如何判断一个点是否属于某一行？

Question

How can I tell if a point belongs to a certain line?

Examples are appreciated, if possible.

Answer 1

In the simplest form, just plug the coordinates into the line equation and check for equality.

Given:

Point p (X=4, Y=5)
Line l (Slope=1, YIntersect=1)

Plug in X and Y:

   Y = Slope * X + YIntersect
=> 5 = 1 * 4 + 1
=> 5 = 5

So yes, the point is on the line.

If your lines are represented in (X1,Y1),(X2,Y2) form, then you can calculate slope with:

 Slope = (y1 - y2) / (x1-x2)

And then get the Y-Intersect with this:

 YIntersect = - Slope * X1 + Y1;

Edit: I fixed the Y-Intersect (which has been X1 / Y1 ...)

You'll have to check that x1 - x2 is not 0 . If it is, then checking if the point is on the line is a simple matter of checking if the Y value in your point is equal to either x1 or x2 . Also, check that the X of the point is not 'x1' or 'x2'.

Answer 2

I just wrote an function which handles a few extra requirements since I use this check in a drawing application:

• Fuzziness - There must be some room for error since the function is used to select lines by clicking on them.
• The line got an EndPoint and a StartPoint, no infinite lines.
• Must handle straight vertical and horizontal lines, (x2 - x1) == 0 causes division by zero in the other answers.

private const double SELECTION_FUZZINESS = 3;

internal override bool ContainsPoint(Point point)
{
    LineGeometry lineGeo = geometry as LineGeometry;
    Point leftPoint;
    Point rightPoint;

    // Normalize start/end to left right to make the offset calc simpler.
    if (lineGeo.StartPoint.X <= lineGeo.EndPoint.X)
    {
        leftPoint   = lineGeo.StartPoint;
        rightPoint  = lineGeo.EndPoint;
    }
    else
    {
        leftPoint   = lineGeo.EndPoint;
        rightPoint  = lineGeo.StartPoint;
    }

    // If point is out of bounds, no need to do further checks.                  
    if (point.X + SELECTION_FUZZINESS < leftPoint.X || rightPoint.X < point.X - SELECTION_FUZZINESS)
        return false;
    else if (point.Y + SELECTION_FUZZINESS < Math.Min(leftPoint.Y, rightPoint.Y) || Math.Max(leftPoint.Y, rightPoint.Y) < point.Y - SELECTION_FUZZINESS)
        return false;

    double deltaX = rightPoint.X - leftPoint.X;
    double deltaY = rightPoint.Y - leftPoint.Y;

    // If the line is straight, the earlier boundary check is enough to determine that the point is on the line.
    // Also prevents division by zero exceptions.
    if (deltaX == 0 || deltaY == 0) 
        return true;

    double slope        = deltaY / deltaX;
    double offset       = leftPoint.Y - leftPoint.X * slope;
    double calculatedY  = point.X * slope + offset;

    // Check calculated Y matches the points Y coord with some easing.
    bool lineContains = point.Y - SELECTION_FUZZINESS <= calculatedY && calculatedY <= point.Y + SELECTION_FUZZINESS;

    return lineContains;            
}

Answer 3

The best way to determine if a point R = (rx, ry) lies on the line connecting points P = (px, py) and Q = (qx, qy) is to check whether the determinant of the matrix

{{qx - px, qy - py}, {rx - px, ry - py}},

namely (qx - px) * (ry - py) - (qy - py) * (rx - px) is close to 0. This solution has several related advantages over the others posted: first, it requires no special case for vertical lines, second, it doesn't divide (usually a slow operation), third, it doesn't trigger bad floating-point behavior when the line is almost, but not quite vertical.

Answer 4

Given two points on the line L0 and L1 and the point to test P .

               (L1 - L0) * (P - L0)
n = (P - L0) - --------------------- (L1 - L0)
               (L1 - L0) * (L1 - L0)

The norm of the vector n is the distance of the point P from the line through L0 and L1 . If this distance is zero or small enough (in the case of rounding errors), the point lies on the line.

