[" Dramper "199],[" bisectors of the sides "AB" and "AC" are "x+y=0" and "y-x],[=0" respectively.If "A-=(5,7)" then find the equation of side "],[BC.]

In triangle ABC, the equation of the right bisectors of the sides AB and AC are x+y=0 and y-x=0, respectively.If A-=(5,7), then find the equation of side BC.

In triangle ABC, the equation of the right bisectors of the sides AB and AC are x+y=0 and y-x=0. respectively. If A-=(5,7) the find the equation of side BC.

In triangle A B C , the equation of the right bisectors of the sides A B and A C are x+y=0 and y-x=0 , respectively. If A-=(5,7) , then find the equation of side B Cdot

In triangle A B C , the equation of the right bisectors of the sides A B and A C are x+y=0 and y-x=0 , respectively. If A-=(5,7) , then find the equation of side B Cdot

In triangle A B C , the equation of the right bisectors of the sides A B and A C are x+y=0 and y-x=0 , respectively. If A-=(5,7) , then find the equation of side B Cdot

In Delta ABC equation of the right bisectors of the sides Ab and AC are x+y = 0 and x - y = 0 respectively .If A = (5,7) then equation of side BC is

The equations of the perpendicular bisectors of the sides AB and AC of triangle ABC are x-y+5=0 and x+2y=0, respectively.If the point A is (1,-2) ,then find the equation of the line BC .

The equations of the perpendicular bisector of the sides AB and perpendicular bisector of the sides AB and AC of a /_ABC are x-y+5=0 and x+2y=0 respectively,if the point Ais (1,-2), then the equation ofthe line BC is

The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are x -y+5=0 and x+2y=0 respectively . If the point A is (1,-2) , find the equation of the line BC.