JEE Mains · Maths · STD 11 - 9. straight line
If the two lines \(x + \left( {a - 1} \right)\,y = 1\) and \(2x + {a^2}y = 1\,\left( {a \in R - \left\{ {0,1} \right\}} \right)\) are perpendicular, then the distance of their point of intersection from the origin is
- A \(\frac{2}{5}\)
- B \(\frac{{\sqrt 2 }}{5}\)
- C \(\frac{2}{{\sqrt 5 }}\)
- D \(\sqrt {\frac{2}{5}} \)
Answer & Solution
Correct Answer
(D) \(\sqrt {\frac{2}{5}} \)
Step-by-step Solution
Detailed explanation
Two lines are perpendicular \(\therefore {m_1}{m_2} = - 1\) \( \Rightarrow \left( {\frac{{ - 1}}{{a - 1}}} \right)\left( {\frac{{ - 2}}{{{a^2}}}} \right) = - 1\) \( \Rightarrow {a^3} - {a^2} + 2 = 0\) \( \Rightarrow \left( {a + 1} \right)\left( {{a^2} - 2a + 2} \right) = 0\)…
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