1.

If the points A (-2,1), B (a, b) and C (4,-1) are collinear and a – b = 1, find the values of a and b.

Answer»

The given points A(−2, 1), B(a, b) and C(4, −1) are collinear. 

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 

∴ −2[b − ( − 1)] + a( − 1 – 1 ) + 4 ( 1 –b ) = 0 

-2b – 2 − 2a + 4 − 4b = 0 

− 2a − 6b = − 2 

a + 3b = 1 …(1) 

Also it is given that a – b = 1 …(2) 

Solving 1 and 2 simultaneously, 

b + 1 + 3b = 1 

4b = 0 

∴ b = 0 

∴ a =1 

Hence, 

the values of a and b are 1 and 0.



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