Find the value of 'a' for which the following points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.

Given that ,

A(a, 3), B(2, 1) and C(5, a) are collinear. Hence,

Slope of AB = Slope of BC

$\therefore \dfrac{1 - 3}{2 - a} = \dfrac{a - 1}{5 - 2} \qquad \text{….[Eq 1]} \\[1em] \Rightarrow \dfrac{-2}{2 - a} = \dfrac{a - 1}{3} \\[1em] \Rightarrow -2 \times 3 = (a - 1) \times (2 - a) \\[1em] \Rightarrow -6 = 2a - a^2 - 2 + a \\[1em] \Rightarrow -6 = 3a - a^2 - 2 \\[1em] \Rightarrow a^2 - 3a - 4 = 0 \\[1em] \Rightarrow a^2 - 4a + a - 4 = 0 \\[1em] \Rightarrow a(a - 4) + 1(a - 4) = 0 \\[1em] \Rightarrow (a + 1)(a - 4) = 0 \\[1em] \Rightarrow a + 1 = 0 \text{ or } a - 4 = 0 \\[1em] \Rightarrow a = -1 \text{ or } a = 4.$

Let us take points A and B for the equation, by two point form the equation of the line will be,

$y - y$1 = \dfrac{y2 - y1}{x2 - x1}(x - x1)

Putting values of points in above formula we get,

$\Rightarrow y - 3 = \dfrac{1 - 3}{2 - a}(x - a) \\[1em]$

Putting a = -1 in above equation,

$\Rightarrow y - 3 = \dfrac{-2}{2 - (-1)}(x - (-1)) \\[1em] \Rightarrow y - 3 = \dfrac{-2}{3}(x + 1) \\[1em] \Rightarrow 3(y - 3) = -2(x + 1) \\[1em] \Rightarrow 3y - 9 = -2x - 2 \\[1em] \Rightarrow 3y + 2x - 9 + 2 = 0 \\[1em] \Rightarrow 3y + 2x - 7 = 0.$

Putting a = 4 in above equation,

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

Hence, the equation of the line is 2x + 3y - 7 = 0 when a = -1 and the equation of the line is x - y - 1 = 0 when a = 4.