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 (x_{1}, y_{1}) and (x_{2}, y_{2}) =

→

→ 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 (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3});

= [x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

∴ 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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