The points $\left(\mathrm{x}+1, 2\right), \left(1, \mathrm{x}+2\right), \left(\frac{1}{\mathrm{x}+1}, \frac{2}{\mathrm{x}+1}\right)$ are collinear. If $\mathrm{x}\ne 0$, then$\left|\mathrm{x}\right|=$

The given points are: $\mathrm{A}\left(\mathrm{x}+1, 2\right), \mathrm{B}\left(1, \mathrm{x}+2\right), \mathrm{C}\left(\frac{1}{\mathrm{x}+1}, \frac{2}{\mathrm{x}+1}\right)$

Three points are collinear if the slope of any two pairs of points is the same.

∴ Slope of AB = Slope of BC

$\frac{\mathrm{x}+2-\left(2\right)}{\left(1-\mathrm{x}\right)-1}=\frac{\frac{2}{\mathrm{x}+1}-\left(\mathrm{x}+2\right)}{\frac{1}{\mathrm{x}+1}-\left(1\right)}$

$\Rightarrow \frac{\mathrm{x}}{-\mathrm{x}}=\frac{\frac{2-\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}{\mathrm{x}+1}}{\frac{1-\mathrm{x}-1}{\mathrm{x}+1}}$

$\Rightarrow -1=\frac{2-\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}{-\mathrm{x}}$

$\Rightarrow 2-\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)=\mathrm{x}$

$\Rightarrow 2-{\mathrm{x}}^2-3\mathrm{x}-2=\mathrm{x}$

$\Rightarrow -{\mathrm{x}}^2-4\mathrm{x}=0$

$\Rightarrow \mathrm{x}\left(\mathrm{x}+4\right)=0$

Given, $\mathrm{x}\ne 0$

Therefore, $\left(\mathrm{x}+4\right)=0$

$\mathrm{x}=-4$

$\therefore \left|\mathrm{x}\right|=\mathbf{4}$