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


16
1