Question

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

• A. $\frac{2}{5}$
• B. $\frac{2}{\sqrt{5}}$
• C. $\frac{\sqrt{2}}{5}$
• D. $\sqrt{\frac{2}{5}}$

Answer 1

Answer: (D)

$\left(\frac{-1}{a-1}\right)\left(\frac{-2}{a^{2}} \right) = - 1 $
$ 2= -\left(a^{2}\right)\left(a-1\right) $
$ a^{3} -a^{2} + 2 =0 $
$ \left(a+1\right)\left(a^{2} - 2a +2 \right) = 0 $
$ \therefore a =- 1 $
$\left.\begin{aligned} L_1 : x -2 y + 1 = 0 \\ L_2 : 2x +y - 1 = 0 \end{aligned}\right\rbrace $
$ 0\left(0,0\right) P\left(\frac{1}{5}, \frac{3}{5}\right) $
$ OP = \sqrt{\frac{1}{25} + \frac{9}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}} $