Prove the following

Let $A \equiv\left(x_{1}, y_{1}\right), B \equiv\left(x_{2}, y_{2}\right)$ and $C \equiv\left(x_{3}, y_{3}\right)$ are the vertices of the $\triangle A B C$.

Gives, $D\left(-\frac{1}{2}, \frac{5}{2}\right), E(7,3)$ and $F\left(\frac{7}{2}, \frac{7}{2}\right)$ be the mid-points of the sides $B C, C A$ and $A B$,

respectively.

Since, $D\left(-\frac{1}{2}, \frac{5}{2}\right)$ is the mid-point of $B C$.

$\therefore$ $\frac{x_{2}+x_{3}}{2}=-\frac{1}{2}$

$\left[\right.$ since, mid-point of a line segment having points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $\left.\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\right]$

and $\frac{y_{2}+y_{3}}{2}=\frac{5}{2}$

$\Rightarrow$ $x_{2}+x_{3}=-1$ $\ldots$ (i)

and $y_{2}+y_{3}=5$...(ii)

As $E(7,3)$ is the mid-point of $C A$.

$\therefore$ $\frac{x_{3}+x_{1}}{2}=7$

and $\frac{y_{3}+y_{1}}{2}=3$

$\Rightarrow$ $x_{2}+x_{1}=14$ ....(iii)

and $y_{3}+y_{1}=6$ ...(iv)

Also, $F\left(\frac{7}{2}, \frac{7}{2}\right)$ is the mid-point of $A B$.

$\therefore$ $\frac{x_{1}+x_{2}}{2}=\frac{7}{2}$

and $\frac{y_{1}+y_{2}}{2}=\frac{7}{2}$

$\Rightarrow$ $x_{1}+x_{2}=7$ $\ldots(\mathrm{v})$

and $y_{1}+y_{2}=7$ ...(vi)

On adding Eqs. (i), (iii) and (v), we get

$2\left(x_{1}+x_{2}+x_{3}\right)=20$

$\Rightarrow \quad x_{1}+x_{2}+x_{3}=10$ ...(vii)

On subtracting Eqs. (i), (iii) and (v) from Eq. (vii) respectively, we get

$x_{1}=11, x_{2}=-4, x_{3}=3$

On adding Eqs. (ii), (iv) and (vi), we get

$2\left(y_{1}+y_{2}+y_{3}\right)=18$

$\Rightarrow$ $y_{1}+y_{2}+y_{3}=9$ ...(viii)

On subtracting Eqs. (ii), (iv) and (vi) from Eq. (viii) respectively, we get

$y_{1}=4, y_{2}=3, y_{3}=2$

Hence, the vertices of $\triangle A B C$ are $A(11,4), B(-4,3)$ and $C(3,2)$.

$\because \quad$ Area of $\triangle A B C=\Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\therefore$ $\Delta=\frac{1}{2}[11(3-2)+(-4)(2-4)+3(4-3)]$

$=\frac{1}{2}[11 \times 1+(-4)(-2)+3(1)]$

$=\frac{1}{2}(11+8+3)=\frac{22}{2}=11$

$\therefore$ Required area of $\triangle A B C=11$