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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.

  • maths -

    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.

  • maths -

    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)

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