Question

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

Answer 1

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 $\left(\frac{15}{2},\frac{{\displaystyle 5}}{{\displaystyle 2}}\right)$.
Now, vertices of triangle ABE rae (6, 1), (8, 2) and $\left(\frac{15}{2},\frac{{\displaystyle 5}}{{\displaystyle 2}}\right)$.
$$\begin{gathered} \Rightarrow Areaofthe∆ABE=\frac{1}{2}\left|\begin{array}{ccc}1& 6& 1\\ 1& 8& 2\\ 1& \frac{15}{2}& \frac{5}{2}\end{array}\right| \\ =\frac{1}{2}\left[1\left(20-15\right)-6\left(\frac{5}{2}-2\right)+1\left(\frac{15}{2}-8\right)\right] \\ =\frac{1}{2}\left[5-\frac{6}{2}-\frac{1}{2}\right] \\ =\frac{3}{4}aq.units \end{gathered}$$