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

\frac{2}{5}

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

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

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

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Solution

The correct option is B $\sqrt{\frac{2}{5}}$
$(\frac{- 1}{a - 1}) (\frac{- 2}{a^{2}}) = - 1$
$2 = - (a^{2}) (a - 1)$
$a^{3} - a^{2} + 2 = 0$
$(a + 1) (a^{2} - 2 a + 2) = 0$
$O (0, 0), P (\frac{1}{5}, \frac{3}{5})$
$O P = \sqrt{\frac{1}{25} + \frac{9}{25}} = \sqrt{\frac{2}{5}}$

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If the two lines x+(a−1)y=1 and 2x+a2y=1(a∈R−{0,1}) are perpendicular, then the distance of their point of intersection from the origin is:

The correct option is B √25(−1a−1)(−2a2)=−12=−(a2)(a−1)a3−a2+2=0(a+1)(a2−2a+2)=0O(0,0),P(15,35)OP=√125+925=√25

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

The correct option is B $\sqrt{\frac{2}{5}}$
$(\frac{- 1}{a - 1}) (\frac{- 2}{a^{2}}) = - 1$
$2 = - (a^{2}) (a - 1)$
$a^{3} - a^{2} + 2 = 0$
$(a + 1) (a^{2} - 2 a + 2) = 0$
$O (0, 0), P (\frac{1}{5}, \frac{3}{5})$
$O P = \sqrt{\frac{1}{25} + \frac{9}{25}} = \sqrt{\frac{2}{5}}$