1.

A(6, 1), B(8, 2) and C(9, 4) are three vertices of parallelogram ABCD. If E is the mid-point of DC, then find the area of DeltaADE.

Answer»

Solution :Let the co-ordinates of point D be `(x_(4), y_(4))`
Now, the mid-point of BD = mid-point of AC
`rArr""((x_(4)+8)/(2), (y_(4)+2)/(2))=((6+9)/(2), (1+4)/(2))`

`rArr""(x_(4)+8)/(2)=(15)/(2)rArr""x_(4)=7`
and `""(y_(4)+2)/(2)=(5)/(2)" "rArr""y_(4)=3`
`therefore""D-= (7, 3)`
Mid-point E of CD `-=((7+9)/(2), (3+4)/(2))-=(8, (7)/(2))`
Now, AREA of`DeltaADE=(1)/(2)[6(3-(7)/(2))+7((7)/(2)-1)+8(1-3)]=(1)/(2)(-3+(35)/(2)-16)=-(3)/(4)`
`therefore` Area of `DeltaADE=(3)/(4)` square units (neglecting negative sign)


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