Find the equation of a line passing through the point P(–2, 1) and parallel to the line joining the points A(4, –3) and B(–1, 5).

Since the required line is parallel to the line segment AB, they must have the same gradient.

Slope of AB = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{5 - (-3)}{-1 - 4} = -\dfrac{8}{5}

Using the point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $-\dfrac{8}{5}$[x - (-2)]

⇒ 5(y - 1) = -8(x + 2)

⇒ 5y - 5 = -8x - 16

⇒ 8x + 5y - 5 + 16 = 0

⇒ 8x + 5y + 11 = 0

Hence, the equation of the required line is 8x + 5y + 11 = 0.