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Step-by-Step Solution
Step 1: Write the equations of the given lines
We have three lines:
L1: $2x + 5y = 10$
L2: $-4x + 3y = 12$
L3: This line passes through point $P(2, 3)$ and intersects L2 at $A$ and L1 at $B$.
The point $P$ divides the segment $AB$ internally in the ratio $1 : 3$.
Step 2: Find the intersection of L1 and L2 (point C)
We first solve the system:
$L_{1}: 2x + 5y = 10$
$L_{2}: -4x + 3y = 12$
To solve, we can use any preferred method (substitution, elimination, etc.). Let us outline a straightforward approach:
From $L_{1}$, express $y$ in terms of $x$:
$$
5y = 10 - 2x \quad \Longrightarrow \quad y = \frac{10 - 2x}{5}.
$$
Substitute into $L_{2}$:
$$
-4x + 3\left(\frac{10 - 2x}{5}\right) = 12.
$$
Simplify to solve for $x$:
$$
-4x + \frac{30 - 6x}{5} = 12 \quad \Longrightarrow \quad -4x + 6 - \frac{6x}{5} = 12.
$$
$$
-4x - \frac{6x}{5} = 12 - 6 \quad \Longrightarrow \quad -\left(4 + \frac{6}{5}\right)x = 6.
$$
$$
-\left(\frac{20 + 6}{5}\right)x = 6 \quad \Longrightarrow \quad -\frac{26}{5}x = 6 \quad \Longrightarrow \quad x = -\frac{6 \cdot 5}{26} = -\frac{30}{26} = -\frac{15}{13}.
$$
Substitute $x = -\frac{15}{13}$ back to find $y$:
$$
y = \frac{10 - 2\left(-\frac{15}{13}\right)}{5} = \frac{10 + \frac{30}{13}}{5} = \frac{\frac{130}{13} + \frac{30}{13}}{5} = \frac{\frac{160}{13}}{5} = \frac{160}{13} \times \frac{1}{5} = \frac{32}{13}.
$$
Hence, the intersection point is
$$
C \left(-\frac{15}{13}, \frac{32}{13}\right).
$$
Step 3: Express coordinates of A on L2 and B on L1
Since $A$ lies on $L_{2}: -4x + 3y = 12,$ we can write
$$
3y = 12 + 4x \quad \Longrightarrow \quad y = \frac{12 + 4x}{3}.
$$
So let us denote $A = \bigl(x_{1},\, \frac{12 + 4x_{1}}{3}\bigr).$
Similarly, since $B$ lies on $L_{1}: 2x + 5y = 10,$ we have
$$
5y = 10 - 2x \quad \Longrightarrow \quad y = \frac{10 - 2x}{5}.
$$
So let us denote $B = \bigl(x_{2},\, \frac{10 - 2x_{2}}{5}\bigr).$
Step 4: Use the ratio of internal division to find x1 and x2
The point $P(2, 3)$ divides the segment $AB$ in the ratio $1 : 3$. By the section formula (internal division):
For the $x$-coordinate,
$$
2 = \frac{1 \cdot x_{2} + 3 \cdot x_{1}}{1 + 3} = \frac{x_{2} + 3x_{1}}{4}.
$$
Thus, we get
$$
x_{2} + 3x_{1} = 8.
$$
For the $y$-coordinate,
$$
3 = \frac{1 \cdot \frac{10 - 2x_{2}}{5} + 3 \cdot \frac{12 + 4x_{1}}{3}}{4}.
$$
However, a simpler approach (often used in solutions) is to write the system in a rearranged form. One standard approach yields two linear equations in $x_{1}$ and $x_{2}$, eventually giving:
$$
3x_{1} + x_{2} = 8,
$$
$$
10x_{1} - x_{2} = -5.
$$
Step 5: Solve for x1 and x2
Add the two equations:
$3x_{1} + x_{2} = 8$
$10x_{1} - x_{2} = -5$
Adding them gives:
$$
(3x_{1} + 10x_{1}) + (x_{2} - x_{2}) = 8 + (-5) \quad \Longrightarrow \quad 13x_{1} = 3.
$$
$$
x_{1} = \frac{3}{13}.
$$
Now substitute $x_{1} = \frac{3}{13}$ into one of the equations, say $3x_{1} + x_{2} = 8$:
$$
3 \left(\frac{3}{13}\right) + x_{2} = 8 \quad \Longrightarrow \quad \frac{9}{13} + x_{2} = 8 \quad \Longrightarrow \quad x_{2} = 8 - \frac{9}{13} = \frac{104 - 9}{13} = \frac{95}{13}.
$$
Step 6: Determine the coordinates of A and B
Hence,
$$
A = \left(x_{1},\frac{12 + 4x_{1}}{3}\right) = \left(\frac{3}{13}, \frac{12 + 4\left(\frac{3}{13}\right)}{3}\right).
$$
Compute $y$-coordinate of $A$:
$$
y_{A} = \frac{12 + \frac{12}{13}}{3} = \frac{\frac{156}{13} + \frac{12}{13}}{3} = \frac{\frac{168}{13}}{3} = \frac{168}{13} \times \frac{1}{3} = \frac{56}{13}.
$$
So
$$
A = \left(\frac{3}{13}, \frac{56}{13}\right).
$$
Similarly,
$$
B = \left(x_{2}, \frac{10 - 2x_{2}}{5}\right) = \left(\frac{95}{13}, \frac{10 - 2\left(\frac{95}{13}\right)}{5}\right).
$$
Compute $y$-coordinate of $B$:
$$
y_{B} = \frac{10 - \frac{190}{13}}{5} = \frac{\frac{130}{13} - \frac{190}{13}}{5} = \frac{-\frac{60}{13}}{5} = -\frac{60}{13} \times \frac{1}{5} = -\frac{12}{13}.
$$
Thus,
$$
B = \left(\frac{95}{13}, -\frac{12}{13}\right).
$$
Step 7: Compute the area of the triangle
The triangle is formed by the lines $L_{1}, L_{2}$, and $L_{3}$. Its vertices are $A$, $B$, and $C$ (where $C$ is the intersection of $L_{1}$ and $L_{2}$). The standard formula for area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:
$$
\text{Area} = \frac{1}{2} \left|\, x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}) \right|.
$$
Substituting $A = \left(\frac{3}{13}, \frac{56}{13}\right)$, $B = \left(\frac{95}{13}, -\frac{12}{13}\right)$, and $C = \left(-\frac{15}{13}, \frac{32}{13}\right)$ into the formula will yield the area. After simplification, one obtains:
$$
\text{Area} = \frac{132}{13}.
$$
Final Answer
The area of the triangle is
$$
\frac{132}{13} \, \text{square units}.
$$
