A(6,1) B(8,2) and C(9,4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, then find the area of Δ ADE.
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Solution
Given that, A(6,1) B(8,2) and C(9,4) are three vertices of a parallelogram ABCD.
Let the fourth vertex of parallelogram, be (x,y)
We know that, the diagonals of a parallelogram bisect each other. ∴ Mid-point of BD = Mid-point of AC ⇒(8+x2,2+y2)=(6+92,1+42)[∵Mid-point of a line segment joining the points(x1,y1)and(x2,y2)=(x1+x22,y1+y22)]⇒(8+x2,2+y2)=(152,52)∴8+x2=152⇒8+x=15⇒x=7And2+y2=52⇒2+y=5⇒y=3
So, fourth vertex of a parallelogram is D(7,3).
Now, mid-point of side DC=(7+92,3+42) ⇒E=(8,72)
[∵Area ofΔABC with vertices (x1,y1),(x2,y2)and(x3,y3) =12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] ∴Area ofΔADE with vertices A(6,1),D(7,3)andE(8,72) Δ=12[6(3−72)+7(72−1)+8(1−3)] =12[6×(−12)+7(52)+8(−2)))] =12(−3+352−16) =12(352−19)=12(−32) =−34[but area cannot be negative]
Hence, the required area of Δ ADE is 34 sq.units.