If the point (0,0), (1,2) and (x, y) are collinear, then find x.

It is given that three points (0,0), (1,2) and (x,y) are collinear and we need to find the value of x.

If three points are collinear, then it can be said that the area of the triangle formed by the points is zero. Mathematically, if points P, Q and R are collinear, then $\mathrm{ar} \left(\mathrm{△}\mathrm{PQR}\right)=0$, i.e., if coordinates of points P,Q and R are (x₁, y₁), (x₂, y₂) and (x₃, x₄) respectively, then

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

If we assume P $\mathrm{P}\left({\mathrm{x}}_1, {\mathrm{y}}_1\right)=(0 ,0)$$\mathrm{Q}\left({\mathrm{x}}_2, {\mathrm{y}}_2\right)=(1, 2)$and $\mathrm{R}\left({\mathrm{x}}_3, {\mathrm{y}}_3\right)=(\mathrm{x}, \mathrm{y})$then on substituting the values in the above relation, we get the value of x as

$\begin{array}{l}\frac{1}{2}\left[{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_1\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right]=0\\ \Rightarrow \frac{1}{2}\left[\right(0\left)\right(2-\mathrm{y})+(1\left)\right(\mathrm{y}-0)+(\mathrm{x}\left)\right(0-2\left)\right]=0\\ \Rightarrow \frac{1}{2}[0+\mathrm{y}+(\mathrm{x}\left)\right(-2\left)\right]=0\\ \Rightarrow \frac{1}{2}[\mathrm{y}-2\mathrm{x}]=0\end{array}$

On multiplying both sides by 2, we get

$\begin{array}{l}\Rightarrow \mathrm{y}-2\mathrm{x}=0\\ \Rightarrow 2\mathrm{x}=\mathrm{y}\\ \Rightarrow \mathrm{x}=\frac{\mathrm{y}}{2}\end{array}$

Hence, the value of x is $\frac{\mathrm{y}}{2}$.