The points (5, -2), (8, -3) and (a, -12) are collinear if the value of a is 

Concept:

If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area (A) of ΔABC is given as; 

\(A=\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

Calculation: 

Here, we have to find the value of a for  which the points (5, -2), (8, -3) and (a, -12) are collinear

Let,

x1 = 5, y1 = -2,

x2 = 8, y2 = -3,

x3 = a, y3 = -12.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a Δ ABC then area of Δ ABC =   \(A= \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

\(⇒ A = \frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{5}}&{{-2}}&1\\ {{8}}&{{-3}}&1\\ {{a}}&{{-12}}&1 \end{array}} \right|\)

2A = 5 (-3 + 12) + 2(8 - a) + 1(-96 + 3a)

2A = 45 + 16 - 2a - 96 + 3a

2A = a - 35  

⇒ A = (a - 35)/2

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ (a - 35)/2 = 0

⇒ a = 35

Hence, option D is the correct answer.

Alternate Method 

Concept:

Three or more points are collinear if the slope of any two pairs of points is the same.

Calculation:

Let, A = (5, -2), B = (8, -3), C = (a, -12)

Now, the slope of AB = Slope of BC = Slope of AC      (∵ points are collinear)

 \(\rm \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{a-8}\\ ⇒ \frac{-1}{3}=\frac{-9}{a-8}\)

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.