The equations of perpendicular bisectors of two sides AB and AC of a triangle ABC are `x+y+1=0` and `x-y+1=0` respectively. If circumradius of `Delta` ABC is 2 units, then the locus of vertex A is

The equations of the perpendicular bisectors of the sides AB&AC of a triangle ABC are x-y+5=0&x+2y=0. find the equation of the bc

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

the equation of perpendicular bisectors of side AB,BC of triangle ABC are x-y=5,x+2y=0 respectively and A(1,-2) then coordinate of C

Let the equations of perpendicular bisectors of sides AC and of Delta ABCbe x+y=3 and x-y=1 respectively Then vertex A is is (0,0) The orthocentre of DeltaABC is

The equation of perpendicular bisectors of the side AB and AC a triangle ABC are x-y+5=0 and x+2y=0 respectively vertex A is (1,-2) Area of triangle ABC is (in sq.units) (A) (36)/(5) (B) (18)/(5) (C) (72)/(5) (D) None of these

The equation of perpendiculars of the sides AB and AC of triangle ABC are x-y-4=0 and 2x-y-5=0 respectively.If the vertex A is (-2,3) and point of intersection of perpendicular bisectors is ((3)/(2),(5)/(2)) find the equation of medians to the sides AB& AC respectively