If the points A(1, 2), B(0, 0)

(c) Let the given points are B = (x1,y1) =  (1,2),

B = (x2,y2) = (0,0) and C3 = (x3,y3)= (a, b).

$\because$ Area of $\triangle A B C \Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\therefore \quad \Delta=\frac{1}{2}[1(0-b)+0(b-2)+a(2-0)]$

$=\frac{1}{2}(-b+0+2 a)=\frac{1}{2}(2 a-b)$

Since, the points $A(1,2), B(0,0)$ and $C(a, b)$ are collinear, then area of $\triangle A B C$ should be equal to zero.

i.e., $\quad$ area of $\triangle A B C=0$

$\Rightarrow \quad \frac{1}{2}(2 a-b)=0$

$\Rightarrow \quad 2 a-b=0$

$\Rightarrow \quad 2 a=b$

Hence, the required relation is 2a = b.