive an object back with some useful info:

• on_projected_line - If pt3 lies anywhere on the line including outside the points.
• on_line - If pt3 lies on the line and between or on pt1 and pt2.
• x_between - If pt3 is between or on the x bounds.
• y_between - If pt3 is between or on the y bounds.
• between - If x_between and y_between are both true.

/**
 * @description Check if pt3 is on line defined by pt1 and pt2.
 * @param {Object} pt1 The first point defining the line.
 * @param {float} pt1.x
 * @param {float} pt1.y
 * @param {Object} pt2 The second point defining the line.
 * @param {float} pt2.x
 * @param {float} pt2.y
 * @param {Object} pt3 The point to test.
 * @param {float} pt3.x
 * @param {float} pt3.y
 */
function pointOnLine(pt1, pt2, pt3) {
    const result = {
        on_projected_line: true,
        on_line: false,
        between_both: false,
        between_x: false,
        between_y: false,
    };

    // Determine if on line interior or exterior
    const x = (pt3.x - pt1.x) / (pt2.x - pt1.x);
    const y = (pt3.y - pt1.y) / (pt2.y - pt1.y);

    // Check if on line equation
    result.on_projected_line = x === y;

    // Check within x bounds
    if (
        (pt1.x <= pt3.x && pt3.x <= pt2.x) ||
        (pt2.x <= pt3.x && pt3.x <= pt1.x)
    ) {
        result.between_x = true;
    }

    // Check within y bounds
    if (
        (pt1.y <= pt3.y && pt3.y <= pt2.y) ||
        (pt2.y <= pt3.y && pt3.y <= pt1.y)
    ) {
        result.between_y = true;
    }

    result.between_both = result.between_x && result.between_y;
    result.on_line = result.on_projected_line && result.between_both;
    return result;
}

console.log("pointOnLine({x: 0, y: 0}, {x: 1, y: 1}, {x: 2, y: 2})")
console.log(pointOnLine({x: 0, y: 0}, {x: 1, y: 1}, {x: 2, y: 2}))

console.log("pointOnLine({x: 0, y: 0}, {x: 1, y: 1}, {x: 0.5, y: 0.5})")
console.log(pointOnLine({x: 0, y: 0}, {x: 1, y: 1}, {x: 0.5, y: 0.5}))

Answer 5

Here is my C# solution. I believe the Java equivalent will be almost identical.

Notes:

1. Method will only return true if the point is within the bounds of the line (it does not assume an infinite line).

2. It will handle vertical or horizontal lines.

3. It calculates the distance of the point being checked from the line so allows a tolerance to be passed to the method.

    /// <summary>
    /// Check if Point C is on the line AB
    /// </summary>
    public static bool IsOnLine(Point A, Point B, Point C, double tolerance)
    {
        double minX = Math.Min(A.X, B.X) - tolerance;
        double maxX = Math.Max(A.X, B.X) + tolerance;
        double minY = Math.Min(A.Y, B.Y) - tolerance;
        double maxY = Math.Max(A.Y, B.Y) + tolerance;
   
        //Check C is within the bounds of the line
        if (C.X >= maxX || C.X <= minX || C.Y <= minY || C.Y >= maxY)
        {
            return false;
        }
   
        // Check for when AB is vertical
        if (A.X == B.X)
        {
            if (Math.Abs(A.X - C.X) >= tolerance)
            {
                return false;
            }
            return true;
        }
   
        // Check for when AB is horizontal
        if (A.Y == B.Y)
        {
            if (Math.Abs(A.Y - C.Y) >= tolerance)
            {
                return false;
            }
            return true;
        }
   
   
        // Check istance of the point form the line
        double distFromLine = Math.Abs(((B.X - A.X)*(A.Y - C.Y))-((A.X - C.X)*(B.Y - A.Y))) / Math.Sqrt((B.X - A.X) * (B.X - A.X) + (B.Y - A.Y) * (B.Y - A.Y));
   
        if (distFromLine >= tolerance)
        {
            return false;
        }
        else
        {
            return true;
        }
    }
   

Answer 6

I think this might help

How to check if a point lies on a line between 2 other points

That solution uses only integers given you only provide integers which removes some pitfalls as well

Answer 7

def DistBetwPoints(p1, p2):
    return math.sqrt( (p2[0] - p1[0])**2 + (p2[1] - p1[1])**2 )

# "Check if point C is between line endpoints A and B"
def PointBetwPoints(A, B, C):
    dist_line_endp = DistBetwPoints(A,B)
    if DistBetwPoints(A,C)>dist_line_endp:       return 1
    elif DistBetwPoints(B,C)>dist_line_endp:     return 1
    else:                                        return 0

Answer 8

An easy way to do that I believe would be the check the angle formed by the 3 points. If the angle ACB is 180 degrees (or close to it,depending on how accurate you want to be) then the point C is between A and B.

