Show that the equation $2x^2 - xy - 3y^2 - 6x + 19 y - 20 = 0$ represents a pair of intersecting lines. Find the equations of the two lines and the angle between them.
Comparing the given equation: $\hspace{1em}$ $2x^2 - xy - 3y^2 - 6x + 19 y - 20 = 0$ $\;\;\; \cdots \; (1)$
with the standard equation: $\hspace{1em}$ $ax^2 + 2 hxy + by^2 + 2 gx + 2fy + c = 0$ $\;$ gives
$a = 2$, $\;\;\;$ $2h = -1 \implies h = \dfrac{-1}{2}$, $\;\;\;$ $b = -3$,
$2 g = - 6 \implies g = -3$, $\;\;\;$ $2f = 19 \implies f = \dfrac{19}{2}$, $\;\;\;$ $c = -20$
Condition for equation $(1)$ to represent a pair of straight lines is if
$abc + 2 fgh - af^2 - bg^2 - ch^2 = 0$
Now, $abc + 2 fgh - af^2 - bg^2 - ch^2$
$= 2 \times \left(-3\right) \times \left(-20\right) + 2 \times \left(\dfrac{19}{2}\right) \times \left(-3\right) \times \left(\dfrac{-1}{2}\right)$
$\hspace{2em}$ $- 2 \times \left(\dfrac{19}{2}\right)^2 - \left(-3\right) \times \left(-3\right)^2 - \left(-20\right) \times \left(\dfrac{-1}{2}\right)^2$
$= 120 + \dfrac{57}{2} - \dfrac{361}{2} + 27 + 5$
$= 152 - \dfrac{304}{2} = 0 $
$\therefore$ $\;$ We have $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
$\implies$ Equation $(1)$ represents a pair of straight lines.
Factorizing the second degree terms in equation $(1)$, we have
$\begin{aligned}
2x^2 - xy - 3y^2 & \equiv 2x^2 + 2xy - 3xy - 3y^2 \\\\
& \equiv \left(2x - 3y\right) \left(x+y\right)
\end{aligned}$
$\therefore$ $\;$ $2x^2 - xy - 3y^2 - 6x + 19 y - 20 \equiv \left(2x - 3y + c_1\right) \left(x + y + c_2\right)$ $\;\;\; \cdots \; (2)$
where $c_1$ and $c_2$ are constants.
Equating the coefficients of the $x$ terms in equation $(2)$ we get,
$-6 = c_1 + 2c_2$ $\;\;\; \cdots \; (3a)$
Equating the coefficients of the $y$ terms in equation $(2)$ we get,
$19 = c_1 - 3c_2$ $\;\;\; \cdots \; (3b)$
Subtracting equations $(3a)$ and $(3b)$ we get,
$25 = -5 c_2$ $\implies$ $c_2 = -5$
Substituting the value of $c_2$ in equation $(3a)$ gives
$-6 = -10 + c_1$ $\implies$ $c_1 = 4$
$\therefore$ $\;$ The separate equations of the lines are
$2x - 3y + 4 = 0$ and $x + y - 5 = 0$
Angle between the lines $= \tan \theta = \left|\dfrac{2 \sqrt{h^2 -ab}}{a+b}\right|$
i.e. $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-1}{2}\right)^2 - 2 \times \left(-3\right)}}{2-3}\right|$
i.e. $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{1}{4}+ 6}}{-1}\right| = 2 \times \sqrt{\dfrac{25}{4}} = 5$
$\implies$ $\theta = \tan^{-1} \left(5\right)$