The symbol * represents the dot product.

Example

P = (5, 5)

L0 = (0, 10)
L1 = (20, -10)

L1 - L0 = (20, -20)
P  - L0 = (5, -5)

              (20, -20) * (5, -5)
n = (5, -5) - --------------------- (20, -20)
              (20, -20) * (20, -20)

              200
  = (5, -5) - --- (20, -20)
              800

  = (5, -5) - (5, -5)

  = (0, 0)

Answer 5

I think Mr.Patrick McDonald put the nearly correct answer and this is the correction of his answer:

public bool IsOnLine(Point endPoint1, Point endPoint2, Point checkPoint)
{
    return (((double)checkPoint.Y - endPoint1.Y)) / ((double)(checkPoint.X - endPoint1.X))
        == ((double)(endPoint2.Y - endPoint1.Y)) / ((double)(endPoint2.X - endPoint1.X));
}

and of course there are many other correct answers especially Mr.Josh but i found this is the best one.

Thankx for evryone.

Answer 6

y = m * x + c

This is the equation of a line. x & y are the co-ordinates. Each line is characterized by its slope (m ) and where it intersects the y-axis (c).

So given m & c for a line, you can determine if the point (x1, y1) is on the line by checking if the equation holds for x = x1 and y = y1

Answer 7

If you have a line defined by its endpoints

PointF pt1, pt2;

and you have a point that you want to check

PointF checkPoint;

then you could define a function as follows:

bool IsOnLine(PointF endPoint1, PointF endPoint2, PointF checkPoint) 
{
    return (checkPoint.Y - endPoint1.Y) / (endPoint2.Y - endPoint1.Y)
        == (checkPoint.X - endPoint1.X) / (endPoint2.X - endPoint1.X);
}

and call it as follows:

if (IsOnLine(pt1, pt2, checkPoint) {
    // Is on line
}

You will need to check for division by zero though.

Answer 8

A 2D line is generally represented using an equation in two variables x and y here is a well known equation

Now imagine your GDI+ line is drawn from (0,0) to (100, 100) then the value of m=(0-100)/(0-100) = 1 thus the equation for your line is y-0=1*(x-0) => y=x

Now that we have an equation for the line in question its easy to test if a point belongs to this line. A given point (x3, y3) belongs to this line if it satisfies the line equation when you substitute x=x3 and y=y3. For example the point (10, 10) belongs to this line since 10=10 but (10,12) does not belong to this line since 12 != 10.

NOTE: For a vertical line the value of the slope (m) is infinite but for this special case you may use the equation for a vertical line directly x=c where c = x1 = x2.

Though I have to say I am not sure if this is the most efficient way of doing this. I will try and find a more efficient way when I have some more time on hand.

Hope this helps.

Answer 9

Equation of the line is:

y = mx + c

So a point(a,b) is on this line if it satisfies this equation ie b = ma + c

Answer 10

As an alternative to the slope/y-intercept method, I chose this approach using Math.Atan2 :

// as an extension method
public static bool Intersects(this Vector2 v, LineSegment s) {
    //  check from line segment start perspective
    var reference = Math.Atan2(s.Start.Y - s.End.Y, s.Start.X - s.End.X);
    var aTanTest = Math.Atan2(s.Start.Y - v.Y, s.Start.X - v.X);

    //  check from line segment end perspective
    if (reference == aTanTest) {
        reference = Math.Atan2(s.End.Y - s.Start.Y, s.End.X - s.Start.X);
        aTanTest = Math.Atan2(s.End.Y - v.Y, s.End.X - v.X);
    }

    return reference == aTanTest;
}

The first check reference determines the arcTan from the start point of the line segment to it's end-point. Then from the start point perspective, we determine the arcTan to the vector v .

If those values are equal, we check from the perspective of the end-point.

Simple and handles horizontal, vertical and all else in between.

Answer 11

Could you be more specific?

What programming language are you talking about?

What environment are you talking about?

What "lines" are you talking about? Text? What point? XY on the screen?