arctan slope of line AB = angle of the line AC from the vertical because of the 90 degree angle between AB and AC.

     So x = 5 + 5 sin arctan (5/14) ≈ 6.68
          y = 15 - 5 cos arctan (5/14) ≈ 10.29

Last edited by irspow (2005-12-06 06:41:07)

Thanks both!

I will play with this and see if I can translate this to delphi pascal code. One thought (maybe I am misunderstanding your words), the 90 degree angle is at point B (AB and BC) not at point A (AB and AC). Anyway if this means your formula should look a bit different then I should be able to change this myself. I did think of the fact that I didn't mention to which side the 90 degree angle point, thanks for letting me know how to deal with that John.

Raoul

That's not how I did it John.  I just observed that if AB had an angle of arctan 14/5 above the horizontal then line C would have that same angle from the y axis or vertical because of the 90 degree angle between them.  Constructing that triangle using C as the hypotenuse is how I used the trigonomic functions.  Notice that the opposite side of this angle is horizontal and thus represents the change in x, hence the use of the sine function for the x position.

Your first problem was because of how I structured the equations.  By definition point A will have a smaller x value then point B.  A simple "if" statement can easily cure that.

     The second problem I hadn't thought about at all.  There is no slope in such a situation.  This would also seem a good place for an "if" statement.  Then you can just pick a location like half of delta y or whatever you feel is a good place for that instance.

      Anyway, I hope your project is coming along now.  Good luck.

Both answers are correct depending which direction you want to go.
I'm glad your computer program is working!!!!  I love to program too.
See image below:

Yes, it seems to be fine.

Well at least safra's program is working correctly now.

Still a long way to go though, but taking things step by step