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copied!<p>Here's Ruby disguised as Pseudo-Code, assuming <code>Point</code> objects each have a <code>x</code> and <code>y</code> field.</p> <pre><code>def GetClosestPoint(A, B, P) a_to_p = [P.x - A.x, P.y - A.y] # Storing vector A->P a_to_b = [B.x - A.x, B.y - A.y] # Storing vector A->B atb2 = a_to_b[0]**2 + a_to_b[1]**2 # **2 means "squared" # Basically finding the squared magnitude # of a_to_b atp_dot_atb = a_to_p[0]*a_to_b[0] + a_to_p[1]*a_to_b[1] # The dot product of a_to_p and a_to_b t = atp_dot_atb / atb2 # The normalized "distance" from a to # your closest point return Point.new( :x => A.x + a_to_b[0]*t, :y => A.y + a_to_b[1]*t ) # Add the distance to A, moving # towards B end </code></pre> <hr> <p>Alternatively:</p> <p>From <a href="http://en.wikipedia.org/wiki/Line-line_intersection" rel="noreferrer"><strong>Line-Line Intersection</strong></a>, at Wikipedia. First, find Q, which is a second point that is to be had from taking a step from P in the "right direction". This gives us four points.</p> <pre><code>def getClosestPointFromLine(A, B, P) a_to_b = [B.x - A.x, B.y - A.y] # Finding the vector from A to B This step can be combined with the next perpendicular = [ -a_to_b[1], a_to_b[0] ] # The vector perpendicular to a_to_b; This step can also be combined with the next Q = Point.new(:x => P.x + perpendicular[0], :y => P.y + perpendicular[1]) # Finding Q, the point "in the right direction" # If you want a mess, you can also combine this # with the next step. return Point.new (:x => ((A.x*B.y - A.y*B.x)*(P.x - Q.x) - (A.x-B.x)*(P.x*Q.y - P.y*Q.x)) / ((A.x - B.x)*(P.y-Q.y) - (A.y - B.y)*(P.y-Q.y)), :y => ((A.x*B.y - A.y*B.x)*(P.y - Q.y) - (A.y-B.y)*(P.x*Q.y - P.y*Q.x)) / ((A.x - B.x)*(P.y-Q.y) - (A.y - B.y)*(P.y-Q.y)) ) end </code></pre> <p>Caching, Skipping steps, etc. is possible, for performance reasons.</p>

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