Question

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.

Answer 1

According to the question,

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

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

We know that, diagonals of a parallelogram bisect each other

Since, mid-point of a line segment joining the points (x₁, y₁) and (x₂, y₂) is given by,

`((x_1 + x_2)/2, (y_1 + y_2)/2)`

Mid-point of BD = Mid-point of AC

`((8 + x)/2, (2 + y)/2) = ((6 + 9)/2, (1 + 4)/2)`

`((8 + x)/2, (2 + y)/2) = (15/2, 5/2)`

So, we have,

`(8 + x)/2 = 15/2`

⇒ 8 + x = 15

⇒ x = 7

And

`(2 + y)/2 = 5/2`

⇒ 2 + y = 5

⇒ y = 3

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

Now,

Mid-point of side

DC = `((7 + 9)/2, (3 + 4)/2)`

E = `(8, 7/2)`

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

= `1/2`[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

∴ Area of ΔADE with vertices A(6, 1), D(7, 3) and `"E"(8, 7/2)`

Δ = `1/2[6(3 - 7/2) + 7(7/2 - 1) + 8(1 - 3)]`

= `1/2[6 xx ((-1)/2) + 7(5/2) + 8(-2)]`

= `1/2(35/2 - 19)`

= `1/2((-3)/2)`

= `(-3)/4` but area can’t be negative

Hence, the required area of ΔADE is `3/4` sq.units.