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Vector Algebra

Question
CBSEENMA12034149

Using vectors, prove that the perpendicular bisectors of the sides of a triangle are concurrent.

Solution

Let ABC be the given triangle and O be the point of intersection of perpendicular bisectors OD and OE of sides BC and CA respectively.
Let F be mid-point of AB. Join O to F.
Take O as origin.

Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top be the position vectors of A, B, C respectively.
            therefore space space OA space equals space straight a with rightwards arrow on top comma space space OB with rightwards arrow on top space equals space straight b with rightwards arrow on top comma space space OC with rightwards arrow on top space equals straight c with rightwards arrow on top
The position vectors of D, E, F are fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction comma space space fraction numerator straight c with rightwards arrow on top plus straight a with rightwards arrow on top over denominator 2 end fraction comma space fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator 2 end fraction respectively.
Since               OD space perpendicular space BC
therefore space space space space space space space OD with rightwards arrow on top. space BC with rightwards arrow on top space equals space 0
rightwards double arrow space space space space open parentheses fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction close parentheses. space left parenthesis straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space space 0 space space space space rightwards double arrow space space 1 half left parenthesis straight c with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space left parenthesis straight c with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space space space 1 half left parenthesis straight c with rightwards arrow on top squared minus straight b with rightwards arrow on top squared right parenthesis space equals space 0
Again, OE space perpendicular space CA
therefore space space space OE with rightwards arrow on top. space CA with rightwards arrow on top space equals space 0 space space space space rightwards double arrow space space space open parentheses fraction numerator straight c with rightwards arrow on top plus straight a with rightwards arrow on top over denominator 2 end fraction close parentheses. space left parenthesis straight a with rightwards arrow on top. space straight c with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space space space 1 half left parenthesis straight a with rightwards arrow on top squared minus straight c with rightwards arrow on top squared right parenthesis space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Adding (1) and (2), we get,
              1 half left parenthesis straight a with rightwards arrow on top space minus straight b with rightwards arrow on top squared right parenthesis space space space space rightwards double arrow space space space 1 half left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space space space open parentheses fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator 2 end fraction close parentheses space. space left parenthesis straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis space equals space stack OF. with rightwards arrow on top space AB with rightwards arrow on top space equals space 0
rightwards double arrow space space space space OF space perpendicular space AB
therefore    perpendicular bisectors meet in a point.