Shorten a line by a number of pixels

Question

I'm drawing a custom diagram of business objects using .NET GDI+. Among other things, the diagram consists of several lines that are connecting the objects.

In a particular scenario, I need to shorten a line by a specific number of pixels, let's say 10 pixels, i.e. find the point on the line that lies 10 pixels before the end point of the line.

Imagine a circle with radius r = 10 pixels, and a line with start point (x1, y1) and end point (x2, y2). The circle is centered at the end point of the line, as in the following illustration.

How do I calculate the point marked with a red circle, i.e. the intersection between circle and line? This would give me the new end point of the line, shortening it by 10 pixels.

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Solution

Thank you for your answers from which I was able to put together the following procedure. I named it LengthenLine, since I find it more natural to pass a negative number of pixels if I want the line shortened.

Specifically, I was trying to put together a function that could draw a line with rounded corners, which can be found here.

public void LengthenLine(PointF startPoint, ref PointF endPoint, float pixelCount)
{
  if (startPoint.Equals(endPoint))
    return; // not a line

  double dx = endPoint.X - startPoint.X;
  double dy = endPoint.Y - startPoint.Y;
  if (dx == 0)
  {
    // vertical line:
    if (endPoint.Y < startPoint.Y)
      endPoint.Y -= pixelCount;
    else
      endPoint.Y += pixelCount;
  }
  else if (dy == 0)
  {
    // horizontal line:
    if (endPoint.X < startPoint.X)
      endPoint.X -= pixelCount;
    else
      endPoint.X += pixelCount;
  }
  else
  {
    // non-horizontal, non-vertical line:
    double length = Math.Sqrt(dx * dx + dy * dy);
    double scale = (length + pixelCount) / length;
    dx *= scale;
    dy *= scale;
    endPoint.X = startPoint.X + Convert.ToSingle(dx);
    endPoint.Y = startPoint.Y + Convert.ToSingle(dy);
  }
}

Answer 1

Find the direction vector, i.e. let the position vectors be (using floats) B = (x2, y2) and A = (x1, y1), then AB = B - A. Normalize that vector by dividing by its length ( Math.Sqrt(x*x + y*y) ). Then multiply the direction vector AB by the original length minus the circle's radius, and add back to the lines starting position:

double dx = x2 - x1;
double dy = y2 - y1;
double length = Math.Sqrt(dx * dx + dy * dy);
if (length > 0)
{
    dx /= length;
    dy /= length;
}
dx *= length - radius;
dy *= length - radius;
int x3 = (int)(x1 + dx);
int y3 = (int)(y1 + dy);

Edit: Fixed the code, aaand fixed the initial explanation (thought you wanted the line to go out from the circle's center to its perimeter :P)

Answer 2

You can use similar triangles. For the main triangle, d is the hypotenuses and the extension of r is the vertical line that meets the right angle. Inside the circle you will have a smaller triangle with a hypotenuses of length r.

r/d = (x2-a0)/(x2-x1) = (y2-b0)/(y2-y1)

a0 = x2 + (x2-x1)r/d

b0 = y2 + (y2-y1)r/d

Answer 3

I'm not sure why you even had to introduce the circle. For a line stretching from (x2,y2) to (x1,y1), you can calculate any point on that line as:

(x2+p*(x1-x2),y2+p*(y1-y2))

where p is the percentage along the line you wish to go.

To calculate the percentage, you just need:

p = r/L

So in your case, (x3,y3) can be calculated as:

(x2+(10/L)*(x1-x2),y2+(10/L)*(y1-y2))

For example, if you have the two points (x2=1,y2=5) and (x1=-6,y1=22), they have a length of sqrt(7² + 17² or 18.38477631 and 10 divided by that is 0.543928293. Putting all those figures into the equation above:

  (x2 + (10/l)      * (x1-x2) , y2 + (10/l)      * (y1-y2))
= (1  + 0.543928293 * (-6- 1) , 5  + 0.543928293 * (22- 5))
= (1  + 0.543928293 * -7      , 5  + 0.543928293 * 17     )
= (x3=-2.807498053,y3=14.24678098)

The distance between (x3,y3) and (x1,y1) is sqrt(3.192501947² + 7.753219015²) or 8.384776311, a difference of 10 to within one part in a thousand million, and that's only because of rounding errors on my calculator.