Let a triangle be bounded by the lines L 1 : 2 x + 5 y = 10 ; L 2 : - 4 x + 3 y = 12 and the line L 3 ,…

Given, point $A$ lies on $L_2 : -4x+3y=12$

Take $x=\alpha , \text{so} y=4+\frac{4}{3}\alpha$, $A\left(\alpha ,4+\frac{4}{3}\alpha \right)$

Points $B$ lies on $L_1 : 2x+5y=10$

Take $x=\beta , \text{so} y=2-\frac{2}{5}\beta$, $B\left(\beta ,2-\frac{2}{5}\beta \right)$

Now point $P$ divides $AB$ internally in the ratio $1:3$

$\Rightarrow P\left(2,3\right)=\mathrm{P}\left(\frac{3\alpha +\beta }{4},\frac{3\left(4+\frac{4}{3}\alpha \right)+1\left(2-\frac{2}{5}\beta \right)}{4}\right)$

$\Rightarrow \alpha =\frac{3}{13},\beta =\frac{95}{13}$

We get, point $A\left(\frac{3}{13},\frac{56}{13}\right),B\left(\frac{95}{13},-\frac{12}{13}\right)$

Vertex $C$ of triangle is the point of intersection of $L_1$ and $L_2$

$\Rightarrow C\left(-\frac{15}{13},\frac{32}{13}\right)$

area $∆ABC=\frac{1}{2}\left|\left|\begin{array}{ccc}\frac{3}{13}& \frac{56}{13}& 1\\ \frac{95}{13}& -\frac{12}{13}& 1\\ -\frac{15}{13}& \frac{32}{13}& 1\end{array}\right|\right|$

$=\frac{1}{2\times {13}^3}||\begin{array}{ccc}3& 56& 13\\ 95& -12& 13\\ -15& 32& 13\end{array}||$

area $∆ABC=\frac{132}{13}$sq. units