Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ is at a distance of $3$ units from the point $(-1, 2)$.

(A) Let the line passing through $(-1, 2)$ make an angle $\theta$ with the positive $x$-axis. The coordinates of any point on this line at a distance $r = 3$ from $(-1, 2)$ are given by $(x, y) = (-1 + r \cos \theta, 2 + r \sin \theta)$.
Since $r = 3$,the point is $(-1 + 3 \cos \theta, 2 + 3 \sin \theta)$.
This point lies on the line $x + y = 4$,so:
$(-1 + 3 \cos \theta) + (2 + 3 \sin \theta) = 4$
$1 + 3(\cos \theta + \sin \theta) = 4$
$3(\cos \theta + \sin \theta) = 3$
$\cos \theta + \sin \theta = 1$
Dividing by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta = \frac{1}{\sqrt{2}}$
$\cos(\theta - 45^{\circ}) = \cos 45^{\circ}$
$\theta - 45^{\circ} = \pm 45^{\circ}$
Case $1$: $\theta - 45^{\circ} = 45^{\circ} \Rightarrow \theta = 90^{\circ}$.
Case $2$: $\theta - 45^{\circ} = -45^{\circ} \Rightarrow \theta = 0^{\circ}$.
Thus,the required directions are $0^{\circ}$ or $90^{\circ}$ with the $x$-axis.