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If the points (2, 1), (−1,4)(-1, 4) and (a, 3) are collinear then the value/(s) of a is/(are) :

Solution

✅ Correct Option: 3

Points are collinear if ∣211−141a31∣=0\begin{vmatrix} 2 & 1 & 1 \\ -1 & 4 & 1 \\ a & 3 & 1 \end{vmatrix} = 0.

Expand: 2(4−3)−1(−1−a)+1(−3−4a)=02(4-3) - 1(-1-a) + 1(-3-4a) = 0.

2+1+a−3−4a=0⇒−3a=0⇒a=02 + 1 + a - 3 - 4a = 0 \Rightarrow -3a = 0 \Rightarrow a = 0.

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