Coordinate Geometry — Maths STD 10 — Question

Let the given points are $A=\left(x_1, y_1\right)=(1,2), B=\left(x_2, y_2\right)=(0,0)$ and $C=\left(x_3, y_3\right)=(a, b)$.
$\because\text{Area of } \triangle\text{ABC }\triangle=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\therefore\triangle=\frac{1}{2}[1(0 - \text{b})+0(\text{b}-2)+\text{a}(2 -0)]$
$\triangle=\frac{1}{2}(\text{-b} + 0 +2\text{a})$
$\triangle=\frac{1}{2}(2\text{a}-\text{b})$
Since, the points $A(1, 2), B(0, 0)$ and $C(a, b)$ are collinear, then area of $\triangle\text{ABC}$ should be equal to zero.
$\text{i.e.,}\text{ area of }\triangle\text{ ABC}=0$
$\Rightarrow\frac{1}{2}(2\text{a} - \text{b})=0 $
$\Rightarrow2\text{a}-\text{b}=0$
$\Rightarrow2\text{a}=\text{b}$
Hence, the required relation is $2a = b.$