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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)

\\ 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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Vakul

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