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I need to find intersection point of two bisectors of KL, KC and CL line.

Coordinates : $C(2a+2, 2b), L(2-2a, 2b)$ and $K \left( \frac{2(a^3 - a^2 + ab^2 + b^2)}{a^2 + b^2}, 2 - \frac{4a}{a^2 + b^2} \right).$

I calculate that the bisector of LC is $x = 2$. Then we must calculate another bisector and find intersection point O.

I already have solution - intersection point is $O(2, 2\frac{a^2 + b^2 - a}{b})$ but I cant compute it by myself. Can you give me some tips or all answer? I think that it is very hard task to compute. I tried use mathematica but it gives me wrong answers. Thanks for every help.

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1 Answer 1

If I understand correctly, you are trying to compute the intersection of the edge bisectors of a triangle given by your three points. Is that right?

Is this homework? Here are some hints:

  1. A line is determined by a point and a direction vector. For two of the edge bisectors, can you write down a formula for this point and vector?

  2. Given two lines $\mathbf{q_1} + s\mathbf{v_1}$ and $\mathbf{q_2} + t\mathbf{v_2}$, their intersection occurs at the value of $s$ and $t$ where $\mathbf{q}_1-\mathbf{q}_2 +s\mathbf{v_1} - t\mathbf{v_2}=0$. What happens when you dot this equation by $\mathbf{v}_2^{\perp} = \{-{v_2}_y, {v_2}_x\}$?

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It's not a homework but own task. It's very hard for me. I'm using vectors to compute bisectors. But my solution isn't corrent ( it must be $(2,2\frac{a^2+b^2-a}{b})$ but i find solution about $(2, (a^5 + a^4b ... +$ very funny numbers$)/(a^2+b^2))$. What's wrong? Mayby the points K,L,C are wrong? Oh, i forgot. a and b > 0. –  John Smith Sep 16 '12 at 22:26

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