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 `triangle`ADE.

Explanation

A(6,1) , B (8,2) and C (9,4) are three vertices of a parallelogram ABCD . Let the fourth vertex of paralleogram be (x,y). We know that. The ,the diagonals os a parallelogram bisect each other.

`:.` Mid - point of BD = Mid point of AC
`rArr((8+x)/(2),(2+y)/(2))=((6+9)/(2),(1+4)/(2))`
[`:'` mid - point of a line segment joining the points `(x_(1),y_(1))and(x_(2),y_(2))=((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))]`
`rArr((8+x)/(2),(2+y)/(2))=((15)/(2),(5)/(2))`
`:. (8+x)/(2)=(15)/(2)`
`rArr8+x=15rArrx=7`
and `(2+y)/(2)=(5)/(2)`
`rArr2+y=5rArry=3`
So ,fourth vertex of a parallelogram is D (7,3).
Now , mid - point of side DC `-=((7+9)/(2),(3+4)/(2))`
E `-=(8,(7)/(2))`
[`:'` area of `triangle`ABC with vertices `(x_(1),y_(1)),(x_(2),y_(2))and(x_(3),y_(3))=(1)/(2)[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+(y_(1)-y_(2))]`
`:.` Area of `triangle` ADE with vertices A (6,1),D (7,3) and `(8,(7)/(2))`,
`triangle =(1)/(2)[6(3-(7)/(2))+7((7)/(2)-1)+8(1-3)]`
`=(1)/(2)[6xx((-1)/(2))+7((5)/(2))+8(-2)]`
`=(1)/(2)(3+(35)/(2)-16)`
`=(1)/(2)((35)/(2)-19)=((-3)/(2))`
`=(-3)/(4)` [but area cannot be negative]
Hence , the required area of `triangle` ADE is `(3)/(4)` sq units.