Question

The perpendicular distance between the lines given by \( (x - 2y + 1)^2 + \text{k(x - 2y + 1) = 0 \) is \( \sqrt{5} \), then k =}

• A. 5
• B. 2
• C. 4
• D. 6

Hint:

Equations of the form \( f(x,y)^2 + k \cdot f(x,y) = 0 \) always represent two parallel lines.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given equation represents a pair of straight lines. Because it's a quadratic equation in terms of a linear expression \( (x - 2y + 1) \), it factors into two parallel lines.
We need to find the equations of these two parallel lines and use the distance formula between them to find \( k \).
Step 2: Key Formula or Approach:
Let \( u = x - 2y + 1 \). Solve the quadratic \( u^2 + ku = 0 \) to get two parallel lines \( L_1 = 0 \) and \( L_2 = 0 \).
The distance between two parallel lines \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \) is \( d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \).
Step 3: Detailed Explanation:
Let \( u = x - 2y + 1 \). The given equation becomes:
\[ u^2 + ku = 0 \] Factor out \( u \):
\[ u(u + k) = 0 \] This implies \( u = 0 \) or \( u = -k \).
Substitute \( u \) back:
Line 1: \( x - 2y + 1 = 0 \)
Line 2: \( x - 2y + 1 + k = 0 \)
These are equations of two parallel lines with \( a = 1 \), \( b = -2 \).
The constant terms are \( c_1 = 1 \) and \( c_2 = 1 + k \).
The perpendicular distance \( d \) between them is given as \( \sqrt{5} \).
Using the distance formula:
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \] \[ \sqrt{5} = \frac{|1 - (1 + k)|}{\sqrt{(1)^2 + (-2)^2}} \] \[ \sqrt{5} = \frac{|1 - 1 - k|}{\sqrt{1 + 4}} \] \[ \sqrt{5} = \frac{|-k|}{\sqrt{5}} \] Multiply both sides by \( \sqrt{5} \):
\[ \sqrt{5} \cdot \sqrt{5} = |-k| \] \[ 5 = |k| \] This means \( k = 5 \) or \( k = -5 \).
Looking at the given options, \( 5 \) is present.
Step 4: Final Answer:
The value of k is 5.