If the equation 6x^2-11xy-10y^2-19y+c=0 represents a pair of straight lines find their quarks. Also find the angle between the two lines.

To determine whether the equation \(6x^2 - 11xy - 10y^2 - 19y + c = 0\) represents a pair of straight lines, we first need to analyze the conditions required for a quadratic equation in two variables to represent two lines. The general form of such an equation is \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). In your case, we can identify \(A = 6\), \(B = -11\), \(C = -10\), \(D = 0\), \(E = -19\), and \(F = c\).To find the value of \(c\) that allows this equation to represent two lines, the determinant formed by the coefficients must be zero. This determinant is given by the formula:\[D = B^2 - 4AC\]Substituting in the values we have:\[D = (-11)^2 - 4 \cdot 6 \cdot (-10)\]\[D = 121 + 240 = 361\]Since \(D\) is greater than zero, it indicates that the equation can represent two distinct lines. Now, to find the slopes (or quarks) of these lines, we can use the relation derived from the quadratic equation for the slopes \(m_1\) and \(m_2\):\[m^2 - \frac{B}{A}m - \frac{C}{A} = 0\]Replacing \(A\), \(B\), and \(C\):\[m^2 - \frac{-11}{6}m - \frac{-10}{6} = 0\]\[m^2 + \frac{11}{6}m + \frac{10}{6} = 0\]To make calculations easier, we can multiply the entire equation by 6 to eliminate fractions:\[6m^2 + 11m + 10 = 0\]Next, we apply the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = 11\), and \(c = 10\).Calculating the discriminant:\[b^2 - 4ac = 11^2 - 4 \cdot 6 \cdot 10 = 121 - 240 = -119\]Since the discriminant is negative, it indicates that there are no real roots, suggesting that the initial assumption that it represents two real lines was incorrect, unless \(c\) is adjusted to make \(D = 0\).To find the angle between the two lines, we can use the formula for the angle \(θ\) between two lines represented by slopes \(m_1\) and \(m_2\):\[\tan(θ) = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|\]Since we have not found \(m_1\) and \(m_2\) as real numbers, we cannot compute the angle unless we set \(D = 0\) by solving for \(c\).If we want to ensure that the equation does represent lines, we can set \(D = 0\):\[361 + c = 0 \implies c = -361\]Now we can substitute \(c = -361\) back into the equation and recalculate the slopes to find \(m_1\) and \(m_2\). After that, we can use the slopes to calculate the angle \(θ\).In summary, the initial equation can represent a pair of straight lines if \(c = -361\). The slopes of the lines can be determined through the quadratic equation method, and the angle can subsequently be calculated using the slopes. If you have any further questions or need more clarification on any part of this, feel free to ask!