Suppose that I have two points in the xy plane: pt1 and pt2init. pt1 is fixed in space, while pt2init is supplied by the user (but its returned value will in general be different). I wish to write a function that will return the coordinates of a point pt2result that is located a user-specified distance d from pt1 along the line connecting pt1 and pt2init and that is closest to pt2init.

I wrote the following function f (and its supporting function dist2D, which finds the distance between two points with x- and y-coordinates):

dist2D[pt1_, pt2_] := 
 Sqrt[(pt2[[1]] - pt1[[1]])^2 + (pt2[[2]] - pt1[[1]])^2]

f[pt1_, pt2_, d_] := Module[{line, x2result, y2result, pt2result},
  line = Normal[LinearModelFit[{pt1, pt2}, xvar, xvar]];
  Solve[{dist2D[{pt1[[1]], pt1[[2]]}, {x2result, y2result}] == d,
    y2result == (line /. xvar -> x2result)},
   {x2result, y2result}];
  pt2result = {x2result, y2result}

But when I run it:

f[{0.77825, 0.551441676}, {0.7075, 0.67398427}]

it does not return anything. Is it possible to do constrained optimization like this?

  • 2
    $\begingroup$ BTW: your dist2D[] is built-in as EuclideanDistance[]. $\endgroup$ May 28, 2012 at 16:42

2 Answers 2



You could use a very simple function:

f[p1_, p2_, d_] := Normalize[p2 - p1] d + p1

This will return a point on the line connecting p1 and p2, a distance d from p1 towards p2.


Using vector arithmetic is usually very convenient for analytic geometry. Fortunately in Mathematica there's no need to separate the components of the vectors (e.g. writing p1[[1]]). We can add two vectors directly. Normalize[p2-p1] will construct the unit vector pointing from p1 in the direction of p2. Multiplying it by d and adding it to p1 gives what you asked for.

Trying it out

p1 = {0, 0};
 Graphics[{PointSize[0.03], Point[p1], 
   Text["\!\(\*SubscriptBox[\(P\), \(1\)]\)", p1, {-2, 2}], Point[p2],
    Text["\!\(\*SubscriptBox[\(P\), \(2\)]\)", p2, {-2, 2}], 
   Line[{p1, p2}], Red, Point[f[p1, p2, d]]}, 
  PlotRange -> {{-1, 1}, {-1, 1}}],
 {{p2, {.5, .5}}, Locator}, {{d, 0.2}, 0, 1}

Mathematica graphics


(Too long for a comment.)

Only simple math is needed; no optimization required at all. Consider the vector equation $\mathbf x(t)=(1-t)\mathbf p_1+t\mathbf p_2$ for the line segment joining $\mathbf p_1$ and $\mathbf p_2$. The distance between $\mathbf p_1$ and $\mathbf x(t)$ is (check this!) $dt$, where $d$ is the length of the segment joining $\mathbf p_1$ and $\mathbf p_2$. You should now be able to reckon out the value of $t$ needed so that it is at a given distance from $\mathbf p_1$.


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