Question

If P$\left( {1 + \dfrac{t}{{\sqrt 2 }},2 + \dfrac{t}{{\sqrt 2 }}} \right)$ 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) $\dfrac{{ - 4\sqrt 2 }}{3}$ <$t$ <$\dfrac{{5\sqrt 2 }}{6}$
B) 0 < $t$ <$\dfrac{{5\sqrt 2 }}{6}$
C) $- 4\sqrt 2$ <$t$ < 0
D) None of these

Answer 1

If a point (x1 , y1) lies between any two given line equations , then the product of the values when that particular point is placed in those two lines must be less than 0 because this should also mean the lines are on opposite side of the point. Substitute the given values in the above condition, we will get the answer.

Complete step-by-step answer:
Given
The lines $x + 2y = 1$ , $2x + 4y = 15$ are parallel lines.
Given a point P$\left( {1 + \dfrac{t}{{\sqrt 2 }},2 + \dfrac{t}{{\sqrt 2 }}} \right)$ on a line lies between those parallel lines.
If a point lies between the two parallel lines the point should obey the following condition.
$x + 2y - 1$>0 and $2x + 4y - 15$ <0
Therefore
$1 + \dfrac{t}{{\sqrt 2 }}$ + 2$\left( {2 + \dfrac{t}{{\sqrt 2 }}} \right)$ $-$ 1 > 0 and 2$\left( {1 + \dfrac{t}{{\sqrt 2 }}} \right)$ + 4$\left( {2 + \dfrac{t}{{\sqrt 2 }}} \right)$ $-$ 15 < 0
1 + 4 $-$ 1 + $\dfrac{{3t}}{{\sqrt 2 }}$ > 0 and 2 + 8 $-$ 15 + $\dfrac{{6t}}{{\sqrt 2 }}$ < 0
$\dfrac{{3t}}{{\sqrt 2 }}$ + 4 > 0 and $\dfrac{{6t}}{{\sqrt 2 }}$ $-$ 5 < 0
$\dfrac{{3t}}{{\sqrt 2 }}$ > $-$ 4 and $\dfrac{{6t}}{{\sqrt 2 }}$ < 5
multiply with $\sqrt 2$ on both sides of both the equations we get
$3t$ > $- 4\sqrt 2$ and $6t$ < $5\sqrt 2$
By dividing first equation with 3 on both sides and second equation with 6 on both sides of the equation we get
$t$ > $\dfrac{{ - 4\sqrt 2 }}{3}$ and $t$ < $\dfrac{{5\sqrt 2 }}{6}$
From this we can say that when the value of $t$ lies between $\dfrac{{ - 4\sqrt 2 }}{3}$ and $\dfrac{{5\sqrt 2 }}{6}$ the given point P$\left( {1 + \dfrac{t}{{\sqrt 2 }},2 + \dfrac{t}{{\sqrt 2 }}} \right)$ will be lying in between the parallel lines $x + 2y = 1$ and $2x + 4y = 15$ .
Therefore the range of $t$ is $\left( {\dfrac{{ - 4\sqrt 2 }}{3},\dfrac{{5\sqrt 2 }}{6}} \right)$
Therefore in the given options option A that is $\dfrac{{ - 4\sqrt 2 }}{3}$ <$t$ <$\dfrac{{5\sqrt 2 }}{6}$ is true.

So, the correct answer is “Option A”.

Note: Make sure to check all the possibilities and do not skip the remaining possibilities as you get one interval as an answer and make sure to check the intervals while taking the intersection of them. Do the calculations properly and go through formulas for a point to be on the same side of two lines or the opposite.