Question

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

• A. $\frac{-4\sqrt{2}}{3}<t<\frac{5\sqrt{2}}{6}$
• B. $0<t<\frac{5\sqrt{2}}{6}$
• C. $-4\sqrt{2}<t<0$
• D. None of these

Answer 1

The correct option is A $\frac{-4\sqrt{2}}{3}<t<\frac{5\sqrt{2}}{6}$

Let, $P$ be on $x+2y=1$, then
$1+\frac{t}{\sqrt{2}}+2(2+\frac{t}{\sqrt{2}})=1$
or $t=\frac{-4\sqrt{2}}{3}$
Let, $P$ be on $2x+4y=15.$ Then,
$2(1+\frac{t}{\sqrt{2}})+4(2+\frac{t}{\sqrt{2}})=15$
or $t=\frac{5\sqrt{2}}{6}$
Since point lies between the lines and $x=t$, then $t\in (\frac{-4\sqrt{2}}{3},\frac{5\sqrt{2}}{6})$