Question

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

• A. $-\frac{4\sqrt{2}}{5}<t<\frac{5\sqrt{2}}{6}$
• B. $-\frac{4\sqrt{2}}{3}<t<\frac{5\sqrt{2}}{6}$
• C. $t<\frac{-4\sqrt{2}}{3}$
• D. $t<\frac{5\sqrt{2}}{6}$

Answer 1

Answer: (B)

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

Given:$P(1+\frac{t}{\sqrt{2}},2+\frac{t}{\sqrt{2}})$ lies on $x+2y=1$

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

$\Rightarrow 1+\frac{t}{\sqrt{2}}+4+\frac{2t}{\sqrt{2}}=1$

$\Rightarrow \frac{3t}{\sqrt{2}}=-4$

$\Rightarrow t=\frac{-4\sqrt{2}}{3}$

$P(1+\frac{t}{\sqrt{2}},2+\frac{t}{\sqrt{2}})$ lies on $2x+4y=15$

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

$\Rightarrow 2+\frac{2t}{\sqrt{2}}+8+\frac{4t}{\sqrt{2}}=15$

$\Rightarrow \frac{6t}{\sqrt{2}}=5$

$\Rightarrow t=\frac{5\sqrt{2}}{6}$

Given:Range of value of $t$ for $t\in (\frac{-4\sqrt{2}}{3},\frac{5\sqrt{2}}{6})$ lies between the parallel lines.

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