Answer 9

I think all the methods here have a pitfall, in that they are not dealing with rounding errors as rigorously as they could. Basically the methods described will tell you if your point is close enough to the line using some straightforward algorithm and that it will be more or less precise.

Why precision is important? Because it's the very problem presented by op. For a computer program there is no such thing as a point on a line, there is only point within an epsilon of a line and what that epsilon is needs to be documented.

Let's illustrate the problem. Using the distance comparison algorithm:

Let's say a segment goes from (0, 0) to (0, 2000), we are using floats in our application (which have around 7 decimal places of precision) and we test whether a point on (1E-6, 1000) is on the line or not.

The distance from either end of the segment to the point is 1000.0000000005 or 1000 + 5E-10, and, thus, the difference with the addition of the distance to and from the point is around 1E-9. But none of those values can be stored on a float with enough precission and the method will return true.

If we use a more precise method like calculating the distance to the closest point in the line, it returns a value that a float has enough precision to store and we could return false depending on the acceptable epsilon.

I used floats in the example but the same applies to any floating point type such as double.

One solution is to use BigDecimal and whichever method you want if incurring in performance and memory hit is not an issue.

A more precise method than comparing distances for floating points, and, more importantly, consistently precise, although at a higher computational cost, is calculating the distance to the closest point in the line.

Shortest distance between a point and a line segment

It looks like I'm splitting hairs but I had to deal with this problem before. It's an issue when chaining geometric operations. If you don't control what kind of precission loss you are dealing with, eventually you will run into difficult bugs that will force you to reason rigorously about the code in order to fix them.

Answer 10

The above answers are unnecessarily complicated. The simplest is as follows.

1. if (x-x1)/(x2-x1) = (y-y1)/(y2-y1) = alpha (a constant), then the point C(x,y) will lie on the line between pts 1 & 2.

2. If alpha < 0.0, then C is exterior to point 1.

3. If alpha > 1.0, then C is exterior to point 2.
4. Finally if alpha = [0,1.0], then C is interior to 1 & 2.

Hope this answer helps.

Answer 11

ATTENTION! Math-only!

You can try this formula. Put your A(x1, y1) and B(x2, y2) coordinates to formula, then you'll get something like

y = k*x + b; // k and b - numbers

Then, any point which will satisfy this equation, will lie on your line. To check that C(x, y) is between A(x1, y1) and B(x2, y2), check this: (x1<x<x2 && y1<y<y2) || (x1>x>x2 && y1>y>y2).

Example

A(2,3) B(6,5)

The equation of line:

(y - 3)/(5 - 3) = (x - 2)/(6 - 2)
(y - 3)/2 = (x - 2)/4
4*(y - 3) = 2*(x - 2)
4y - 12 = 2x - 4
4y = 2x + 8
y = 1/2 * x + 2; // equation of line. k = 1/2, b = 2;

Let's check if C(4,4) lies on this line.

2<4<6 & 3<4<5 // C between A and B

Now put C coordinates to equation:

4 = 1/2 * 4 + 2
4 = 2 + 2 // equal, C is on line AB

PS: as @paxdiablo wrote, you need to check if line is horizontal or vertical before calculating. Just check

y1 == y2 || x1 == x2

Answer 12

if (distance(A, C) + distance(B, C) == distance(A, B))
    return true; // C is on the line.
return false;    // C is not on the line.

or just:

return distance(A, C) + distance(B, C) == distance(A, B);

The way this works is rather simple. If C lies on the AB line, you'll get the following scenario:

A-C------B

and, regardless of where it lies on that line, dist(AC) + dist(CB) == dist(AB). For any other case, you have a triangle of some description and 'dist(AC) + dist(CB) > dist(AB)':

A-----B
 \   /
  \ /
   C

In fact, this even works if C lies on the extrapolated line:

C---A-------B

provided that the distances are kept unsigned. The distance dist(AB) can be calculated as:

  ___________________________
 /           2              2
V (A.x - B.x)  + (A.y - B.y)

Keep in mind the inherent limitations (limited precision) of floating point operations. It's possible that you may need to opt for a "close enough" test (say, less than one part per million error) to ensure correct functioning of the equality.