If two lines \(x+(a-1) y=1\) and \(2 x+\mathrm{a}^2 y=1(\mathrm{a} \in \mathrm{R}-\{0,1\})\) are perpendicular, then the distance of their point of intersection from the origin is

Given equations of lines
\(\begin{aligned}
& x+(a-1) y=1 \\
& 2 x+a^2 y=1
\end{aligned}\)
Slope of \(x+(a-1) y=1\) is \(\frac{-1}{a-1}\)
and slope of \(2 x+a^2 y=1\) is \(\frac{-2}{a^2}\)
Given lines are perpendicular
\(\begin{aligned}
& \therefore \quad \text { Product of slope }=-1 \\
& \therefore \quad \frac{-1}{(a-1)} \times \frac{-2}{a^2}=-1 \\
& \Rightarrow \frac{2}{\mathrm{a}^2(\mathrm{a}-1)}=-1 \\
& \Rightarrow \mathrm{a}^3-\mathrm{a}^2=-2 \\
& \Rightarrow \mathrm{a}^3-\mathrm{a}^2+2=0 \\
& \Rightarrow(\mathrm{a}+1)\left(\mathrm{a}^2-2 \mathrm{a}+2\right)=0 \\
& \Rightarrow \mathrm{a}=-1, \mathrm{a}^2-2 \mathrm{a}+2=0 \\
& \Rightarrow a^2-2 a+2 \neq 0 \ldots[\because a \in \dot{R}-\{0,1\}] \\
& \therefore \quad a=-1
\end{aligned}\)
\(\therefore \quad\) Equations will be
\(\begin{aligned}
& x-2 y=1 \\
& 2 x+y=1
\end{aligned}\)
\(\therefore \quad\) Point of intersection is \(\left(\frac{3}{5}, \frac{-1}{5}\right)\)
Distance of point of intersection from origin
\(\begin{aligned}
& =\sqrt{\left(0-\frac{3}{5}\right)^2+\left(0-\left(\frac{-1}{5}\right)\right)^2} \\
& =\sqrt{\frac{9}{25}+\frac{1}{25}} \\
& =\sqrt{\frac{10}{25}}=\sqrt{\frac{2}{5}}
\end{aligned}\)