A variable point `(1+(lamda)/(sqrt(2)),2+(lamda)/(sqrt(2)))`lies in between two parallel lines `x+2y=1` and `2x+4y=15` then the range of `lamda` is given by

Options

A. `0lt lamdalt (5sqrt(2))/6`
B. `-(4sqrt(2))/5lt lamdalt (5sqrt(2))/6`
C. `-(4sqrt(2))/5ltlamdalt0`
D. None of these

Answer

To find the range of \(\lambda\) for the point \((1 + \frac{\lambda}{\sqrt{2}}, 2 + \frac{\lambda}{\sqrt{2}})\) to lie between the two parallel lines \(x + 2y = 1\) and \(2x + 4y = 15\), we can follow these steps:

### Step 1: Identify the equations of the lines
The two lines given are:
1. \(L_1: x + 2y = 1\)
2. \(L_2: 2x + 4y = 15\)

### Step 2: Convert the second line to the same form as the first line
We can simplify the second line:
\[ 2x + 4y = 15 \implies x + 2y = \frac{15}{2} \]
Thus, the two lines are:
1. \(L_1: x + 2y = 1\)
2. \(L_2: x + 2y = \frac{15}{2}\)

### Step 3: Find the distance between the two parallel lines
The formula for the distance \(d\) between two parallel lines of the form \(Ax + By = C_1\) and \(Ax + By = C_2\) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 1\), \(B = 2\), \(C_1 = 1\), and \(C_2 = \frac{15}{2}\):
\[ d = \frac{\left|\frac{15}{2} - 1\right|}{\sqrt{1^2 + 2^2}} = \frac{\left|\frac{15}{2} - \frac{2}{2}\right|}{\sqrt{5}} = \frac{\left|\frac{13}{2}\right|}{\sqrt{5}} = \frac{13}{2\sqrt{5}} \]

### Step 4: Calculate the perpendicular distance from the point to the first line
The point is given as \((1 + \frac{\lambda}{\sqrt{2}}, 2 + \frac{\lambda}{\sqrt{2}})\). The distance \(d_1\) from this point to the line \(L_1: x + 2y = 1\) is given by:
\[ d_1 = \frac{|(1 + \frac{\lambda}{\sqrt{2}}) + 2(2 + \frac{\lambda}{\sqrt{2}}) - 1|}{\sqrt{1^2 + 2^2}} = \frac{|1 + \frac{\lambda}{\sqrt{2}} + 4 + \frac{2\lambda}{\sqrt{2}} - 1|}{\sqrt{5}} \]
This simplifies to:
\[ d_1 = \frac{|4 + \frac{3\lambda}{\sqrt{2}}|}{\sqrt{5}} \]

### Step 5: Set up inequalities for \(\lambda\)
For the point to lie between the two lines, the distance \(d_1\) must be less than the distance \(d\):
\[ d_1 < d \implies \frac{|4 + \frac{3\lambda}{\sqrt{2}}|}{\sqrt{5}} < \frac{13}{2\sqrt{5}} \]
Multiplying both sides by \(\sqrt{5}\):
\[ |4 + \frac{3\lambda}{\sqrt{2}}| < \frac{13}{2} \]

### Step 6: Solve the absolute inequality
This gives us two inequalities:
1. \(4 + \frac{3\lambda}{\sqrt{2}} < \frac{13}{2}\)
2. \(4 + \frac{3\lambda}{\sqrt{2}} > -\frac{13}{2}\)

#### Solving the first inequality:
\[ 4 + \frac{3\lambda}{\sqrt{2}} < \frac{13}{2} \]
\[ \frac{3\lambda}{\sqrt{2}} < \frac{13}{2} - 4 = \frac{5}{2} \]
\[ 3\lambda < \frac{5\sqrt{2}}{2} \implies \lambda < \frac{5\sqrt{2}}{6} \]

#### Solving the second inequality:
\[ 4 + \frac{3\lambda}{\sqrt{2}} > -\frac{13}{2} \]
\[ \frac{3\lambda}{\sqrt{2}} > -\frac{13}{2} - 4 = -\frac{21}{2} \]
\[ 3\lambda > -\frac{21\sqrt{2}}{2} \implies \lambda > -\frac{7\sqrt{2}}{2} \]

### Step 7: Combine the results
Thus, the range of \(\lambda\) is:
\[ -\frac{7\sqrt{2}}{6} < \lambda < \frac{5\sqrt{2}}{6} \]

### Final Answer
The range of \(\lambda\) is:
\[ \lambda \in \left(-\frac{7\sqrt{2}}{6}, \frac{5\sqrt{2}}{6}\right) \]