Question

Area of the triangle bounded by the lines given by the equations: \[ 12x^2 - 20xy + 7y^2 = 0 \quad \text{and} \quad x + y - 1 = 0 \]

• A. \( \frac{8}{29} \)
• B. \( \frac{8}{39} \)
• C. \( \frac{4}{29} \)
• D. \( \frac{4}{39} \)

Hint:

When dealing with second-degree equations forming two lines, factorize and solve for intersection points before using the area formula.

Answer 1

The Correct Option is D

To solve the problem, we need to find the area of the triangle bounded by the lines given by the equations: \[ 12x^2 - 20xy + 7y^2 = 0 \quad \text{and} \quad x + y - 1 = 0. \] Step 1: Factorize the quadratic equation The equation \( 12x^2 - 20xy + 7y^2 = 0 \) represents a pair of lines. To factorize it, we treat it as a quadratic in \( x \): \[ 12x^2 - 20xy + 7y^2 = 0. \] Using the quadratic formula \( x = \frac{20y \pm \sqrt{(20y)^2 - 4 \cdot 12 \cdot 7y^2}}{2 \cdot 12} \), we get: \[ x = \frac{20y \pm \sqrt{400y^2 - 336y^2}}{24} = \frac{20y \pm \sqrt{64y^2}}{24} = \frac{20y \pm 8y}{24}. \] Thus, the two lines are: \[ x = \frac{28y}{24} = \frac{7y}{6} \quad \text{and} \quad x = \frac{12y}{24} = \frac{y}{2}. \] So, the equations of the lines are: \[ x = \frac{7y}{6} \quad \text{and} \quad x = \frac{y}{2}. \] Step 2: Find the points of intersection The lines \( x = \frac{7y}{6} \) and \( x = \frac{y}{2} \) intersect the line \( x + y - 1 = 0 \). We solve for the points of intersection. 1. Intersection of \( x = \frac{7y}{6} \) and \( x + y - 1 = 0 \): \[ \frac{7y}{6} + y - 1 = 0 \implies \frac{13y}{6} = 1 \implies y = \frac{6}{13}, \quad x = \frac{7}{6} \cdot \frac{6}{13} = \frac{7}{13}. \] So, the point is \( \left( \frac{7}{13}, \frac{6}{13} \right) \). 2. Intersection of \( x = \frac{y}{2} \) and \( x + y - 1 = 0 \): \[ \frac{y}{2} + y - 1 = 0 \implies \frac{3y}{2} = 1 \implies y = \frac{2}{3}, \quad x = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}. \] So, the point is \( \left( \frac{1}{3}, \frac{2}{3} \right) \). 3. Intersection of \( x = \frac{7y}{6} \) and \( x = \frac{y}{2} \): \[ \frac{7y}{6} = \frac{y}{2} \implies 7y = 3y \implies y = 0, \quad x = 0. \] So, the point is \( (0, 0) \). Step 3: Compute the area of the triangle The vertices of the triangle are: \[ A = \left( \frac{7}{13}, \frac{6}{13} \right), \quad B = \left( \frac{1}{3}, \frac{2}{3} \right), \quad C = (0, 0). \] The area of the triangle is given by: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \] Substitute the coordinates: \[ \text{Area} = \frac{1}{2} \left| \frac{7}{13}\left( \frac{2}{3} - 0 \right) + \frac{1}{3}\left( 0 - \frac{6}{13} \right) + 0\left( \frac{6}{13} - \frac{2}{3} \right) \right|. \] Simplify: \[ \text{Area} = \frac{1}{2} \left| \frac{7}{13} \cdot \frac{2}{3} - \frac{1}{3} \cdot \frac{6}{13} \right| = \frac{1}{2} \left| \frac{14}{39} - \frac{6}{39} \right| = \frac{1}{2} \left| \frac{8}{39} \right| = \frac{4}{39}. \] Final Answer: \[ \boxed{\frac{4}{39}} \]