A straight line passes through the points P(–1, 4) and Q(5, –2). It intersects x-axis and y-axis at the points A and B respectively and M is the mid-point of AB. Find :

(i) the equation of the line

(ii) the co-ordinates of A and B

(iii) the co-ordinates of M

(i) Given points, P(-1, 4) and Q(5, -2)

$\Rightarrow \text{ Slope of PQ } = \dfrac{y$2 - y1}{x2 - x1} \\[1em] = \dfrac{-2 - 4}{5 - (-1)} \\[1em] = \dfrac{-6}{6} = -1.

By point-slope form,

Equation of the line PQ, y - y₁ = m(x - x₁)

⇒ y - 4 = -1[x - (-1)]

⇒ y - 4 = -1[x + 1]

⇒ y - 4 = -x - 1

⇒ x + y = -1 + 4

⇒ x + y - 3 = 0.

Hence, equation of line is x + y - 3 = 0.

(ii) For point A (on x-axis), y = 0.

So, putting y = 0 in the equation of PQ, we have

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0).

For point B (on y-axis), x = 0.

So, putting x = 0 in the equation of PQ, we have

⇒ 0 + y = 3

⇒ y = 3

∴ B = (0, 3).

Hence, co-ordinates of A = (3, 0) and B = (0, 3).

(iii) M is the mid-point of AB.

∴ M = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

= $\Big(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\Big)$

= $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$

Hence, mid-point of AB = $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$.