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

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(D) Given equations of lines: $x+(a-1)y=1$ and $2x+a^2y=1$. Slope of $x+(a-1)y=1$ is $m_1 = \frac{-1}{a-1}$. Slope of $2x+a^2y=1$ is $m_2 = \frac{-2}{

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

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(D) Given equations of lines: $x+(a-1)y=1$ and $2x+a^2y=1$. Slope of $x+(a-1)y=1$ is $m_1 = \frac{-1}{a-1}$. Slope of $2x+a^2y=1$ is $m_2 = \frac{-2}{a^2}$. Since the lines are perpendicular,$m_1 	imes m_2 = -1$. $\frac{-1}{a-1} 	imes \frac{-2}{a^2} = -1$ $\frac{2}{a^2(a-1)} = -1$ $2 = -a^3 + a^2$ $a^3 - a^2 + 2 = 0$. Factoring the cubic equation: $(a+1)(a^2-2a+2) = 0$. Since $a^2-2a+2 = (a-1)^2 + 1 > 0$ for all real $a$,we must have $a+1=0$,so $a=-1$. Substituting $a=-1$ into the line equations: $x+(-1-1)y=1 \Rightarrow x-2y=1$. $2x+(-1)^2y=1 \Rightarrow 2x+y=1$. Solving the system: $x-2y=1$ $(i)$ $2x+y=1$ (ii) Multiply (ii) by $2$: $4x+2y=2$. Adding $(i)$ and (ii): $5x=3 \Rightarrow x=\frac{3}{5}$. Substituting $x=\frac{3}{5}$ into (ii): $2(\frac{3}{5})+y=1 \Rightarrow y=1-\frac{6}{5} = -\frac{1}{5}$. The point of intersection is $(\frac{3}{5}, -\frac{1}{5})$. The distance from the origin $(0,0)$ is $\sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$.

If 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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