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), intersects `L_(2)` at A and `L_(1)` at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3 , then the area of the triangle is equal to:

Options

A. `(110)/(13)`
B. `(132)/(13)`
C. `(142)/(13)`
D. `(151)/(13)`

Answer

To solve the problem, we need to find the area of the triangle formed by the lines \( L_1 \), \( L_2 \), and \( L_3 \). Let's go through the steps systematically.

### Step 1: Find the intersection points of lines \( L_1 \) and \( L_2 \)
The equations of the lines are:
- \( L_1: 2x + 5y = 10 \)
- \( L_2: -4x + 3y = 12 \)

To find the intersection point \( C \) of \( L_1 \) and \( L_2 \), we can solve these equations simultaneously.

From \( L_1 \):
\[ 5y = 10 - 2x \implies y = 2 - \frac{2}{5}x \]

Substituting \( y \) in \( L_2 \):
\[ -4x + 3(2 - \frac{2}{5}x) = 12 \]
\[ -4x + 6 - \frac{6}{5}x = 12 \]
Multiplying through by 5 to eliminate the fraction:
\[ -20x + 30 - 6x = 60 \]
\[ -26x = 30 \implies x = -\frac{30}{26} = -\frac{15}{13} \]

Now substituting \( x \) back to find \( y \):
\[ y = 2 - \frac{2}{5}(-\frac{15}{13}) = 2 + \frac{6}{13} = \frac{26}{13} + \frac{6}{13} = \frac{32}{13} \]

Thus, the intersection point \( C \) is:
\[ C = \left(-\frac{15}{13}, \frac{32}{13}\right) \]

### Step 2: Find the equation of line \( L_3 \)
Line \( L_3 \) passes through point \( P(2, 3) \) and intersects \( L_1 \) at point \( B \) and \( L_2 \) at point \( A \).

Let the equation of \( L_3 \) be in the slope-intercept form:
\[ y - 3 = m(x - 2) \]
or
\[ y = mx - 2m + 3 \]

### Step 3: Find the intersection points \( A \) and \( B \)
To find point \( A \) (intersection with \( L_2 \)):
Substituting \( y \) in \( L_2 \):
\[ -4x + 3(mx - 2m + 3) = 12 \]
\[ -4x + 3mx - 6m + 9 = 12 \]
\[ (3m - 4)x = 12 + 6m - 9 \]
\[ (3m - 4)x = 3 + 6m \]
\[ x = \frac{3 + 6m}{3m - 4} \]

Substituting \( x \) back to find \( y \):
\[ y = m\left(\frac{3 + 6m}{3m - 4}\right) - 2m + 3 \]

Now, for point \( B \) (intersection with \( L_1 \)):
Substituting \( y \) in \( L_1 \):
\[ 2x + 5(mx - 2m + 3) = 10 \]
\[ 2x + 5mx - 10m + 15 = 10 \]
\[ (2 + 5m)x = 10m - 5 \]
\[ x = \frac{10m - 5}{2 + 5m} \]

Substituting \( x \) back to find \( y \):
\[ y = m\left(\frac{10m - 5}{2 + 5m}\right) - 2m + 3 \]

### Step 4: Use the section formula to find the coordinates of \( A \) and \( B \)
Since point \( P(2, 3) \) divides \( AB \) in the ratio \( 1:3 \):
Using the section formula:
\[ x_P = \frac{3x_A + 1x_B}{3 + 1} \quad \text{and} \quad y_P = \frac{3y_A + 1y_B}{3 + 1} \]

### Step 5: Calculate the area of triangle \( ABC \)
Using the formula for the area of a triangle given vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]