The three given points are A(a, 1), B(1, −1) and C(11, 4).
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= \(\frac{1}2\) |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Given that area of ∆ABC = 0
∴ 0 = \(\frac{1}2\) |a(-1 – 4) + 1(4 - 1) + 11(1 – (-1))|
∴ 0 = \(\frac{1}2\) |-5a + 3 + 22|
∴ -5a + 3 + 22 = 0
a = 5
Hence the value of a is 5