Correct option is (A) 5
Consider the given points
(k, 1), (1, -1) and (11, 4)
Since, these points are collinear means that the area of triangle must be zero.
So,
\(\frac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| = 0\)
where \((x_1, \ y_1), (x_2, \ y_2), (x_3, y_3)\) are the points
Therefore,
\(\frac{1}{2} |k(-1-4)+1(4-1)+11(1-(-1))| = 0\)
\(|k(-5)+1(3)+11(2)| = 0\)
-5k + 3 + 22 = 0
-5k + 25 = 0
-5k = -25
k = \(\frac{25}{5}\)
\(\therefore k = 5\)
