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Question
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
Solution
Let (x,y) be the coordinates of D and ( x ',y')be thee 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 ⇒ x' = 8`
`y' =(3+4)/2 ⇒ y' = 7/2 `
Thus, the coordinates of E are `(8,7/2)`
`"let" A (x_1, y_1) = A(6,1) ,E (x_2,y_2) = E (8,7/2) and D(x_3,y_3) = D(7,3) Now`
`"Area" (Δ ABC) = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)]`
`=1/2[6(7/2-3)+8(3-1)+7(1-7/2)]`
`=1/2[3/2]`
`=3/4` sq. units
Hence, the area of the triangle ΔADE is `3/ 4`sq. units
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