If and are the mid - points of sides of ΔABC, then find the area of the ΔABC.

Let the vertices of ∆ABC are;

A = (x₁, y₁)

B = (x₂, y₂)

C = (x₃, y₃)

Given,

D, E and F are the mid - points of the sides BC, CA and AB respectively.

Where Coordinates are;

D =

E = (7, 3)

F =

By mid - point formula of a line segment having points (x₁, y₁) and (x₂, y₂) =

So,

D is the mid - point of BC

∴

And

x₂ + x₃ = - 1 …..(i)

And

y₂ + y₃ = 5 … (ii)

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

∴

And

∴ x₃ + x₁ = 14 .. (iii)

And

Y₃ + y₁ = 6 …. (iv)

Also,

F is the mid - point of AB

∴

And

∴ 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 Eq. (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 Eq. (viii) respectively,

We get;

y₁ = 4

y₂ = 3

y₃ = 2

Hence, the vertices of ΔABC are A (11, 4), B ( - 4, 3) and C (3, 2)

So,

∵ Area of ∆ABC = 1/2 [x₁ (y₂ – y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

∆ = [11(3 – 2) + ( - 4)(2 – 4) + (4 – 3)]

= [11 × 1 + ( - 4)( - 2) + 3(1)]

= [11 + 8 + 3] = 22/2 = 11

∴ Required area of ΔABC = 11