posted by Anamika .
the equation of the perpendicular bisectors of sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0
respectively. if the point is A (1,-2), find the equation of line BC.
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 x-y+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 2x-y-4=0
x-y+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 2x-y-4=0 intersect at Q=(8/5,-4/5).
Since Q-A=(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
y-6 = (-14/23)(x+7)