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