Question

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

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

Answer 1

Answer: (B), (D)

The correct options are
B $t>-\frac{4\sqrt{2}}{3}$
D $t<\frac{5\sqrt{2}}{6}$

Let $L_1:x+2y-1=0$

and $L_2:2x+4y-15=0$

$L_1=0,L_2=0$ are parallel as in the figure.

We require that $P(1+\frac{t}{\sqrt{2}},2+\frac{t}{\sqrt{2}})$ lies in between these lines.

For this we need $O(0,0)$ and $P$ to lie on opposite sides of $x+2y-1=0$ and on the same side of $2x+4y-15=0$

When we substitute $O(0,0)$ in $L_1$, we get negative value.

So,substituting $P$ in $L_1$ must give a positive value and substituting $P$ in $L_2$ must give a negative value

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

$\Rightarrow \frac{t}{\sqrt{2}}+4+\frac{2t}{\sqrt{2}}>0$ and $2+\frac{2t}{\sqrt{2}}+8+\frac{4t}{\sqrt{2}}-15<0$

$\Rightarrow 4+\frac{3t}{\sqrt{2}}>0$ and $\frac{6t}{\sqrt{2}}-5<0$

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

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