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If the length of the intercept made on the line \(y=a x\) by the lines \(y=2\) and \(y=6\) is less than 5
Options:
Solution:
2094 Upvotes
Verified Answer
The correct answer is:
\(a < \frac{-4}{3}\) or \(a>\frac{4}{3}\)
The point of intersection of lines \(y=a x\) and \(y=2\) is \(A\left(\frac{2}{a}, 2\right)\) and of the lines \(y=a x\) and \(y=6\) is \(B\left(\frac{6}{a}, 6\right)\), so
\(\begin{aligned}
& A B=\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^2+(6-2)^2} < 5 \quad \text{(given)} \\
& \Rightarrow \quad \frac{16}{a^2}+16 < 25 \\
& \Rightarrow \quad a^2 > \frac{16}{9} \Rightarrow a \in\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right) \\
\end{aligned}\)
(given)
Hence, option (d) is correct.
\(\begin{aligned}
& A B=\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^2+(6-2)^2} < 5 \quad \text{(given)} \\
& \Rightarrow \quad \frac{16}{a^2}+16 < 25 \\
& \Rightarrow \quad a^2 > \frac{16}{9} \Rightarrow a \in\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right) \\
\end{aligned}\)
(given)
Hence, option (d) is correct.
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