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    <p>This is a simple explanation of Il-Bhima's solution. The trick is to notice that what you want is to project that point orthogonally on the line, move it by that much, and then move it once again, in the same direction.</p> <p>For these types of problems, it's easier to work with a slightly more redundant representation for a line. Instead of <code>y = m x + b</code>, let's represent the line by a point <code>p</code> that is on the line and a vector <code>d</code> in the line's direction. Let's call this point <code>p = (0, b)</code>, the vector <code>d = (1, m)</code>, and your input point will be <code>p1</code>. The projected point on the line will be <code>pl</code> and your output point <code>p2</code>, then, is <code>p1 + 2 * (pl - p1) = 2 * pl - p1</code></p> <p>The formula you need here is the projection of a vector <code>v</code> onto a line which goes through the origin in direction <code>d</code>. It is given by <code>d * &lt;v, d&gt; / &lt;d, d&gt;</code> where <code>&lt;a, b&gt;</code> is the dot product between two vectors.</p> <p>To find <code>pl</code>, we have to move the whole problem so that the line goes through the origin by subtracting <code>p</code> from <code>p1</code>, using the above formula, and moving it back. Then, <code>pl = p + (d * &lt;p - p1, d&gt; / &lt;d, d&gt;)</code>, so <code>pl_x = p_x + (b * p1_x) / (1 + m * m)</code>, <code>pl_y = p_y + (m * p1_x) / (1 + m * m)</code>, and then use <code>p2 = 2 * pl - p1</code> to get the final values.</p>
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    1. POAlgorithm for reflecting a point across a line
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    1. USCarlos Scheidegger
    1. USCarlos Scheidegger
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    1. CO+1 That's a great explanation!
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      1. USIl-Bhima
 

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