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    <p>For each pair of adjacent vertices A,B:</p> <ol> <li><p>construct a vector from A to B, call it p</p></li> <li><p>now construct a vector from A to your test point X call it q</p></li> <li><p>the dot product formula for a pair of vectors is p.q = |p||q|cosC where C is the angle between the vectors.</p></li> <li><p>so if p.q/|p||q| == 1 then the points AX and AB are co-linear. Working on a computer, you will want 1 - p.q/|p||q| &lt; some_small_value depending on how accurate you want to be.</p></li> <li><p>also need to check that |q| &lt; |p| (ie X is closer to A than B)</p></li> </ol> <p>if 4&amp;5 are true your point is on the border.</p> <p>Edit</p> <p>The other way I think I've seen this done is to take your test point X, and construct a line through X perpendicular to the line between A and B. Find where this line and the line A->B cross. Work out the distance from X to this crossing point, if that is sufficiently small you consider the point as being on the line.</p> <p>Edit -- fun little exercise!</p> <p>Posted some code that was wrong earlier due to me misremembering some maths. Had a play in Pythonista on the train home and came up with this which seems to basically work. Have left the maths proof out as editing posts on iPad is painful!</p> <p>Not much testing done, no testing for division by zero etc, caveat user.</p> <pre><code> # we determine the point of intersection X between # the line between A and B and a line through T # that is perpendicular to the line AB (can't draw perpendicular # in ascii, you'll have to imagine that angle between AB and XT is 90 # degrees. # # B # / #. X # / \ # / T # A # once we know X we can work out the closest the line AB # comes to T, if that distance is 0 (or small enough) # we can consider T to be on the line import math # work out where the line through test point t # that is perpendicular to ab crosses ab # # inputs must be 2-tuples or 2-element lists of floats (x,y) # returns (x,y) of point of intersection def intersection_of_perpendicular(a,b,t): if a[0] == b[0]: return (a[0],t[1]) if a[1] == b[1]: return (t[0],a[1]) m = (a[1] - b[1])/(a[0] - b[0]) #slope of ab x_inter = (t[1] - a[1] + m*a[0] + (1/m)*t[0])*m/(m**2 + 1) y_inter = m*(x_inter - a[0]) + a[1] y_inter2 = -(1/m)*(x_inter - t[0]) + t[1] #print '...computed ',m,(x_inter, y_inter), y_inter2 return (x_inter, y_inter) # basic Pythagorean formula for distance between two points def distance(a,b): return math.sqrt( (a[0]-b[0])**2 + (a[1]-b[1])**2 ) # check if a point is within the box defined by a,b at # diagonally opposite corners def point_in_box(a,b,t): xmin = min(a[0],b[0]) xmax = max(a[0],b[0]) ymin = min(a[1],b[1]) ymax = max(a[1],b[1]) x_in_bounds = True if xmax != xmin: x_in_bounds = xmin &lt;= t[0] &lt;= xmax y_in_bounds = True if ymax != ymin: y_in_bounds = ymin &lt;= t[1] &lt;= ymax return x_in_bounds and y_in_bounds # determine if point t is within 'tolerance' distance # of the line between a and b # returns Boolean def is_on_line_between(a,b,t,tolerance=0.01): intersect = intersection_of_perpendicular(a,b,t) dist = distance(intersect, t) in_bounds = point_in_box(a,b,t) return in_bounds and (dist &lt; tolerance) a = (0,0) b = (2,2) t = (0,2) p = intersection_of_perpendicular(a,b,t) bounded = point_in_box(a,b,t) print 'd ',distance(p,t), ' p ',p, bounded a = (0,2) b = (2,2) t = (1,3) p = intersection_of_perpendicular(a,b,t) bounded = point_in_box(a,b,t) print 'd ',distance(p,t),' p ',p, bounded a = (0.0,2.0) b = (2.0,7.0) t = (1.7,6.5) p = intersection_of_perpendicular(a,b,t) bounded = point_in_box(a,b,t) on = is_on_line_between(a,b,t,0.2) print 'd ',distance(p,t),' p ',p, bounded,on </code></pre>
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    1. COthanks. Could i ask to convert in a Python code because it's not very clear :D!
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    2. CObit busy right now, might have a go at it tonight if i have time.
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    3. COthanks I am curious to test your idea, Thanks again
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