Find the value(s) of k so that PQ will be parallel to RS. Given :

(i) P(2, 4), Q(3, 6), R(8, 1) and S(10, k)

(ii) P(3, -1), Q(7, 11), R(-1, -1) and S(1, k)

(iii) P(5, -1), Q(6, 11), R(6, -4k) and S(7, k²)

Since, PQ is parallel to RS. So, slope of PQ will be equal to slope of RS.

(i) P(2, 4), Q(3, 6), R(8, 1) and S(10, k)

Slope of PQ = Slope of RS

$\therefore \dfrac{6 - 4}{3 - 2} = \dfrac{k - 1}{10 - 8} \\[1em] \Rightarrow \dfrac{2}{1} = \dfrac{k - 1}{2} \\[1em] \Rightarrow 4 = k - 1 \\[1em] \Rightarrow k = 4 + 1 = 5.$

Hence, k = 5.

(ii) P(3, -1), Q(7, 11), R(-1, -1) and S(1, k)

Slope of PQ = Slope of RS

$\therefore \dfrac{11 - (-1)}{7 - 3} = \dfrac{k - (-1)}{1 - (-1)} \\[1em] \Rightarrow \dfrac{12}{4} = \dfrac{k + 1}{2} \\[1em] \Rightarrow 3 = \dfrac{k + 1}{2} \\[1em] \Rightarrow k + 1 = 6 \\[1em] \Rightarrow k = 5.$

Hence, k = 5.

(iii) P(5, -1), Q(6, 11), R(6, -4k) and S(7, k²)

Slope of PQ = Slope of RS

$\therefore \dfrac{11 - (-1)}{6 - 5} = \dfrac{k^2 - (-4k)}{7 - 6} \\[1em] \Rightarrow \dfrac{12}{1} = \dfrac{k^2 + 4k}{1} \\[1em] \Rightarrow 12 = k^2 + 4k \\[1em] \Rightarrow k^2 + 4k - 12 = 0 \\[1em] \Rightarrow k^2 + 6k - 2k - 12 = 0 \\[1em] \Rightarrow k(k + 6) - 2(k + 6) = 0 \\[1em] \Rightarrow (k - 2)(k + 6) = 0 \\[1em] \Rightarrow k - 2 = 0 \text{ or } k + 6 = 0 \\[1em] \Rightarrow k = 2 \text{ or } k = -6.$

Hence, k = 2 or -6.