The slope of a line joining P(6, k) and Q(1 - 3k, 3) is $\dfrac{1}{2}$. Find :

(i) k

(ii) mid-point of PQ, using the value of 'k' found in (i).

(i) By formula,

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

Substituting values we get,

$\Rightarrow \text{Slope of PQ} = \dfrac{3 - k}{1 - 3k - 6} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{3 - k}{-3k - 5} \\[1em] \Rightarrow -3k - 5 = 2(3 - k) \\[1em] \Rightarrow -3k - 5 = 6 - 2k \\[1em] \Rightarrow -3k + 2k = 6 + 5 \\[1em] \Rightarrow -k = 11 \\[1em] \Rightarrow k = -11.$

Hence, k = -11.

(ii) P = (6, k) = (6, -11).

Q = (1 - 3k, 3) = (1 - 3(-11), 3) = (1 + 33, 3) = (34, 3).

Mid-point of PQ = $\Big(\dfrac{6 + 34}{2}, \dfrac{-11 + 3}{2}\Big) = \Big(\dfrac{40}{2}, \dfrac{-8}{2}\Big)$ = (20, -4).

Hence, mid-point of PQ = (20, -4).