Question

A(3,-4), B(5,-2), C(-1,8) are the vertices of ΔABC. D, E, F are the mid-points of sides BC, CA and AB respectively. Find area of ΔABC. Using coordinates of D, E, F, find area of ΔDEF. Hence show that the ABC = 4(DEF).

Answer 1

A = (3, - 4) , B = (5, -2) and C = (-1, 8)

we know, area of triangle = $\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

area of triangle, ABC = 1/2[3( -2 - 8) + 5(8 + 4) -1(-4 + 2)]

= 1/2[3 × - 10 + 5 × 12 + 2 ]

= 1/2[ -30 + 60 + 2]

= 1/2 × 32 = 16 --------(1)

now, midpoint of BC is D.
so, D = {(5 - 1)/2 , (-2 + 8)/2 } = (2, 3)

midpoint of CA is E.
so, E = {(3 - 1)/2, (-4 + 8)/2} = (1 , 2)

midpoint of AB is F.
so, F = {(3 + 5)/2, (-4 -2)/2} = (4, -3)


area of triangle , DEF = 1/2[2(2 + 3) + 1(-3 - 3) + 4(3 - 2)]

= 1/2[10 - 6 + 4 ]

= 1/2 × 8 = 4 ------(2).

from equations (1) and (2),
ar(∆ABC) = 4 × ar(∆DEF)