MathematicsIf the equations of the perpendicular bisectors of the sides AB and AC of a ΔABC are x−y+5=0 and x+2y=0respectively and if A is (1, -2) then the equation of the perpendicular bisector of the side BC is

If the equations of the perpendicular bisectors of the sides AB and AC of a ΔABCarexy+5=0andx+2y=0respectively and if A is (1, -2) then the equation of the perpendicular bisector of the side BC is

  1. A

    3x+3y+5=0

  2. B

    9x23y+40=0

  3. C

    6x+15y=5

  4. D

    23x14y+100=0

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    Solution:

    Ifxy+5=0isrbisectorofABthenBisImageofA(1,2)=(x1,  y1)h11=k+21=2(1+2+5)1+1h11=k+21=8B(h  k)=(7,  6)Ifx+2y=0isrbisectorofACthencisImageofA(1,2)h11=k+22=2(14)1+4h11=k+22=65

    h=65+1115,k=1252,25C(h  k)=(115,  25)slopeofBC¯=256115+7=2846=1423MidpointofBC=(11572,  25+62)(125,  165)EquationofrbisectorofBC¯isy165=2314(x+125)5y16=2314(5x+12)70y224=115x+276115x70y+500=0

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