Question

If `D((-1)/2, 5/2), E(7, 3)` and `F(7/2, 7/2)` are the midpoints of sides of ∆ABC, find the area of the ∆ABC.

Answer 1

Let A = (x₁, y₁), B = (x₂, y₂) and C = (x₃, y₃) are the vertices of the ∆ABC.

Givens, `D(- 1/2, 5/2), E(7, 3)` and `F(7/2, 7/2)` be the mid-points of the sides BC, CA and AB, respectively.

Since, `D(- 1/2, 5/2)` is the mid-point of BC.

∴ `(x_2 + x_3)/2 = - 1/2`

`["Since, mid-point of a line segment having points"  (x_1, y_1)  "and"  (x_2, y_2)  "is"  ((x_1 + x_2)/2, (y_1 + y_2)/2)]`

And `(y_2 + y_3)/2 = 5/2`

⇒ x₂ + x₃ = – 1  ...(i)

And y₂ + y₃ = 5  ...(ii)

As E(7, 3) is the mid-point of CA.

∴ `(x_3 + x_1)/2` = 7

And `(y_3 + y_1)/2` = 3

⇒ x₃ + x₁ = 14  ...(iii)

And y₃ + y₁ = 6   ...(iv)

Also, `F(7/2, 7/2)` is the mid-point of AB.

∴ `(x_1 + x_2)/2 = 7/2`

And `(y_1 + y_2)/2 = 7/2`

⇒ x₁ + x₂ = 7   ...(v)

And y₁ + y₂ = 7  ...(vi)

On adding equations (i), (iii) and (v), we get

2(x₁ + x₂ + x₃) = 20

⇒ x₁ + x₂ + x₃ = 10  ...(vii)

On subtracting equations (i), (iii) and (v) from equation (vii) respectively, we get

x₁ = 11, x₂ = – 4, x₃ = 3

On adding equations (ii), (iv) and (vi), we get

2(y₁ + y₂ + y₃) = 18

⇒ y₁ + y₂ + y₃ = 9  ...(viii)

On subtracting equations (ii), (iv) and (vi) from equation (viii) respectively, we get

y₁ = 4, y₂ = 3, y₃ = 2

Hence, the vertices of ∆ABC are A(11, 4), B(– 4, 3) and C(3, 2)

∵ Area of ∆ABC = ∆ = `1/2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]`

∴  ∆ = `1/2[11(3 - 2) + (-4)(2 - 4) + 3(4 - 3)]`

= `1/2[11 xx 1 + (-4)(-2) + 3(1)]`

= `1/2(11 + 8 + 3)`

= `22/2`

= 11

∴ Required area of ∆ABC = 11