1. Given lines

$L_1: 2x+5y=10, \qquad L_2:-4x+3y=12$

A third line $L_3$ passes through $P(2,3)$, meets $L_2$ at $A$ and $L_1$ at $B$.

Also, $P$ divides $AB$ internally in the ratio $1:3$.

2. Use section formula

Since $P$ divides $AB$ internally in the ratio $1:3$, we interpret

$AP:PB=1:3.$

If $A(x_1,y_1)$ and $B(x_2,y_2)$, then by section formula,

$P=\left(\frac{1\cdot x_2+3\cdot x_1}{1+3},\;\frac{1\cdot y_2+3\cdot y_1}{1+3}\right).$

So,

$4P = B+3A.$

Since $P=(2,3)$,

$4P=(8,12).$

Hence,

$B+3A=(8,12).$

Let

$A=(x_1,y_1), \quad B=(x_2,y_2).$

Because $A$ lies on $L_2$ and $B$ lies on $L_1$,

$-4x_1+3y_1=12 \quad ...(1)$ $2x_2+5y_2=10 \quad ...(2)$

and from $B+3A=(8,12)$,

$x_2+3x_1=8 \quad ...(3)$ $y_2+3y_1=12 \quad ...(4)$

3. Solve for $A$ and $B$

From (3) and (4),

$x_2=8-3x_1, \qquad y_2=12-3y_1.$

Substitute into (2):

$2(8-3x_1)+5(12-3y_1)=10$ $16-6x_1+60-15y_1=10$ $6x_1+15y_1=66$ $2x_1+5y_1=22 \quad ...(5)$

Now solve (1) and (5):

$-4x_1+3y_1=12$ $2x_1+5y_1=22$

Multiply the second by $2$:

$4x_1+10y_1=44$

Add with the first:

$13y_1=56 \Rightarrow y_1=\frac{56}{13}.$

Then from (5):

$2x_1+5\cdot \frac{56}{13}=22$ $2x_1=22-\frac{280}{13}=\frac{286-280}{13}=\frac{6}{13}$ $x_1=\frac{3}{13}.$

Therefore,

$A=\left(\frac{3}{13},\frac{56}{13}\right).$

Now,

$x_2=8-3\cdot \frac{3}{13}=\frac{104-9}{13}=\frac{95}{13},$ $y_2=12-3\cdot \frac{56}{13}=\frac{156-168}{13}=-\frac{12}{13}.$

Hence,

$B=\left(\frac{95}{13},-\frac{12}{13}\right).$

4. Find the third vertex of the triangle

The triangle is bounded by $L_1, L_2, L_3$.

Its vertices are:

• $A = L_2 \cap L_3$
• $B = L_1 \cap L_3$
• $C = L_1 \cap L_2$

Find $C$ by solving:

$2x+5y=10$ $-4x+3y=12$

Multiply first by $2$:

$4x+10y=20$

Add with second:

$13y=32 \Rightarrow y=\frac{32}{13}.$

Then

$2x+5\cdot \frac{32}{13}=10$ $2x=10-\frac{160}{13}=\frac{130-160}{13}=-\frac{30}{13}$ $x=-\frac{15}{13}.$

So,

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

5. Compute area of triangle $ABC$

Use determinant formula:

$\text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$

Take

$A\left(\frac{3}{13},\frac{56}{13}\right), \quad B\left(\frac{95}{13},-\frac{12}{13}\right), \quad C\left(-\frac{15}{13},\frac{32}{13}\right).$

Then

$y_2-y_3=-\frac{12}{13}-\frac{32}{13}=-\frac{44}{13}$ $y_3-y_1=\frac{32}{13}-\frac{56}{13}=-\frac{24}{13}$ $y_1-y_2=\frac{56}{13}-\left(-\frac{12}{13}\right)=\frac{68}{13}$

So,

$\text{Area}=\frac12\left|\frac{3}{13}\cdot \left(-\frac{44}{13}\right)+\frac{95}{13}\cdot \left(-\frac{24}{13}\right)+\left(-\frac{15}{13}\right)\cdot \frac{68}{13}\right|$

$=\frac12\left|\frac{-132-2280-1020}{169}\right|$ $=\frac12\cdot \frac{3432}{169}$ $=\frac{1716}{169}$

Since $169=13^2$ and $1716=132\cdot 13$,

$\text{Area}=\frac{132}{13}.$

6. Check with options

$\boxed{\frac{132}{13}}$

So the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They agree.