Let a triangle be bounded by the lines L_1:2x+5y=10; L_2:-4x+3y=12 and the line L_3, which passes through the point P(2,3), intersect L_2 at A and…

Given, point $A$A lies on ${L}_{2}:-4x+3y=12$L2​:−4x+3y=12

Take $x=\alpha ,\text{so}y=4+\frac{4}{3}\alpha$x=α,soy=4+34​α, $A(\alpha ,4+\frac{4}{3}\alpha )$A(α,4+34​α)

Points $B$B lies on ${L}_{1}:2x+5y=10$L1​:2x+5y=10

Take $x=\beta ,\text{so}y=2-\frac{2}{5}\beta$x=β,soy=2−52​β, $B(\beta ,2-\frac{2}{5}\beta )$B(β,2−52​β)

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

$$\Rightarrow P(2,3)=P(\frac{3\alpha +\beta }{4},\frac{3(4+\frac{4}{3}\alpha )+1(2-\frac{2}{5}\beta )}{4})$$⇒P(2,3)=P(43α+β​,43(4+34​α)+1(2−52​β)​)

$$\Rightarrow \alpha =\frac{3}{13},\beta =\frac{95}{13}$$⇒α=133​,β=1395​

We get, point $A(\frac{3}{13},\frac{56}{13}),B(\frac{95}{13},-\frac{12}{13})$A(133​,1356​),B(1395​,−1312​)

Vertex $C$C of triangle is the point of intersection of ${L}_{1}$L1​ and ${L}_{2}$L2​

$$\Rightarrow C(-\frac{15}{13},\frac{32}{13})$$⇒C(−1315​,1332​)

area $\Delta ABC=\frac{1}{2}||\begin{matrix}\frac{3}{13} & \frac{56}{13} & 1 \\ \frac{95}{13} & -\frac{12}{13} & 1 \\ -\frac{15}{13} & \frac{32}{13} & 1\end{matrix}||$ΔABC=21​∣∣133​1395​−1315​​1356​−1312​1332​​111​∣∣

$$=\frac{1}{2\times {13}^{3}}||\begin{matrix}3 & 56 & 13 \\ 95 & -12 & 13 \\ -15 & 32 & 13\end{matrix}||$$=2×1331​∣∣395−15​56−1232​131313​∣∣

area $\Delta ABC=\frac{132}{13}$ΔABC=13132​sq. units