If `theta` is the angle between the lines given by the equation `6x^2+5x y-4y^2+7x+13 y-3=0` , then find the equation of the line passing through the point of intersection of these lines and making an angle `theta` with the positive x-axis.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Identify the given equation
We are given the equation of the pair of straight lines:
\[ 6x^2 + 5xy - 4y^2 + 7x + 13y - 3 = 0 \]

### Step 2: Differentiate the equation
To find the point of intersection of the lines represented by the equation, we need to differentiate the equation with respect to \(x\) and \(y\).

1. Differentiate with respect to \(x\):
\[ \frac{\partial F}{\partial x} = 12x + 5y + 7 = 0 \quad \text{(Equation 1)} \]

2. Differentiate with respect to \(y\):
\[ \frac{\partial F}{\partial y} = 5x - 8y + 13 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we solve the two equations simultaneously:

From Equation 1:
\[ 12x + 5y + 7 = 0 \implies 5y = -12x - 7 \implies y = -\frac{12}{5}x - \frac{7}{5} \]

Substituting \(y\) in Equation 2:
\[ 5x - 8\left(-\frac{12}{5}x - \frac{7}{5}\right) + 13 = 0 \]
\[ 5x + \frac{96}{5}x + \frac{56}{5} + 13 = 0 \]
Multiply through by 5 to eliminate the fraction:
\[ 25x + 96x + 56 + 65 = 0 \]
\[ 121x + 121 = 0 \implies x = -1 \]

Substituting \(x = -1\) back into the equation for \(y\):
\[ y = -\frac{12}{5}(-1) - \frac{7}{5} = \frac{12}{5} - \frac{7}{5} = \frac{5}{5} = 1 \]

Thus, the point of intersection is:
\[ (-1, 1) \]

### Step 4: Find the angle \( \theta \)
Using the formula for the angle between two lines given by the coefficients of the quadratic equation:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
Where \(a = 6\), \(b = -4\), and \(h = \frac{5}{2}\).

Calculating:
\[ \tan \theta = \frac{2\sqrt{\left(\frac{5}{2}\right)^2 - (6)(-4)}}{6 - 4} = \frac{2\sqrt{\frac{25}{4} + 24}}{2} = \sqrt{\frac{25 + 96}{4}} = \sqrt{\frac{121}{4}} = \frac{11}{2} \]

### Step 5: Write the equation of the line
The slope \(m\) of the line making an angle \(\theta\) with the positive x-axis is:
\[ m = \frac{11}{2} \]

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(x_1 = -1\), \(y_1 = 1\), and \(m = \frac{11}{2}\):
\[ y - 1 = \frac{11}{2}(x + 1) \]
Multiplying through by 2 to eliminate the fraction:
\[ 2y - 2 = 11x + 11 \]
Rearranging gives:
\[ 11x - 2y + 13 = 0 \]

### Final Answer
The equation of the line passing through the point of intersection and making an angle \(\theta\) with the positive x-axis is:
\[ 11x - 2y + 13 = 0 \]