To determine the value of \(\lambda\) such that the angles between the pair of straight lines represented by the equation:
\(x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0\)
are equal to \(\tan^{-1}(1/3)\), we begin with understanding the general form for the equation of a pair of lines:
\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\)
In this case, comparing with our equation, we have:
The formula for the tangent of the angle between the lines is given by:
\(|\tan \theta| = \frac{2\sqrt{h^2 - ab}}{a + b}\)
Substituting the given values, \(|\tan \theta| = \frac{1}{3}\) and using the expression:
\(\frac{2\sqrt{\left(-\frac{3}{2}\right)^2 - (1 \cdot \lambda)}}{1 + \lambda} = \frac{1}{3}\)
Simplifying inside the square root,
\(\frac{2\sqrt{\frac{9}{4} - \lambda}}{1 + \lambda} = \frac{1}{3}\)
Cross-multiply to clear the fraction:
\(6\sqrt{\frac{9}{4} - \lambda} = 1 + \lambda\)
Isolating the square root gives:
\(6\sqrt{\frac{9}{4} - \lambda} = 1 + \lambda\)
Divide both sides by 6:
\(\sqrt{\frac{9}{4} - \lambda} = \frac{1 + \lambda}{6}\)
Square both sides to eliminate the square root:
\(\frac{9}{4} - \lambda = \left(\frac{1 + \lambda}{6}\right)^2\)
Expanding the right-hand side, we have:
\(\left(\frac{1 + \lambda}{6}\right)^2 = \frac{(1 + \lambda)^2}{36}\)
Equating the equations:
\(\frac{9}{4} - \lambda = \frac{(1 + \lambda)^2}{36}\)
Clearing the fraction gives:
\(9 \times 36 = 4 \times ((1 + \lambda)^2 + 36 \lambda)\)
Solve for \(\lambda\):
By solving this equation, it ultimately yields \(\lambda = 2\).
Therefore, the correct answer is
2
.