Two lines are given by `(x-2y)^2 + k (x-2y) = 0` . The value of `k`, so that the distance between them is `3`, is:

Options

A. (a) `sqrt(5)`
B. (b) `2sqrt(5)`
C. (c) `3sqrt(5)`
D. (d) `4sqrt(5)`

Answer

To solve the problem, we need to find the value of \( k \) such that the distance between the two lines represented by the equation \( (x - 2y)^2 + k(x - 2y) = 0 \) is equal to 3.

### Step-by-Step Solution:

1. **Rewrite the Equation:**
The given equation is \( (x - 2y)^2 + k(x - 2y) = 0 \). We can factor this equation:
\[ (x - 2y)((x - 2y) + k) = 0 \]
This gives us two lines:
\[ L_1: x - 2y = 0 \quad \text{and} \quad L_2: x - 2y + k = 0 \]

2. **Identify the Coefficients:**
The lines can be expressed in the standard form \( ax + by + c = 0 \):
- For \( L_1 \): \( 1x - 2y + 0 = 0 \) (where \( a = 1, b = -2, c = 0 \))
- For \( L_2 \): \( 1x - 2y + k = 0 \) (where \( a = 1, b = -2, c = k \))

3. **Distance Between Two Parallel Lines:**
The formula for the distance \( d \) between two parallel lines \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \) is given by:
\[ d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} \]
Here, \( c_1 = 0 \) and \( c_2 = k \). Thus, the distance \( d \) between the lines \( L_1 \) and \( L_2 \) is:
\[ d = \frac{|k - 0|}{\sqrt{1^2 + (-2)^2}} = \frac{|k|}{\sqrt{1 + 4}} = \frac{|k|}{\sqrt{5}} \]

4. **Set the Distance Equal to 3:**
We are given that the distance between the lines is 3:
\[ \frac{|k|}{\sqrt{5}} = 3 \]

5. **Solve for \( k \):**
To find \( |k| \), we can cross-multiply:
\[ |k| = 3\sqrt{5} \]
Therefore, \( k \) can be either \( 3\sqrt{5} \) or \( -3\sqrt{5} \).

6. **Final Answer:**
Since the options provided include \( 3\sqrt{5} \), we conclude that the value of \( k \) such that the distance between the lines is 3 is:
\[ k = 3\sqrt{5} \]