Show that the equation $4x^2 + 4 xy + y^2 - 6x - 3y - 4 = 0$ represents a pair of parallel lines. Find the distance between them.
Comparing the given equation $\;$ $4x^2 + 4 xy + y^2 - 6x - 3y - 4 = 0$ $\;\;\; \cdots \; (1)$
with the standard equation $\;$ $ax^2 + 2 hxy + by^2 + 2 gx + 2 fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$\begin{aligned}
a = 4, & \hspace{1em} 2 f = -3 \implies f = \dfrac{-3}{2} \\\\
b = 1, & \hspace{1em} 2 g = -6 \implies g = -3 \\\\
c = -4, & \hspace{1em} 2 h = 4 \implies h = 2
\end{aligned}$
Two straight lines represented by the standard equation $(2)$ are parallel if $\;\;$ $af^2 = bg^2$
Now, $a f^2 = 4 \times \left(\dfrac{-3}{2}\right)^2 = 9$ $\;\;\; \cdots \; (3a)$
and $bg^2 = 1 \times \left(-3\right)^2 = 9$ $\;\;\; \cdots \; (3b)$
$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$ $\;$ $af^2 = bg^2$
$\implies$ Equation $(1)$ represents a pair of parallel lines.
Distance between the pair of parallel lines $= d = 2 \times \sqrt{\dfrac{g^2 - ac}{a \left(a + b\right)}}$
i.e. $d = 2 \times \sqrt{\dfrac{\left(-3\right)^2 - 4 \times \left(-4\right)}{4 \left(4 + 1\right)}} = 2 \times \sqrt{\dfrac{9 + 16}{4 \times 5}} = \sqrt{5}$
$\therefore$ $\;$ Distance between the pair of parallel lines represented by equation $(1)$ is $d = \sqrt{5}$ units.