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

(i) the equation of the line.

(ii) the co-ordinates of points A and B.

(iii) the co-ordinates of point M.

(i) Slope of PQ = $\dfrac{-2 - 4}{5 - (-1)} = -\dfrac{6}{6}$ = -1.

By point-slope form,

Equation of PQ is :

⇒ 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.

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

(ii) Given,

PQ intersects x-axis at A and y-axis at B.

At x-axis, y co-ordinate = 0.

Substituting y = 0 in equation we get,

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0)

At y-axis, x co-ordinate = 0.

Substituting x = 0 in equation we get,

⇒ 0 + y = 3

⇒ y = 3.

∴ B = (0, 3)

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

(iii) By formula,

Mid-point = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values we get,

$\text{M} = \Big(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\Big) \\[1em] = \Big(\dfrac{3}{2}, \dfrac{3}{2}\Big).$

Hence, co-ordinates of M = $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$.