Find the value of k such that the points P(k, 1), Q(2, –5) and R(k - 2, –3) are collinear.

Given,

Points P, Q and R are collinear.

Thus, the slope of PQ equal to the slope of QR.

By formula,

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

Substituting values we get,

$\Rightarrow \text{Slope of PQ} = \dfrac{-5 - 1}{2 - k} \\[1em] = \dfrac{-6}{2 - k} \\[1em] \Rightarrow \text{Slope of QR} = \dfrac{-3 -(-5)}{(k - 2) - 2} \\[1em] = \dfrac{-3 + 5}{k - 4} \\[1em] = \dfrac{2}{k - 4}.$

Slope of PQ = Slope of QR

$\Rightarrow \dfrac{-6}{2 - k} = \dfrac{2}{k - 4}$

⇒ -6(k - 4) = 2(2 - k)

⇒ -6k + 24 = 4 - 2k

⇒ 24 - 4 = -2k + 6k

⇒ 20 = 4k

⇒ k = $\dfrac{20}{4}$

⇒ k = 5

Hence, k = 5.