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Mid-points of the sides AB and AC of a triangle ABC are ( 3 , 5 ) and ( -3 , -3 ) respectively,then the length of the side BC is

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    let A = (h,k). Then

    B=(h+2(3-h),k+2(5-k))=(6-h,10-k)
    C=(h+2(-3-h),k+2(-3,k))=(-6-h,-6-k)

    BC^2 = 12^2+16^2 = 340
    BC=√340

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