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This is an extraordinarily simple question but I can't find the answer so I'm bracing myself for the ridicule.

Given a vector: enter image description here

(in this case going from 2,3 to 9,9) I want to start at 2,3 and travel up the vector a distance of length l.

How do I calculate the new point (x,y)? Thanks!

Here's the code (after implementing Dmitry's equation):

 double MetersMoved = ((Dinosaurs[i].Speed * 1000) / (60 * 60)) * TimeStepInSecs;
 double fi = Math.Atan2((Dinosaurs[i].Goal.Y - Dinosaurs[i].Head.Y),(Dinosaurs[i].Goal.X - Dinosaurs[i].Head.X));
 Dinosaurs[i].Head.X = Dinosaurs[i].Head.X + MetersMoved * Math.Cos(fi);
 Dinosaurs[i].Head.Y = Dinosaurs[i].Head.Y + MetersMoved * Math.Sin(fi);

The dinosaurs are aligning themselves on the correct vector but not advancing (one should be moving 2 pixels and the other about 7).

Dmitry's answer is correct.

zetar
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  • What have you tried? What have you googled? Done anything with trigonometry? `Math.Cos`? `Math.Sin`? `Math.Atan2`? – Chris Sinclair Jul 18 '13 at 13:36
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    x_new = x + l * Math.Cos(fi); y_new = y + l * Math.Sin(fi); where fi is angle, in your case Math.Atan2(9 - 3, 9 - 2) – Dmitry Bychenko Jul 18 '13 at 13:38
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    Looks like the solution is here: http://stackoverflow.com/questions/1800138/given-a-start-and-end-point-and-a-distance-calculate-a-point-along-a-line?rq=1 – zetar Jul 18 '13 at 13:38
  • That implementation that I linked to isn't correctg. What is (fi) in your equation? – zetar Jul 18 '13 at 13:49

2 Answers2

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You can find the length of the current segment and calculate the proportion of which you want to expand (or contract) the segment.

(2,3) to (9,9)

(9-2)^2 + (9-3)^2 = (c)^2

7^2 + 6^2 = (c)^2

define a such that (a^2)(c)^2 = l^2

(a^2)(7^2 + 6^2) = (a^2)(c)^2 = l^2

(7a)^2 + (6a)^2 = (ca)^2 = l^2 => l = ca => a = l/c

You can then extend the x and y coordinate acordingly

(2+7*l/c, 3+6*l/c) is your new endpoint.

Point p1 = new Point(2.0, 3.0);
Point p2 = new Point(9.0, 9.0);
double l = 25.0;

Point vector = new Point(p2.X - p1.X, p2.Y - p1.Y);
double c = Math.Sqrt(vector.X * vector.X + vector.Y * vector.Y);
double a = l / c;

Point p3 = new Point(p1.X + vector.X * a, p1.Y + vector.Y * a);

//Writes "25"
Console.WriteLine(Math.Sqrt((p3.X - p1.X) * (p3.X - p1.X) + (p3.Y - p1.Y) * (p3.Y - p1.Y)));

Point is just a struct I to hold a double X and double Y.

Cemafor
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The solution (trigonometry) could be like this:

  // Origin point (2, 3)
  Double origin_x = 2.0;
  Double origin_y = 3.0;  

  // Point you are moving toward
  Double to_x = 9.0;
  Double to_y = 9.0;

  // Let's travel 10 units 
  Double distance = 10.0;

  Double fi = Math.Atan2(to_y - origin_y, to_x - origin_x);

  // Your final point
  Double final_x = origin_x + distance * Math.Cos(fi);
  Double final_y = origin_y + distance * Math.Sin(fi);
Dmitry Bychenko
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    There's absolutely no need at all for trig. No need for Atan2. It just obscures the simple vector operations. Just form the vector with `to-origin` and then re-normalise to be the desired length (you'll need Euclidean length calc). Finally add that onto `origin`. – David Heffernan Jul 18 '13 at 13:56