Q.Show that $x^2 - y^2 + x - 3y - 2 = 0$ represents a pair of straight lines. Also find the angle between them.

The equation factors as $(x-y-1)(x+y+2)=0$, confirming it is a pair of straight lines with slopes $1$ and $-1$ — which are perpendicular, so the angle between them is $90^\circ$.

Writing $x^2-y^2+x-3y-2=0$ in the general second-degree form $ax^2+2hxy+by^2+2gx+2fy+c=0$: here $a=1$, $b=-1$, $h=0$, $g=\dfrac12$, $f=-\dfrac32$, $c=-2$.

Condition check (a pair of lines requires $abc+2fgh-af^2-bg^2-ch^2=0$):

$abc = (1)(-1)(-2)=2$

$2fgh = 2\left(-\dfrac32\right)\left(\dfrac12\right)(0) = 0$

$af^2 = (1)\left(\dfrac94\right) = \dfrac94$

$bg^2 = (-1)\left(\dfrac14\right) = -\dfrac14$

$ch^2 = (-2)(0) = 0$

Sum: $2+0-\dfrac94-\left(-\dfrac14\right)-0 = 2-\dfrac94+\dfrac14 = 2-2 = 0$ ✓ — condition satisfied, so this is indeed a pair of straight lines.

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