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

Let P be on $x + 2y = 1$, then, $1+\frac{t}{\sqrt{2}}+2\left(2+\frac{t}{\sqrt{2}}\right)=1\text{ or }t=\frac{-4\sqrt{2}}{3}$

Let P be on $2x + 4y = 15$, then $2\left(1+\frac{t}{\sqrt{2}}\right)+4\left(2+\frac{t}{\sqrt{2}}\right)=15\text{ or }t=\frac{5\sqrt{2}}{3}$

$\therefore t\in \left(\frac{-4\sqrt{2}}{3},\frac{5\sqrt{2}}{3}\right)$

Origin, P lies opposite side to the first line and same side to the second line