# If the two lines  $x+(a-1)y=1\:\:\:and\:\:\:2x+a^{2}y=1\:(a\:\epsilon \:R-\left \{ 0,1 \right \})$   are perpendicular , then the distance of their point of intersection from the origin is : Option 1) $\sqrt{\frac{2}{5}}$ Option 2) $\frac{2}{5}$ Option 3) $\frac{2}{\sqrt{5}}$ Option 4) $\frac{\sqrt{2}}{5}$

Answers (1)
V Vakul

$\\ x+(a-1)y=1 \\\\\: 2x+a^{2}y =1$

$\\ m_{1}=\frac{1}{-(a-1)} \\\\\: m_{2}= - \frac{2}{a^{2}}$

$\\ m_{1}m_{2}=-1\\\\\:\frac{-1}{(a-1)}\:\frac{-2}{a^{2}}=-1$

$\\2=a^{2}(a-1)\\\\\:a^{3}-a^{2}-2=0$

$\\=(a+1)(a^{2}-2a+2)=0\\\\\:a=-1$

$\\\therefore lines \:\:are=x-2y=1\\\\\:\:\:\:\:\:\:2x+y=1$

$\\2x+y-2(x-2y)=1-2(1)\\\\\:2x+y-2x+4y=1-2=-1$

$\\5y=-1\\\\\:y=\frac{-1}{5}$

$\\x-2y=1\\\\\:x-2\left ( \frac{-1}{5} \right )=1\\\\\:x=1-\frac{2}{5}=\frac{3}{5}$

$distance=\sqrt{x^{2}+y^{2}}=\sqrt{\left ( \frac{1}{5} \right )^{2}+\left ( \frac{3}{5} \right )^{2}}=\frac{\sqrt{10}}{5}=\sqrt{\frac{2}{5}}$

Option 1)

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

Option 2)

$\frac{2}{5}$

Option 3)

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

Option 4)

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

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