A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. IF E is the mid - point of DC, then find the area of ΔADE.


Given,

A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD;


Let the fourth vertex of parallelogram be (x, y),


We know that, the diagonals of a parallelogram bisect each other



Mid - point of BD = Mid - point of AC



Mid - point of a line segment joining the points (x1, y1) and (x2, y2) =




8 + x = 15 x = 7


And



2 + y = 5 y = 3


So, fourth vertex of a parallelogram is D (7, 3)


Now,


Mid - point of side


DC =


E =


Area of ΔABC with vertices (x1, y1), (x2, y2) and (x3, y3);


= [x1(y2 - y3) + x2(y3 – y1) + x3(y1 – y2)]


Area of ΔADE with vertices A (6, 1), D (7, 3) and E


∆ =





=( - 3)/4 but area can’t be negative


Hence, the required area of ΔADE is sq. units


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