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

Question

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

Examples are appreciated, if possible.

Answer 1

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?

Answer 2

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 3

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:

 Y-Intersect = - Slope * X1 / Y2;

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.

Answer 4

Equation of the line is:

y = mx + c

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

Answer 5

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 6

Given to 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 7

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 8

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.