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If $P (\frac{1 + t}{\sqrt{2}}, \frac{2 + t}{\sqrt{2}})$ be any point on a line, then the range of values of t for which the point P lies between the parallel lines $x + 2 y = 1 a n d 2 x + 4 y = 15$ is

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$- \frac{4 \sqrt{2}}{3} < t < \frac{5 \sqrt{2}}{6}$

$0 < t < \frac{5 \sqrt{2}}{6}$

$- \frac{4 \sqrt{2}}{5} < t < 0$

$- \frac{4 \sqrt{2}}{5} < t < 2$

answer is A.

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Detailed Solution

Let P be on $x + 2 y = 1$

$\Rightarrow$ $1 + \frac{t}{\sqrt{2}} + 2 (2 + \frac{t}{\sqrt{2}}) = 1 \Rightarrow t = - \frac{4 \sqrt{2}}{3}$

Let P be on $2 x + 3 y = 15$

$\Rightarrow$ $2 (1 + \frac{t}{\sqrt{2}}) + 4 (2 + \frac{t}{\sqrt{2}}) = 15 \Rightarrow t = \frac{5 \sqrt{2}}{6}$

$\frac{- 4 \sqrt{2}}{3} < t < \frac{5 \sqrt{2}}{6} .$

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