Prove that the equation $\;$ $x^2 - 5xy + 4y^2 + x + 2y - 2 = 0$ $\;$ represents two straight lines.
Given equation is: $\;$ $x^2 - 5xy + 4y^2 + x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general second degree equation
$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$a = 1, \; h = \dfrac{-5}{2}, \; b = 4, \; g = \dfrac{1}{2}, \; f = 1, \; c = -2$
Equation $(2)$ represents a pair of straight lines if
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;$ i.e. $\;$ $\begin{vmatrix}
a & h & g \\
h & b & f \\
g & f & c
\end{vmatrix} = 0$
Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in the expression $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2$ $\;$ we have,
$1 \times 4 \times \left(-2\right) + 2 \times 1 \times \dfrac{1}{2} \times \left(\dfrac{-5}{2}\right) - 1 \times 1^2 - 4 \times \left(\dfrac{1}{2}\right)^2 - \left(-2\right) \times \left(\dfrac{-5}{2}\right)^2$
$= -8 - \dfrac{5}{2} -1 -1 + \dfrac{25}{2}$
$= -10 + 10 = 0$
i.e. $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
$\implies$ Equation $(1)$ represents a pair of straight lines.