Question

The area of the triangle formed by the pair of lines \( 23x^2 - 48xy + 3y^2 = 0 \) with the line \( 2x + 3y + 5 = 0 \) is:

• A. \( \frac{1}{13\sqrt{3}} \)
• B. \( \frac{25}{13\sqrt{3}} \)
• C. \( \frac{7}{13\sqrt{5}} \)
• D. \( \frac{9}{25\sqrt{3}} \)

Hint:

To find the area of a triangle formed by a pair of lines and an external line, use the formula: \[ \text{Area} = \frac{1}{2} d^2 \tan \theta \] where \( d \) is the perpendicular distance from the origin and \( \theta \) is the angle between the two lines.

Answer 1

The Correct Option is B

Step 1: Understanding the Given Pair of Lines
The given equation of the pair of lines is: \[ 23x^2 - 48xy + 3y^2 = 0 \] This represents two straight lines passing through the origin. Step 2: Finding the Angle Between the Lines
The general form of the second-degree homogeneous equation representing a pair of lines is: \[ Ax^2 + 2Hxy + By^2 = 0 \] Comparing with \( 23x^2 - 48xy + 3y^2 = 0 \), we have: \[ A = 23, \quad H = -24, \quad B = 3 \] The angle \( \theta \) between the two lines is given by: \[ \tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A+B} \right| \] Substituting values: \[ \tan \theta = \left| \frac{2\sqrt{(-24)^2 - (23)(3)}}{23+3} \right| \] \[ = \left| \frac{2\sqrt{576 - 69}}{26} \right| = \left| \frac{2\sqrt{507}}{26} \right| = \left| \frac{\sqrt{507}}{13} \right| \] Step 3: Finding Perpendicular Distance
The given line equation is: \[ 2x + 3y + 5 = 0 \] The perpendicular distance from the origin to this line is: \[ d = \frac{|5|}{\sqrt{2^2 + 3^2}} = \frac{5}{\sqrt{13}} \] Step 4: Finding the Area of Triangle
The area of the triangle formed by the intersection of the pair of lines and the given line is given by: \[ \text{Area} = \frac{1}{2} d^2 \tan \theta \] \[ = \frac{1}{2} \times \left(\frac{5}{\sqrt{13}}\right)^2 \times \frac{\sqrt{507}}{13} \] \[ = \frac{1}{2} \times \frac{25}{13} \times \frac{\sqrt{507}}{13} \] \[ = \frac{25\sqrt{507}}{2 \times 169} = \frac{25}{13\sqrt{3}} \] Thus, the correct answer is \( \frac{25}{13\sqrt{3}} \).