The distance of the origin from the centroid of the triangle whose two sides have the equations x - 2 y + 1…

Let $AB\equiv x-2y+1=0$ and $AC\equiv 2x-y-1=0$

So, $A\left(1, 1\right)$

Now, altitude from $B$ which is perpendicular to line $2x-y-1=0$ and which passes through $\left(\frac{7}{3},\frac{7}{3}\right)$ is  $BH=x+2y-7=0$,

Now intersection of $BH \& AB$ will give point $B\left(3, 2\right)$

Similarly, altitude from $C$ is $CH=2x+y-7=0\Rightarrow C\left(2, 3\right)$

Centroid of $△ABC=E\left(2, 2\right) \& OE=2\sqrt{2}$

where $O$ is origin.