Point A and B have co-ordinates (7, -3) and (1, 9) respectively. Find :

(i) the slope of AB.

(ii) the equation of perpendicular bisector of the line segment AB.

(iii) the value of 'p' if (-2, p) lies on it.

(i) By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

Substituting values we get,

$\text{Slope of AB } = \dfrac{9 - (-3)}{1 - 7} \\[1em] = \dfrac{12}{-6} \\[1em] = -2.$

Hence, slope of AB = -2.

(ii) Let O be the mid-point of AB.

By mid-point formula,

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

O = $\Big(\dfrac{7 + 1}{2}, \dfrac{-3 + 9}{2}\Big) = \Big(\dfrac{8}{2}, \dfrac{6}{2}\Big)$ = (4, 3).

⇒ Slope of AB × Slope of perpendicular bisector = -1 (As product of slopes of perpendicular lines = -1)

⇒ -2 × Slope of perpendicular bisector = -1

⇒ Slope of perpendicular bisector = $\dfrac{1}{2}$.

By point-slope form,

Equation of perpendicular bisector is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $\dfrac{1}{2}$(x - 4)

⇒ 2(y - 3) = 1(x - 4)

⇒ 2y - 6 = x - 4

⇒ x - 2y - 4 + 6 = 0

⇒ x - 2y + 2 = 0.

Hence, the equation of perpendicular bisector of AB is x - 2y + 2 = 0.

(iii) Since, (-2, p) lies on perpendicular bisector. So, it will satisfy the equation.

⇒ -2 - 2p + 2 = 0

⇒ -2p = 0

⇒ p = 0.

Hence, p = 0.