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Vector Algebra

Question
CBSEENMA12033970

Prove by vector method that the line segment joining the mid-points of the diagonals of trapezium is parallel to the parallel sides and equal to help of there difference.

Solution
Let ABCD be a trapezium and 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 space straight d with rightwards arrow on top be the position vectors of A, B, C, D respectively. Let P be the mid-point of diagonal AC and Q be mid-point of diagonal BD.
therefore  position vectors of straight P with rightwards arrow on top space straight Q with rightwards arrow on top space are space fraction numerator straight a with rightwards arrow on top space plus space straight c with rightwards arrow on top over denominator 2 end fraction comma space fraction numerator straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top over denominator 2 end fraction respectively
                   PQ with rightwards arrow on top space space equals straight P. straight V. space of space straight Q space minus space straight P. straight V. space of space straight P
                                  equals space fraction numerator straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top over denominator 2 end fraction space minus fraction numerator straight a with rightwards arrow on top space plus space straight c with rightwards arrow on top over denominator 2 end fraction space equals space 1 half left parenthesis straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top space minus space straight a with rightwards arrow on top space minus space straight c with rightwards arrow on top right parenthesis
equals space 1 half open square brackets left parenthesis straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis space minus space left parenthesis straight c with rightwards arrow on top space minus space stack straight d right parenthesis with rightwards arrow on top close square brackets
equals space 1 half open square brackets AB with rightwards arrow on top space minus space DC with rightwards arrow on top close square brackets

Now,        AB || DC         rightwards double arrow space space space space space AB with rightwards arrow on top space equals space straight lambda. space DC with rightwards arrow on top
therefore space space space space PQ with rightwards arrow on top space equals space 1 half open square brackets straight lambda. space stack DC space with rightwards arrow on top space minus space DC with rightwards arrow on top close square brackets space equals space 1 half left parenthesis straight lambda minus 1 right parenthesis space DC with rightwards arrow on top
therefore space space space PQ thin space vertical line vertical line thin space DC space space or space space space AB.
therefore space space space PQ space is space parallel space to space the space parallel space sides.
Again,    open vertical bar AB with rightwards arrow on top close vertical bar space minus space open vertical bar DC with rightwards arrow on top close vertical bar space equals space straight lambda space open vertical bar DC with rightwards arrow on top close vertical bar space minus space open vertical bar DC close vertical bar space equals space left parenthesis straight lambda minus 1 right parenthesis thin space DC with rightwards arrow on top
Now,  PQ with rightwards arrow on top space equals space 1 half left parenthesis straight lambda minus 1 right parenthesis space DC with rightwards arrow on top space space space rightwards double arrow space space space space space open vertical bar PQ with rightwards arrow on top close vertical bar space equals space 1 half left parenthesis straight lambda minus 1 right parenthesis space open vertical bar DC with rightwards arrow on top close vertical bar
rightwards double arrow space space space space space open vertical bar PQ with rightwards arrow on top close vertical bar space equals space 1 half open parentheses open vertical bar AB with rightwards arrow on top close vertical bar space minus space open vertical bar DC with rightwards arrow on top close vertical bar close parentheses
∴   PQ is half of the difference of parallel sides.
Hence the result.