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

Let the vertices of ∆ABC are;

A = (x_{1}, y_{1})

B = (x_{2}, y_{2})

C = (x_{3}, y_{3})

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

So,

D is the mid - point of BC

∴

And

x_{2} + x_{3} = - 1 …..(i)

And

y_{2} + y_{3} = 5 … (ii)

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

∴

And

∴ x_{3} + x_{1} = 14 .. (iii)

And

Y_{3} + y_{1} = 6 …. (iv)

Also,

F is the mid - point of AB

∴

And

∴ x_{1} + x_{2} = 7 … (v)

And

y_{1} + y_{2} = 7 …(vi)

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

We get,

2(x_{1} + x_{2} + x_{3}) = 20

x_{1} + x_{2} + x_{3} = 10………(vii)

On subtracting Equations (i), (iii) and (v) from Eq. (vii) respectively,

We get;

x_{1} = 11,

x_{2} = - 4

x_{3} = 3

On adding Equations (ii), (iv) and (vi),

We get;

2(y_{1} + y_{2} + y_{3}) = 18

y_{1} + y_{2} + y_{3} = 9…..(viii)

On subtracting Equations (ii), (iv) and (vi) from Eq. (viii) respectively,

We get;

y_{1} = 4

y_{2} = 3

y_{3} = 2

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

So,

∵ Area of ∆ABC = 1/2 [x_{1} (y_{2} – y_{3} ) + x_{2} (y_{3} - y_{1} ) + x_{3} (y_{1} - y_{2} )]

∆ = [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

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