Let a triangle be bounded by the lines L _ 1 : 2x + 5y = 10 ; L _ 2...

Given, point $A$ lies on ${L}_{2}: - 4x + 3y = 12$

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

Points $B$ lies on ${L}_{1}: 2x + 5y = 10$

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

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

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

$⇒ \alpha = \dfrac{3}{13}, \beta = \dfrac{95}{13}$

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

Vertex $C$ of triangle is the point of intersection of ${L}_{1}$ and ${L}_{2}$

$⇒ C\left(- \dfrac{15}{13}, \dfrac{32}{13}\right)$

area ∆ ABC = \dfrac{1}{2}\left|\left|\begin{cases}\dfrac{3}{13} & \dfrac{56}{13} & 1 \\ \dfrac{95}{13} & - \dfrac{12}{13} & 1 \\ - \dfrac{15}{13} & \dfrac{32}{13} & 1\end{cases}\right|\right|

= \dfrac{1}{2 \times {13}^{3}}\left||\begin{cases}3 & 56 & 13 \\ 95 & - 12 & 13 \\ - 15 & 32 & 13\end{cases}\right||

area $∆ ABC = \dfrac{132}{13}$sq. units