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(sorry, low level so I can't attach the image in the post)

  • a is given.
  • b is given.
  • c is given.
  • For X, I don't have a clue where he is.

I was thinking about rect / vector instersection, projection, and other things but still I couldn't come with a solution. If it's computer light, the better.

Sorry for not making clear before: ab and bx are not always orthogonal.

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If "ab and bx are not always orthogonal" you have to specify the angle between them or give an extra condition, if you want do determine x. –  Américo Tavares Mar 2 '11 at 23:37
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2 Answers 2

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From the drawing it seems that the segments $ab$ and $bx$ are orthogonal. So $x$ is the intersection of $ac$ and the line orthogonal to $ab$ that passes through $b$.

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Sorry about not making that clear, but ab and bx are not always orthogonal. –  Veehmot Mar 2 '11 at 23:31
    
In the ended, I did it like you told me, because it doesn't matter anyway if it's orthogonal or not. You got the answer for being the fist one who told me this. Thanks for your time. –  Veehmot Mar 4 '11 at 1:22
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You have a triangle with vertices $a$, $b$ and $x$ which looks right-angled at $b$; you also have a point on the hypotenuse $c$. So you need to draw the perpendicular line to $ab$ at $b$ and extend the line $ac$; where the two lines intersect must be $x$ (unless $ac$ is perpendicular to $ab$).

If you want a more detailed solution, than you could give us some information about how you are expressing the locations of $a$, $b$ and $c$.

EDIT in light of comment. If you know nothing except the location of $a$, $b$ and $c$, not even the angle at $b$, then $x$ could be anywhere on the line $ac$ (possibly required to be beyond $c$)

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Thanks for your answer! –  Veehmot Mar 4 '11 at 1:23
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