Find $\;$ $p$ $\;$ and $\;$ $q$ $\;$ if the equation $\;$ $6x^2 + 5xy - py^2 + 7x + qy - 5 = 0$ $\;$ represents a pair of perpendicular lines.
Comparing the equation $\hspace{1em}$ $6x^2 + 5xy - py^2 + 7x + qy - 5 = 0$ $\;\;\; \cdots \; (1)$
with the standard equation $\hspace{1em}$ $ax^2 + 2 hxy + by^2 + 2gx + 2 fy + c = 0$ $\;$ gives
$\begin{aligned}
a = 6, & \hspace{1em} 2 f = q \implies f = \dfrac{q}{2} \\\\
b = - p, & \hspace{1em} 2 g = 7 \implies g = \dfrac{7}{2} \\\\
c = -5, & \hspace{1em} 2 h = 5 \implies h = \dfrac{5}{2}
\end{aligned}$
Given: The two lines represented by equation $(1)$ are perpendicular to each other.
Now, a pair of lines are perpendicular when $\;\;\;$ $a + b = 0$
Substituting the values of $a$ and $b$ we have,
$6 - p = 0$ $\implies$ $p = 6$
Condition for equation $(1)$ to represent a pair of lines is: $\hspace{1em}$ $abc + 2 fgh - af^2 - bg^2 - ch^2 = 0$
Substituting the values of $a$, $b$, $c$, $f$, $g$ and $h$ we have,
$6 \times \left(-6\right) \times \left(-5\right) + 2 \times \left(\dfrac{q}{2}\right) \times \left(\dfrac{7}{2}\right) \times \left(\dfrac{5}{2}\right)$
$\hspace{3em}$ $- 6 \times \left(\dfrac{q}{2}\right)^2 - \left(-6\right) \times \left(\dfrac{7}{2}\right)^2 - \left(-5\right) \times \left(\dfrac{5}{2}\right)^2 = 0$
i.e. $180 + \dfrac{35q}{4} - \dfrac{6q^2}{4} + \dfrac{294}{4} + \dfrac{125}{4} = 0$
i.e. $6 q^2 - 35 q - 1139 = 0$
i.e. $q = \dfrac{35 \pm \sqrt{35^2 + 4 \times 6 \times 1139}}{2 \times 6}$
i.e. $q = \dfrac{35 \pm \sqrt{28561}}{12} = \dfrac{35 \pm 169}{12}$
$\implies$ $q = 17$ $\;$ or $\;$ $q = \dfrac{-67}{6}$