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 $b$.

Three points $A(-2,1),\ B(a,b)$ and $C(4,-1)$ are collinear so the area of the triangle is equal to zero

Here the equation $(1)$ is $\to a-b=1$

Here $x_1=-2\ y_1=1$

$x_1 =a\ y_1=b$

$x_3=4\ y_3=-1$

Area of the triangle $\triangle ABC=\dfrac {1}{2} [x_1 (y_2 -y_3)+x_2 (y_3 -y_1)+x_3 (y_1 -y_2)]=0$

$\Rightarrow \ \dfrac {1}{2} [-2(b+1)+a(-1 -1)+4(1-b)]=0$

$\Rightarrow \ \dfrac {1}{2}[-2b-2-2b+4-4b]=0$

$\Rightarrow \ \dfrac {1}{2} [-6b+2-2b]=0$

$\Rightarrow \ -3b+1-a=0$

$\Rightarrow \ -a-3b=-1$......Equation $2$

Equation $(1)\times 1\longrightarrow a-b=1$

Equation $(2)\times 1\to \dfrac {-a-2b=-1}{-4b=0\\ \Rightarrow b=0}$

Putting the value of $b=0$ in Equation $1:-$

$a-b=1$

$\Rightarrow \ a=1$

So the value of $a=1, b=0$