Answer :

The given points A(−1, −4), B(b, c) and C(5, −1) are collinear.


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


1[c − (− 1)]+b[− 1 − ( − 4)] + 5( − 4 c) = 0


c 1 + 3b 20 5c = 0


3b 6c = 21


b 2c = 7 …(1)


Also it is given that 2b + c = 4 …(2)


Solving 1 and 2 simultaneously, we get,


2(7 + 2c) + c = 4


14 + 4c + c = 4


5c = − 10


c = − 2


b = 3


Hence, value of b and c are 3 and -2 respectively


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