If the two lines x + (a - 1)y = 1 and 2x + a ^ 2 y = 1, (a ∈ R - \...

If two lines ${a}_{1}x + {b}_{1}y = {c}_{1}$ and ${a}_{2}x + {b}_{2}y = {c}_{2}$ are perpendicular, then ${a}_{1}{a}_{2} + {b}_{1}{b}_{2} = 0 .$

Since, given lines $x + \left(a - 1\right)y = 1$ and $2x + {a}^{2}y = 1$ are perpendicular, hence, we get

$1 ⋅ 2 + \left(a - 1\right) ⋅ {a}^{2} = 0$

$⇒ {a}^{3} - {a}^{2} + 2 = 0$

$⇒ \left(a + 1\right) ⋅ \left({a}^{2} - 2a + 2\right) = 0$

$⇒ a = - 1$ or ${a}^{2} - 2a + 2 = 0$

But, the discriminant of the quadratic equation is ${\left(- 2\right)}^{2} - 4 \times 1 \times 2 = 4 - 8 = - 4,$ hence, no real roots exist.

Thus, the only possible value of $a$ is $- 1 .$

And, hence the lines are $x - 2y = 1$ and $2x + y = 1 .$

Point of intersection of these two lines is \left(\dfrac{3}{5}, - \dfrac{1}{5}\right)$</p><p>We know that, the distance of a point \left(x, y\right)$ from origin is $\sqrt{{x}^{2} + {y}^{2}}$
Hence, distance of the point $\left(\dfrac{3}{5}, - \dfrac{1}{5}\right)$ from origin $= \sqrt{\dfrac{9}{25} + \dfrac{1}{25}}$$= \sqrt{\dfrac{2}{5}} .$