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Question

The three vertices of a parallelogram ABCD are A(3,4),B(1,3) and C(6,2). Find the coordinates of vertex D and find the area of ABCD.

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Solution

Given A(3,4),B(1,3),C(6,2)

O is M.P of AC=(362,4+22)=(32,1)

O=(32,1)

Similarly, O is MP of BD

32=1+a2, 1=3+62

a=2,b=+1

D=(2,1)

Area of parallelogram ABCD = Ar of ABC + Ar ADC

Area of ABC=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

=12[3(32)1(2+4)6(4+3)]

=12[156+6]=|15|2=152

Area of ADC=12[3(12)2(2+4)6(41)]

=12[312+30]=152

Area of parallelogram ABCD=152+152=15

1348420_1231181_ans_466d0503cd5c4c9d91135ed354656178.PNG

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