##### 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

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