Find the value of $k$ in order that the equation $\;$ $2x^2 + xy - y^2 + kx + 6y - 9 = 0$ $\;$ may represent a pair of straight lines.
Given equation is: $\;$ $2x^2 + xy - y^2 + kx + 6y - 9 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general second degree equation
$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$a = 2, \; h = \dfrac{1}{2}, \; b = -1, \; g = \dfrac{k}{2}, \; f = 3, \; c = -9$
Equation $(2)$ represents a pair of straight lines if
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;\;\; \cdots \; (3)$
Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in equation $(3)$ gives
$2 \times \left(-1\right) \times \left(-9\right) + 2 \times 3 \times \dfrac{k}{2} \times \dfrac{1}{2} - 2 \times 3^2 - \left(-1\right) \times \left(\dfrac{k}{2}\right)^2 - \left(-9\right) \times \left(\dfrac{1}{2}\right)^2 = 0$
i.e. $\;$ $18 + \dfrac{3k}{2} - 18 + \dfrac{k^2}{4} + \dfrac{9}{4} = 0$
i.e. $\;$ $k^2 + 6k + 9 = 0$
i.e. $\;$ $\left(k + 3\right)^2 = 0$
$\implies$ $k = -3$