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in Coordinate Geometry by (28.9k points)
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For what value of a the point (a, 1),(1, -1) and(11, 4) are collinear?

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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

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