Answer :

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)


= |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|


Given that area of ∆ABC = 0


0 = |a(-1 – 4) + 1(4 - 1) + 11(1 – (-1))|


0 = |-5a + 3 + 22|


-5a + 3 + 22 = 0


a = 5


Hence the value of a is 5


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