Step-by-step Solution:

We are given the equation of a pair of lines: \[ 6x^2 - xy - 2y^2 = 0 \] This is a homogeneous second-degree equation in the form: \[ Ax^2 + 2Hxy + By^2 = 0 \] We can rewrite the equation in this standard form: \[ 6x^2 - xy - 2y^2 = 0 \quad \Rightarrow \quad 6x^2 + (-1)xy + (-2)y^2 = 0 \] So, \[\] \( A = 6 \) \[\] \( 2H = -1 \Rightarrow H = -\frac{1}{2} \) \[\] \( B = -2 \) \[\] Step 1: Use the formula for slope difference \[\] If the equation is of the form: \[ Ax^2 + 2Hxy + By^2 = 0 \] Then the slopes \( m_1 \) and \( m_2 \) of the two lines satisfy: \[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \text{slope difference} \] But to directly find slope difference, we can instead solve using: \[\] Step 2: Use the quadratic in \( m \) (slope) \[\] Assume the pair of lines are \( y = mx \). Substitute into the equation: \[ 6x^2 - x(mx) - 2(m x)^2 = 0 \Rightarrow 6x^2 - m x^2 - 2m^2 x^2 = 0 \Rightarrow x^2(6 - m - 2m^2) = 0 \] Since \( x \neq 0 \), we solve: \[ 6 - m - 2m^2 = 0 \Rightarrow 2m^2 + m - 6 = 0 \] Use the quadratic formula: \[ m = \frac{-1 \pm \sqrt{1^2 + 4×2×6}}{2×2} = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4} \] So, \[ m_1 = \frac{6}{4} = \frac{3}{2}, \quad m_2 = \frac{-8}{4} = -2 \] Step 3: Find the difference \[ |m_1 - m_2| = \left| \frac{3}{2} - (-2) \right| = \left| \frac{3}{2} + 2 \right| = \frac{7}{2} \]