maths
posted by Anamika .
the equation of the perpendicular bisectors of sides AB and AC of a triangle ABC are xy+5=0 and x+2y=0
respectively. if the point is A (1,2), find the equation of line BC.

Hint:
Find the intersection of the two perpendicular bisectors, which gives the circumcentre O.
The distance AO=r is the radius of the circumscribed circle.
The circle with radius r, and centred at O will intersect AB and AC at B and C respectively. 
side AB is a line ┴ to xy+5=0, so it has slope 1. It is thus x+y+1=0
side AC is ┴ to x+2y=0, so it has slope 2. It is thus 2xy4=0
xy+5=0 and x+y+1=0 intersect at P=(3,2).
Since (3,2)A = (4,4), B=P+(4,4) = (7,6)
x+2y=0 and 2xy4=0 intersect at Q=(8/5,4/5).
Since QA=(3/5,6/5), C=Q+(3/5,6/5) = (11/5,2/5)
So, now you have B and C, and the line through those points is
y6 = (14/23)(x+7)