Show that the equation $\;$ $9x^2 - 12xy + 4y^2 + 15x - 10y + 4 = 0$ $\;$ represents a pair of parallel straight lines.
Given equation is: $\;$ $9x^2 - 12xy + 4y^2 + 15x - 10y + 4 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general second degree equation
$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$a = 9, \; h = -6, \; b = 4, \; g = \dfrac{15}{2}, \; f = -5, \; c = 4$
Equation $(2)$ represents a pair of parallel straight lines if
$h^2 = ab$ $\;$ and $\;$ $bg^2 = af^2$
Now, $\;$ $h^2 = \left(-6\right)^2 = 36$, $\;\;$ $a \times b = 9 \times 4 = 36$
i.e. $\;$ $h^2 = ab$
and $\;$ $bg^2 = 4 \times \left(\dfrac{15}{2}\right)^2 = 225$, $\;\;$ $af^2 = 9 \times \left(-5\right)^2 = 225$
i.e. $\;$ $bg^2 = af^2$
$\implies$ Equation $(1)$ represents a pair of parallel straight lines.