Prove that the equation $\;$ $x^2 - 3xy + 2y^2 + 4x - 5y + 3 = 0$ $\;$ represents two straight lines.
Given equation is: $\;$ $x^2 - 3xy + 2y^2 + 4x - 5y + 3 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general second degree equation
$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$a = 1, \; h = \dfrac{-3}{2}, \; b = 2, \; g = 2, \; f = \dfrac{-5}{2}, \; c = 3$
Equation $(2)$ represents a pair of straight lines if
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in the expression $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2$ $\;$ we have,
$1 \times 2 \times 3 + 2 \times \left(\dfrac{-5}{2}\right) \times 2 \times \left(\dfrac{-3}{2}\right) - 1 \times \left(\dfrac{-5}{2}\right)^2 - 2 \times 2^2 - 3 \times \left(\dfrac{-3}{2}\right)^2$
$= 6 + 15 - \dfrac{25}{4} - 8 - \dfrac{27}{4}$
$= 13 - \dfrac{52}{4} = 0$
i.e. $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
$\implies$ Equation $(1)$ represents a pair of straight lines.