Question

If $P(1+\frac{t}{\sqrt{2}},2+\frac{t}{\sqrt{2}})$ be any points on a line then the range of 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. $-\frac{4\sqrt{2}}{5}<t<0$
• D. none of these

Answer 1

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

If the point $P(1+\frac{t}{\sqrt{2}},2+\frac{t}{\sqrt{2}})$ lies between line than

$1+\frac{t}{\sqrt{2}}+2(2+\frac{t}{\sqrt{2}})-1>0$

$\Rightarrow t>\frac{-4\sqrt{2}}{3}$
and
$2(1+\frac{t}{\sqrt{2}})+4(2+\frac{t}{\sqrt{2}})-15<0$

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

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