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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.

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#### Solution

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

Since the given points are collinear, the area of the triangle ABC is 0*.*

`=>1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0`

Here, `x_1=-2, y_1=1, x_2=a, y_2=b and x_3=4, y_3=-1`

`therefore 1/2[-2(b+1)+a(-1-1)+4(1-b)]=0`

-2b-2-2a+4-4b=0

2a+6b=2

a+3b=1 ........(1)

Given :

a-b=1 ..............(2)

Subtracting equation (1) from (2), we get:

4b=0

b=0

Substituting *b* = 0 in (2), we get:

a=1

Thus, the values of *a* and *b* are 1 and 0, respectively.

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