Question

$\text{A(6,1) B(8,2)}$and $\text{C(9,4)}$ are the three vertices of a parallelogram $\text{ABCD}$. If $\text{E}$ is the mid point of $\text{DC}$. Find the area of the $△\text{ADE}$ ?

Answer 1

Finding area of the triangle $\text{ADE}$ :

The three vertices of $△\text{ABC}$ are $\text{A(6,1) B(8,2) C(9,4)}$

So we have ${\text{A(x}}_1{\text{,y}}_1{\text{)=(6,1) B(x}}_2{\text{,y}}_2{\text{)=(8,2) C(x}}_3{\text{,y}}_3\text{)=(9,4)}$

Let the fourth vertex ${\text{D(x}}_4{\text{,y}}_4\text{)=(x,y)}$

We know that the diagonals of a parallelogram bisect each other so

mid-point of $\text{AC}$$=$mid-point of $\text{BD}$

$$\begin{gathered} \Rightarrow (\frac{6+9}{2},\frac{1+4}{2})=(\frac{8+\text{x}}{2},\frac{2+\text{y}}{2}) \\ \Rightarrow (\frac{15}{2},\frac{5}{2})=(\frac{8+\text{x}}{2},\frac{2+\text{y}}{2}) \end{gathered}$$

By comparing the coefficients on both sides we get,

$$\begin{gathered} \frac{15}{2}=\frac{8+\text{x}}{2},\frac{5}{2}=\frac{2+\text{y}}{2} \\ 15=8+\text{x,5=2+y} \\ \text{x=15-8=7 , y=5-2=3} \end{gathered}$$

So the coordinates of ${\text{D(x}}_4{\text{,y}}_4\text{)=(7,3)}$

The mid-point of side $\text{DC}$ $=(\frac{9+7}{2},\frac{3+4}{2})=(8,\frac{7}{2)}$

So the coordinates of mid-point of $\text{DC}$ such that the coordinates of point $\text{E}$ are $=(8,\frac{7}{2})$

Since the area of the $△\text{ABC}$ with vertices ${\text{A(x}}_1{\text{,y}}_1{\text{) B(x}}_2{\text{,y}}_2{\text{) C(x}}_3{\text{,y}}_3\text{)}$ is

$\Rightarrow \frac{1}{2}\left[{\text{x}}_1{\text{(y}}_2{\text{-y}}_3{\text{)+x}}_2{\text{(y}}_3{\text{-y}}_1{\text{)+x}}_3{\text{(y}}_1{\text{-y}}_2\text{)}\right]$

So area of the $△\text{ADE}$ with vertices $\text{A(6,1) D(7,3) E(8,}\frac{7}{2}\text{)}$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left[6(3-\frac{7}{2})+7(\frac{7}{2}-1)+8(1-3)\right] \\ \Rightarrow \frac{1}{2}\left[6\left(\frac{6-7}{2}\right)+7\left(\frac{7-2}{2}\right)+8(-2)\right] \\ \Rightarrow \frac{1}{2}\left[6\left(\frac{-1}{2}\right)+7\left(\frac{5}{2}\right)-16\right] \\ \Rightarrow \frac{1}{2}\left[\left(\frac{-6}{2}\right)+\frac{35}{2}-16\right] \\ \Rightarrow \frac{1}{2}\left[\left(\frac{-6+35}{2}\right)-16\right] \\ \Rightarrow \frac{1}{2}\left[(\frac{29}{2}-16)\right] \\ \Rightarrow \frac{1}{2}\left[\frac{29-32}{2}\right] \\ \Rightarrow \frac{1}{2}\left[\frac{-3}{2}\right]=\frac{-3}{4} \end{gathered}$$

Area of the triangle can never be negative so area of the triangle is $\frac{3}{4}$.

Hence, the required answer is $\frac{3}{4}$.