\[A\left( 6, 1 \right) , B(8, 2) \text{ and } C(9, 4)\] are three vertices of a parallelogram *ABCD* . If E is the mid-point of *DC* , find the area of \[∆\] *ADE*.

#### Solution

Three vertices are given, then D can be calulated and it comes out to be (7, 3).

Since, E is midpoint of BD.

Therefore, coordinates of E are \[\left( \frac{15}{2}, \frac{5}{2} \right)\] .

Now, vertices of triangle ABE rae (6, 1), (8, 2) and \[\left( \frac{15}{2}, \frac{5}{2} \right)\] .

\[\Rightarrow \text{ Area of the ∆ ABE } = \frac{1}{2}\begin{vmatrix}1 & 6 & 1 \\ 1 & 8 & 2 \\ 1 & \frac{15}{2} & \frac{5}{2}\end{vmatrix}\]

\[ = \frac{1}{2}\left[ 1\left( 20 - 15 \right) - 6\left( \frac{5}{2} - 2 \right) + 1\left( \frac{15}{2} - 8 \right) \right]\]

\[ = \frac{1}{2}\left[ 5 - \frac{6}{2} - \frac{1}{2} \right]\]

\[ = \frac{3}{4} \text{ aq . units } \]