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copied!<p>Find the slope a1 of AB by dividing the y-difference with the x-difference; then draw a perpendicular line (with slope a2 = -1/a1, you need to solve for the offset (b2) by putting P's coordinates into y = a2*x + b2); then you have two lines (i.e. two linear equations), and you need to solve the intersection. That will be your closest point.</p> <p>Do the math right, and the function will be pretty trivial to write.</p> <p>To elaborate a bit:</p> <pre><code>Original line: y = a1 * x + b1 a1 = (By - Ay) / (Bx - Ax) <-- b1 = Ay - a1 * Ax <-- Perpendicular line: y = a2 * x + b2 a2 = -1/a1 <-- b2 = Py - a2 * Px <-- Now you have P which lies on both lines: y = a1 * x + b1 y = a2 * x + b2 --------------- subtract: 0 = (a1 - a2) * Px + (b1 - b2) x = - (b1 - b2) / (a1 - a2) <-- y = a1 * x + b1 <-- </code></pre> <p>Hope I didn't mess up somewhere :) <strong>UPDATE</strong> Of course I did. Serve me right for not working things out on paper first. I deserved every downvote, but I'd've expected someone to correct me. Fixed (I hope).</p> <p>Arrows point the way.</p> <p><strong>UPDATE</strong> Ah, the corner cases. Yeah, some languages don't handle infinities well. I did say the solution was language-free...</p> <p>You can check the special cases, they're quite easy. The first one is when the x difference is 0. That means the line is vertical, and the closest point is on a horizontal perpendicular. Thus, <code>x = Ax, y = Px</code>.</p> <p>The second one is when y difference is 0, and the opposite is true. Thus, <code>x = Px, y = Ay</code></p>

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