Find the area of ΔABC, the midpoints of whose sides AB, BC and CA are D(3, -1), E(5, 3) and F(1, -3) respectively.


The figure is as shown above.


x1 + x2 = 2 × 3 = 6 …..(1)


x1 + x3 = 2 × 1 = 2 …..(2)


x2 + x3 = 2 × 5 = 10 …..(3)


Equation 1 – Equation 2 gives us


x2 – x3 = 4 …..(4)


Equation 3 + Equation 4,


2x2 = 14 x2 = 7


x1 = -1 and x3 = 3


Similarly,


y1 + y2 = 2 × -1 = -2 …..(5)


y1 + y3 = 2 × -3 = -6 …..(6)


y2 + y3 = 2 × 3 = 6 …..(7)


Equation 5 – Equation 6 gives us


y2 – y3 = 4 …..(8)


Equation 7 + Equation 8,


2y2 = 10 y2 = 5


y1 = -7 and y3 = 1


Area of Δ ABC




= 8 square units


Therefore, the area of Δ ABC is 8 square units.


1