mjhjmtu

I have a list of 2D points such as in the image.

coord = 0, 0, 10, 0, 20, 0, 30, 0, 25, 10, 0, 10, 0, 
5;

I would like to determine the longest "edge" length in a way that if 2 segments are in same line, they would be considered as part of 1 edge. For example, segments 1-2, 2-3, 3-4 are continuous and in the same line, so we consider it as 1 edge connecting Vertex 1 and Vertex 4. In this example, the longest edge length would be the distance from vertex 1 to vertex 4. How can I determine the longest edge length in Mathematica?

Update: The function in the original answer does not work for arbitrary polygons. The following seems to work

ClearAll[nonCollinearHull]
nonCollinearHull = Module[coords = #, 
 angles = ArcTan @@@ (Subtract @@@ Partition[#, 2, 1 , 1, 1]),
 rotation, lengths,
rotation = LengthWhile[Reverse[angles], # == angles[[1]] &];
lengths = Length /@ Split[RotateRight[angles, rotation]];
TakeList[RotateRight[coords, rotation], lengths][[All, 1]]] &;

Examples:

coord = 0, 0, 10, 0, 20, 0, 30, 0, 25, 10, 0, 10, 0, 5;
lines = Line /@ Partition[nonCollinearHull[coord], 2, 1];
longest = Last@SortBy[lines, N@ArcLength[#] &];
Graphics[EdgeForm[Gray], FaceForm, Polygon@coord, Blue, 
 PointSize[Large], Point@coord, Opacity[.5, Green], 
 AbsolutePointSize[15], Point[nonCollinearHull[coord]], 
 Thickness[.03], CapForm["Round"], Opacity[.5], Red, longest]

Using

SeedRandom[123]
coord2 = DeleteDuplicates[#[[FindShortestTour[#][[2]]]] & @
 DeleteDuplicates@RandomInteger[10, 50, 2];

we get

And with

SeedRandom[123]
coord3 = DeleteDuplicates[#[[FindShortestTour[#][[2]]]] &@ 
 DeleteDuplicates@RandomInteger[20, 200, 2]];

Alternatively, you can use MaximalBy to define longest:

SeedRandom[777777]
coord = MapIndexed[#2[[1]], # &, Accumulate[RandomInteger[-2, 2, 50]]];
lines = Line /@ Partition[nonCollinearHull[coord], 2, 1];
longest = MaximalBy[lines, N@ArcLength[#] &];
Graphics[EdgeForm[Gray], FaceForm, Line@coord, Blue, 
 PointSize[Large], Point@coord, Opacity[.5, Green], 
 AbsolutePointSize[15], Point[nonCollinearHull[coord]], 
 Thickness[.03], CapForm["Round"], Opacity[.5], Red, longest]

Original answer:

Using the function noncollinearF from this answer:

ClearAll[noncollinearF]
noncollinearF[verts_] := Function[k, Nor @@ (RegionMember[ConvexHullMesh[#], k] & /@ 
 Subsets[Complement[verts, k], 2])]

lines = Line /@ Partition[Pick[#, noncollinearF[#] /@ #], 2, 1, 1, 1]& @ coord;
longest = Last@SortBy[lines, N@ArcLength[#] &]; 
Graphics[EdgeForm[Gray], FaceForm, Polygon@coord, 
 Blue, PointSize[Large], Point@coord, 
 Thickness[.03], CapForm["Round"], Opacity[.5], Red, longest]