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  1. PO
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    <p>If you assume a and b are the opposite corners of a rectangle</p> <pre><code>a = (xa, ya) b = (xb, yb) </code></pre> <p>then the top right rectangle point is <code>c1 = (max(xa,xb), max(ya,yb))</code> and the bottom left rectangle point is <code>c2 = (min(xa,xb), min(ya,yb))</code></p> <p>Assuming that <code>xa != xb</code> and <code>ya != yb</code></p> <pre><code> (xa, ya) A C1 (max(xa, xb), max(ya, yb)) o----------o |\ | | \ | | \ | | \ | | \ | | \ | | \ | | \ | | \ | o----------o (min(xa, xb), min(ya, yb)) C2 B (xb, yb) </code></pre> <p>If your diagonal is going the other way (to test this see if xa > xb) you need to swap min for max on the x</p> <pre><code>(min(xa, xb), max(ya, yb)) C3 A' o----------o | /| | / | | / | | / | | / | | / | | / | | / | | / | o----------o B' C4 (max(xa, xb), min(ya, yb)) </code></pre> <p>And if you're interested, the full set of solutions actually lies on the circle:</p> <p><img src="https://i.stack.imgur.com/PY9qO.png" alt="Set of solutions"></p> <p>To compute this, suppose we have two points <code>A = (xa, ya)</code> and <code>B = (xb, yb)</code>. Then the center point of this circle is <code>c = (0.5 (xa + xb), 0.5 (ya + yb))</code> - just the midpoint of the A and B. The radius of the circle is <code>r = sqrt( (xb - xa)^2 + (yb - ya)^2) / 2</code> - using pythagoras' theorem to get the length of the line and halving it. Then any point on the circle can be defined by <code>p = c + (rcos(u), rsin (u))</code> for some angle <code>u</code>. There are 2 angles which give you the points <code>p = A</code> and <code>p = B</code> so these values of <code>u</code> are not good solutions. You can write out the equation and solve it for these 2 points to give you the values of u which you cannot use.</p>
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    1. POUsing the coordinates of two of a triangle's vertices to calculate the coordinates of the third
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    1. USMatt Esch
    1. USMatt Esch
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    1. COThe length of (a,c) is fixed, there are only two solutions, see the image attached to question.
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      1. USViesturs
    2. COThe question actually changed after a number of comments were made. The original question assumed there was 1 line joining 2 points. and the second and third line were unknown. If 2 sides of a right angle triangle are known, trivial pythagoras solves the problem.
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      1. USMatt Esch
 

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