Show that the equation $\;$ $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$ $\;$ represents a pair of parallel straight lines and find the perpendicular distance between them.
Given equation is: $\;$ $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general second degree equation
$Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives
$A = 1, \; H = 1, \; B = 1, \; G = -4a, \; F = -4a, \; C = -9a^2$
Equation $(2)$ represents a pair of parallel straight lines if
$H^2 = AB$ $\;$ and $\;$ $BG^2 = AF^2$
Now, $\;$ $H^2 = 1^2 = 1$, $\;\;$ $A \times B = 1 \times 1 = 1$
i.e. $\;$ $H^2 = AB$
and $\;$ $BG^2 = 1 \times \left(-4a\right)^2 = 16a^2$, $\;\;$ $AF^2 = 1 \times \left(-4a\right)^2 = 16a^2$
i.e. $\;$ $BG^2 = AF^2$
$\implies$ Equation $(1)$ represents a pair of parallel straight lines.
Consider the terms $\;\;$ $x^2 + 2xy + y^2$ $\;\;$ from equation $(1)$.
Now, $\;\;$ $x^2 + 2xy + y^2 = \left(x + y\right) \left(x + y\right)$
$\therefore \;$ Equation $(1)$ can be written as
$\left(x + y + \ell\right) \left(x + y + m\right) = 0$ $\;\;\; \cdots \; (2) \;$ where $\ell$ and $m$ are constants
Since equations $(1)$ and $(2)$ represent the same pair of straight lines,
comparing the coefficients of the $x$ terms gives
$\ell + m = -8a$ $\;\;\; \cdots \; (3a)$
comparing the constant terms gives
$\ell \times m = -9a^2$ $\;\;\; \cdots \; (3b)$
Now,
$\begin{aligned}
\left(\ell - m\right)^2 & = \left(\ell + m\right)^2 - 4 \ell m \\\\
& = \left(-8a\right)^2 - 4 \times \left(-9a^2\right) \\\\
& = 64a^2 - 36a^2 = 28a^2
\end{aligned}$
$\therefore \;$ $\left|\ell - m\right| = 2a \sqrt{7}$ $\;\;\; \cdots \; (4)$
Equation $(1)$ is represented by the two parallel lines
$x + y + \ell = 0$ $\;\;\; \cdots \; (5a)$ $\;$ and $\;$ $x + y + m = 0$ $\;\;\; \cdots \; (5b)$
Distance between the parallel lines given by equations $(5a)$ and $(5b)$ is
$= \dfrac{\left|\ell - m\right|}{\sqrt{\left(\text{coefficient of x}\right)^2 + \left(\text{coefficient of y}\right)^2}}$
$= \dfrac{2a \sqrt{7}}{\sqrt{1^2 + 1^2}}$
$= \dfrac{2a \sqrt{7}}{\sqrt{2}} = a \sqrt{14}$