# find equation of triangle sides in cartesian system

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?

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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 point –  dato datuashvili Jun 19 '12 at 12:32
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. –  Henning Makholm Jun 19 '12 at 12:34
i can find symmetry point of given point about some line,but reflecting i dont know –  dato datuashvili Jun 19 '12 at 12:34
@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. –  Henning Makholm Jun 19 '12 at 12:37
ok i know coordinates of point A,but what about B or C? –  dato datuashvili Jun 19 '12 at 12:46

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 bisector –  dato datuashvili Jun 19 '12 at 12:25
Indeed, this appears to lead nowhere. –  Henning Makholm Jun 19 '12 at 12:34
now it works,thanks very much –  dato datuashvili Jun 19 '12 at 13:17
@Mohamed thanks for helping,you are welcome –  dato datuashvili Jun 19 '12 at 13:21