No trig needed, this is linear algebra. The two lines have the form y = mAx + bA and y = mBx + bB. The point where both are satisfied is the intersection you want.

Solving for x, we get x = (bB - bA)/(mA - mB). Similarly, y = (mBbA - mAbB) / (mB - mA).

That is with the numeric values

  1. mA = (Dy - A) / Dx
  2. mB = (Cy - B) / Cx
  3. bA = A
  4. bB = B
I hope I haven't made a mess of this :) It will simplify things a bit that Dx and Cx are equal.

After Compline,
Zaxo

In reply to Re: A little schoolboy math (but my schooldays were a long time ago). by Zaxo
in thread A little schoolboy math (but my schooldays were a long time ago). by BrowserUk

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