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it is know that one vertex of triangle is located at point $A(2,-4)$ and equation of angle bisector of two another angle is given
1.$x+y-2=0$

2.$x-3*y-6=0$ we have to find equation of sides of triangle i have found point where this two line intersect ,got $(3,-1)$,not dont know how to get corrdinates of B and C vertices,i know that bisector cuts angle into two equal parts,also i know theorem of angle bisector,but how can i use it to find coordinates?

3 Answers 3

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If you reflect the point $A$ about the bisector of the angle at $B$ you get a point somewhere on $BC$. You get another point on $BC$ by reflecting about the other bisector. Now you know two points on $BC$...

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    how can it help me?or how can i reflect point2012-06-19
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    There's probably some slick algebraic way to reflect a point about the line which escapes me at the moment, but you can always construct the perpendicular through the point and find the place on the perpendicular that is as far beyond the line as the line is from the given point.2012-06-19
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    i can find symmetry point of given point about some line,but reflecting i dont know2012-06-19
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    @dato: do you distinguish between "symmetry point of a given point about some line" and "reflecting the point about the line"? I can't imagine two different meanings for them.2012-06-19
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    ok i know coordinates of point A,but what about B or C?2012-06-19
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    That's what my answer is about! If there is something that is unclear about the answer, ask about that instead of just restating the question like the answer was not there at all. That's rude. **Think:** If you know the line BC and know a bisector through B, how would you find B itself?2012-06-19
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    sorry i just wanted to make sure that i have understood everything,i dont know equation of BC,just bisector2012-06-19
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    it should be intersection of this two line yes?2012-06-19
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Yes find the third bissector is without utility. I replace this by explain mors the Henning Makholm solution: If we name $D_B$ and $D_C$ the bissectors and $A'=S_{D_B} (A)$ and $A''=S_{D_C}(A)$, then $A'$ and $A''$ are in line $(BC)$, so $B$ is the point intersection of $D_B$ and $(A'A'')$ and $C$ is the point intersection of $D_C$ and this $(A'A'')$. You need translate this using coordinates .

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    i am not looking equation for third bisector2012-06-19
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    Indeed, this appears to lead nowhere.2012-06-19
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    now it works,thanks very much2012-06-19
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    @Mohamed thanks for helping,you are welcome2012-06-19
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Find the image of A in both bisectors.These points will lie on the base.Now find the equation of the base from these two points.Now, solve the base with the given angle bisectors to get the points of other vertices and then finally find the equations of the remaining two sides.

Hope this helps.