Question

The area of the triangle formed by the lines represented by \( 3x + y + 15 = 0 \) and \( 3x^2 + 12xy - 13y^2 = 0 \) is:

• A. \( \frac{15 \sqrt{3}}{2} \)
• B. \( 15 \sqrt{3} \)
• C. \( \frac{15 \sqrt{3}}{4} \)
• D. \( \frac{15}{\sqrt{3}} \)

Hint:

When calculating the area of a triangle formed by lines, use the formula for the area of a triangle with three vertices, and find the points of intersection of the lines to get the coordinates of the vertices.

Answer 1

The Correct Option is A

We are given the following two equations of lines:
1. \( 3x + y + 15 = 0 \)
2. \( 3x^2 + 12xy - 13y^2 = 0 \)
We are required to find the area of the triangle formed by these lines.
Step 1: The equation \( 3x + y + 15 = 0 \) represents a straight line. Let us first rewrite it in the standard form of a line: \[ y = -3x - 15 \] Step 2: The equation \( 3x^2 + 12xy - 13y^2 = 0 \) represents a pair of lines, which can be factored to find the lines formed. To factor the equation: \[ 3x^2 + 12xy - 13y^2 = 0 \] We can factor it as: \[ y(3x + 13y) = 0 \] So, the two lines represented by this equation are: 1. \( y = 0 \) 2. \( 3x + 13y = 0 \), or \( y = -\frac{3}{13}x \) 
Step 3: We now have the three lines:
1. \( 3x + y + 15 = 0 \)
2. \( y = 0 \)
3. \( y = -\frac{3}{13}x \)
Step 4: To find the area of the triangle formed by these lines, we can use the formula for the area of a triangle with three vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) as: \[ {Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] The points of intersection of these lines give the vertices of the triangle.
- The intersection of \( y = 0 \) and \( 3x + y + 15 = 0 \) gives the point \( (-5, 0) \).
- The intersection of \( y = 0 \) and \( y = -\frac{3}{13}x \) gives the point \( (0, 0) \).
- The intersection of \( 3x + y + 15 = 0 \) and \( y = -\frac{3}{13}x \) gives the point \( \left( -\frac{65}{13}, -\frac{3}{13} \times -\frac{65}{13} \right) \), or \( \left( -5, 15 \right) \).
Step 5: The area of the triangle is given by the formula: \[ {Area} = \frac{1}{2} \left| -5(0 - 15) + 0(15 - 0) + (-5)(0 - 0) \right| = \frac{1}{2} \left| -5(-15) \right| = \frac{1}{2} \times 75 = \frac{15 \sqrt{3}}{2} \] Thus, the area of the triangle formed by the lines is \( \frac{15 \sqrt{3}}{2} \).