A(6,1), B(8,2), C(9,4) are three vertices of a parallelogram ACBD If E is midpoint of DC, find the area - Maths - Coordinate Geometry

Question

A(6,1), B(8,2), C(9,4) are three vertices of a
parallelogram ACBD If E is midpoint of DC, find the area of
âˆ†ADE

Lalit Mehra · Accepted Answer

Given, ACBD is parallelogram AB and CD are the diagonal intersecting in E Let the coordinates of D be (x, y) We know that, diagonals of parallelogram bisect each other âˆ´ Mid point of CD = Mid point of AB $\Rightarrow (\frac{x + 9}{2}, \frac{y + 4}{2}) = (\frac{6 + 8}{2}, \frac{1 + 2}{2}) \Rightarrow (\frac{x + 9}{2}, \frac{y + 4}{2}) = (\frac{14}{2}, \frac{3}{2}) \Rightarrow \frac{x + 9}{2} = \frac{14}{2} and \frac{y + 4}{2} = \frac{3}{2} \Rightarrow x + 9 = 14 and y + 4 = 3 \Rightarrow x = 5 and y = 1$ Coordinates of D = (5, 1) Coordinates of E $= (\frac{6 + 8}{2}, \frac{1 + 3}{2}) = (\frac{14}{2}, \frac{3}{2}) = (7, \frac{3}{2})$ Area of Î”ADE $= | \frac{1}{2} [6 (1 - \frac{3}{2}) + 5 (\frac{3}{2} - 1) + 7 (1 - 1)] | = | \frac{1}{2} [6 \times (\frac{- 1}{2}) + 5 \times (\frac{1}{2})] | = | \frac{1}{2} (- 3 + \frac{5}{2}) | = | \frac{1}{2} \times (\frac{- 1}{2}) | = | \frac{- 1}{4} | = \frac{1}{4}$

A(6,1), B(8,2), C(9,4) are three vertices of a parallelogram ACBD. If E is midpoint of DC, find the area of ∆ADE.