If the two lines x + (a - 1)y = 1 and 2x + a^ | vedclass

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(D) Two lines are perpendicular,so the product of their slopes is $-1$.Slope of $L_1: x + (a - 1)y = 1$ is $m_1 = -\frac{1}{a - 1}$.Slope of $L_2: 2x

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$\sqrt{\frac{2}{5}}$

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(D) Two lines are perpendicular,so the product of their slopes is $-1$.Slope of $L_1: x + (a - 1)y = 1$ is $m_1 = -\frac{1}{a - 1}$.Slope of $L_2: 2x + a^2y = 1$ is $m_2 = -\frac{2}{a^2}$.Since $m_1 m_2 = -1$,we have $\left(-\frac{1}{a - 1}\right) \left(-\frac{2}{a^2}\right) = -1$.$\Rightarrow \frac{2}{a^2(a - 1)} = -1$ $\Rightarrow a^3 - a^2 + 2 = 0$.Factoring the cubic equation: $(a + 1)(a^2 - 2a + 2) = 0$.Since $a^2 - 2a + 2 = (a - 1)^2 + 1 > 0$,the only real solution is $a = -1$.Substituting $a = -1$ into the line equations:$L_1: x + (-1 - 1)y = 1 \Rightarrow x - 2y = 1$.$L_2: 2x + (-1)^2y = 1 \Rightarrow 2x + y = 1$.Solving the system:$x - 2y = 1$ $(1)$$2x + y = 1$ $(2)$Multiplying $(2)$ by $2$: $4x + 2y = 2$.Adding $(1)$ and $(2)$: $5x = 3 \Rightarrow x = \frac{3}{5}$.Substituting $x = \frac{3}{5}$ into $(2)$: $2(\frac{3}{5}) + y = 1 \Rightarrow y = 1 - \frac{6}{5} = -\frac{1}{5}$.The point of intersection is $P(\frac{3}{5}, -\frac{1}{5})$.The distance from the origin $(0, 0)$ is $OP = \sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$.

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

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