Step -1 : Find equation

Since A, O, C are linear they will lie on same line.

Given points are A(1, 2), O(0, 0) and C(a, b)

Therefore

area ΔAOC = 0

$\therefore \frac{1}{2}[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)]=0$

Here x₁ = 1, x₂ = 0 and x₃ = a

and y₁ = 2, y₂ = 0 and y₃ = b

Step - 2 : Find the relation

$\frac{1}{2}[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)]=0$

$\Rightarrow \frac{1}{2}[1(0-\mathrm{b})+0(\mathrm{b}-2)+\text{a}(2-0)]=0$

⇒ – b + 2a = 0

⇒ 2a = b

Hence the correct answer is option C