Let θ be the angle formed by the required direction with the positive direction of x-axis, then the equation of line passes through a point (–1, 2) and in the positive direction is $\frac{x-(-1)}{\cos\theta}$ = $\frac{y-2}{\sin\theta}$ = a Clearly |a| is the distance of (x, y) from (–1, 2). Any point on this line is (x, y) = (a cosθ – 1, a sinθ + 2) Since $\frac{x+1}{\cos\theta}$ = a ⇒ x = a cos θ - 1 and $\frac{y-2}{\sin\theta}$ = a ⇒ y = a sin θ + 2 We have given that when this point lies on x + y = 4, then |a| = 3, ∴ a cosθ – 1 + a sinθ + 2 = 4, |a| = 3 ⇒ a (cos θ + sin θ) = 3 , $a^2$ = 9 ⇒ $(\cos\theta + \sin\theta)^2$ = $\frac{9}{a^2}, a^2$ = 9 ⇒ $\cos^2\theta + \sin^2\theta$ + 2 sin θ cos θ = $\frac{9}{9}$ = 1 ⇒ 1 + sin2θ = 1 ⇒ sin2θ = 0 ⇒ 2θ = 0° or 180° ⇒ θ = 0° or 90° Hence the required line is either parallel to x-axis or parallel to y-axis.