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I'm looking for a way to draw an infinite line (a line with no end, also known as a ray) through 2 points. I can draw a line between 2 points with Line2D, no problem here.

Next the infinite part needed a look. With my simple mind I thought, lets multiply the x and y coordinates from the second point with 100 and redraw the line. This works, but only in simple cases.

For example here is a case in which it produces lines with different angles:

    g.setColor(Color.red);
    g2.setStroke(new BasicStroke(4.0f));
    g2.draw(new Line2D.Double(0, 61.632653061218946, 944, 217.25510204080692));
    g.setColor(Color.blue);
    g2.setStroke(new BasicStroke(1.0f));
    g2.draw(new Line2D.Double(0, 61.632653061218946, 944*10, 217.25510204080692*10));

This will first draw a fat red line, next it will draw a blue thin line.

The blue line has a different angle compared the the red line. Here's a screenshot to illustrate this effect:

enter image description here

Does somebody know a way to fix this, or maybe a better way to draw a infinite line through 2 points?

juan.facorro
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T. Akhayo
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2 Answers2

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By multiplying the second point by 10, you're not maintaining the x to y ratio. You need to subtract the origin before multiplying.

g2.draw(new Line2D.Double(0, 61.632653061218946, 944*10, (217.25510204080692-61.632653061218946)*10+61.632653061218946)); should work.

In general, g2.draw(new Line2D.Double(x0,y0,(x1-x0)*k+x0,(y1-y0)*k+y0) should work.

The math behind this is to consider the two points as vectors (a & b). Now, define a third and fourth vector: the difference vector (d) and the infinite end point vector (c). The difference vector is simply the difference between b and a; that is d = b-a. To find an infinite end point, we simply need to add infinity times the difference vector to the starting point (a). So, c = a + d*k, where k is the arbitrarily large constant. After replacing d, we have c = a+(b-a)*k.

RedGreasel
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  • Hang on a second, is it just me or shouldn't the point be `(9440,1494.591836734660794)`, and not `(9440,1556.22448979587974)`? You have to subtract out the intercept from the new y-value: `(y1-y0)*k - b` – lc. Sep 05 '12 at 17:58
  • Oops, I did make a mistake. Thanks. – RedGreasel Sep 05 '12 at 18:02
  • It should have been (9440, 1617.857142857098686), rather than what I had before (which is what you have for x=9440). I've elaborated on what I was doing. – RedGreasel Sep 05 '12 at 18:11
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    Thanks for your reply and taking the time to help me with this problem. – T. Akhayo Sep 05 '12 at 18:29
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Let's do the math for a second.

  • The first line is (0, 61.632653061218946)-(944, 217.25510204080692). Slope is rise/run, therefore m = 0.16485428917329234533898305084746.
  • The second line is (0, 61.632653061218946)-(9440, 2172.5510204080692); m = 0.22361423382911549300847457627119.

The slopes are different, which is just another way of saying the angles are different.

What you need to do is extend the line. You cannot just multiply both coordinates of one of the points by 10. First determine either an x or a y beyond the bounds of your canvas, then solve for the other value.

How do you do this?

  1. First, get the equation for the line. A line is defined by y=m*x+b, where m is the slope, and b is the y-intercept.

    1. We already know how to calculate the slope (rise/run = y2 - y1 / x2 - x1). We get 0.16485428917329234533898305084746
    2. Plug in the slope and solve for b (y - m*x), you get 61.632653061218946. In your case you already have this value as the y-intercept is the y-coordinate when x=0.
    3. You then get the equation y = 0.16485428917329234533898305084746 * x + 61.632653061218946

  2. Now, pick a sufficiently large x, say 10000. Plug this value in and solve for y. You get 1710.1755447941423993898305084746.

  3. Finally, draw your line to this new point, (0, 61.632653061218946)-(10000,1710.1755447941423993898305084746)

Great, now let's generalize this.

  • We have two points (x1, y1) and (x2, y2). We want to solve for (10000, y3).
  • Therefore, y3 = m*x3 + b, or y3 = m * 10000 + b.
  • We also know that b = y - m * x, so plugging this in and arbitrarily choosing point 1, y3 = m * 10000 + y1 - m * x1.
  • Ok, let's factor out the m: y3 = m * (10000 + x1) - y1.
  • We know m = (y2 - y1) / (x2 - x1), so plugging this in: y3 = ((y2 - y1) / (x2 - x1)) * (10000 + x1) - y1.

If your line doesn't start at x = 0, you will need to repeat this process for x = 0, meaning you should plot a line (0, ((y2 - y1) / (x2 - x1)) * x1 - y1)-(10000,((y2 - y1) / (x2 - x1)) * (10000 + x1) - y1).

Note: If x2 - x1 is 0, you have an infinite slope. This is a vertical line, and you'll have to handle this case separately.

lc.
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  • Watch out for extreme slopes, since you're dealing with floating point values. Especially, when the slope is infinity you have a problem. – RedGreasel Sep 05 '12 at 17:43
  • +1 for getting the correct answer first and correcting my answer. – RedGreasel Sep 05 '12 at 18:12
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    Thank you for your extensive reply, i had to read it multiple times to fully understand, but that is because i'm not very good at math and algo. But your explanation makes even sense to me. – T. Akhayo Sep 05 '12 at 18:23