Question

The equation \( 2x^2 - 3xy - 2y^2 = 0 \) represents two lines \( L_1 \) and \( L_2 \). The equation \( 2x^2 - 3xy - 2y^2 - x + 7y - 3 = 0 \) represents another two lines \( L_3 \) and \( L_4 \). Let \( A \) be the point of intersection of lines \( L_1 \) and \( L_3 \), and \( B \) be the point of intersection of lines \( L_2 \) and \( L_4 \). The area of the triangle formed by the lines \( AB \), \( L_3 \), and \( L_4 \) is: .

• A. \( \frac{3}{10} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{15}{2} \)
• D. \( \frac{5}{2} \)

Hint:

For triangle areas from three points, use the determinant method.

Answer 1

The Correct Option is A

Step 1: Find Intersection Points
Solving equations of given lines to find vertices of the triangle. Step 2: Using Area Formula
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] \[ = \frac{3}{10} \] Thus, the correct answer is \( \frac{3}{10} \).