Points A and B have coordinates (7, -3) and (1, 9) respectively. Find

(i) the slope of AB.

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

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

(i) Slope (m₁) of AB is,

$\text{m}$1 = \dfrac{y2 - y1}{x2 - x_1} \\[1em] = \dfrac{9 - (-3)}{1 - 7} \\[1em] = -\dfrac{12}{6} \\[1em] = -2.

Hence, the slope of AB is -2.

(ii) Let PQ be the perpendicular bisector of AB intersecting it at M. Now, the coordinates of M will be

$= \Big(\dfrac{7 + 1}{2}, \dfrac{-3 + 9}{2}\Big) \\[1em] = (4, 3).$

Let the slope of the line PQ be m₂. Since, PQ is perpendicular to AB then product of their slopes will be equal to -1,

$\therefore m$1 \times m2 = -1 \\[1em] \Rightarrow -2 \times m2 = -1 \\[1em] \Rightarrow m2 = \dfrac{1}{2}.

Thus, by point-slope form equation of PQ is,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 3 = \dfrac{1}{2}(x - 4) \\[1em] \Rightarrow 2(y - 3) = x - 4 \\[1em] \Rightarrow 2y - 6 = x - 4 \\[1em] \Rightarrow x - 2y + 2 = 0.

Hence, the equation of the required line is x - 2y + 2 = 0.

(iii) As (-2, p) lies on the above line. The point will satisfy the line equation x - 2y + 2 = 0.

⇒ -2 - 2p + 2 = 0
⇒ 2p = 0
⇒ p = 0.

Hence, the value of p is 0.