Find the value of k for which the points A (-1,3), B (2, k) and C (5, -1) are collinear.

Given points are $A(-1, 3)$, $B(2, k)$, and $C(5, -1)$.

For three points to be collinear, the slope between any two pairs of points must be equal.

For points $A(-1, 3)$ and $B(2, k)$:

Slope of $AB = \dfrac{k - 3}{2 - (-1)}$

$= \dfrac{k - 3}{3}$

For points $B(2, k)$ and $C(5, -1)$:

Slope of $BC = \dfrac{-1 - k}{5 - 2}$

$= \dfrac{-1 - k}{3}$

For collinearity, slope of $AB =$ slope of $BC$:

$\dfrac{k - 3}{3} = \dfrac{-1 - k}{3}$

$k - 3 = -1 - k$

$k + k = -1 + 3$

$2k = 2$

$k = 1$

Therefore, the value of $k = 1$.