A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE.

Question

Philoid · Accepted Answer

Let (x, y) be the coordinates of D and ( x^’, y' ) be the coordinates of E. since the diagonals of a parallelogram bisect each other at the same point, therefore

(x + 8)/2 = (6 + 9)/2

X = 7

(y + 2)/2 = (1 + 4)/2

Y = 3

Thus, the coordinates of D are (7,3)

E is the midpoint of DC,

therefore

x^’ = (7 + 9)/2 = 8

y^’ = (3 + 4)/2 = 7/2

Thus, the coordinates of E are ( 8,7/ 2)

Let A(x₁,y₁) = A(6,1), E(x₂,y₂) = (8,7/2) and D(x₃,y₃) = D(7,3)

Now Area

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(6(7/2–3) + 8(3–1) + 7(1–7/2))

= 1/2(3/2)

= 3/4 sq unit

Hence, the area of the triangle ΔADE is 3/4 sq. units.