Question

Two lines are given by \(x^2-4xy+4y^2+kx-2ky = 0\), then the value of k so that the distance between them is 3 is

• A. \(\sqrt{5}\)
• B. \(3\sqrt{3}\)
• C. \(\sqrt{3}\)
• D. \(3\sqrt{5}\)

Hint:

Factorise the equation as a perfect square plus a linear term.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given second degree equation represents two lines. If the quadratic part is a perfect square, the two lines are parallel.

Step 2: Key Formula or Approach:
\(x^2 - 4xy + 4y^2 = (x - 2y)^2\) and \(kx - 2ky = k(x - 2y)\).

Step 3: Detailed Explanation:
The equation becomes \((x - 2y)^2 + k(x - 2y) = 0\), i.e. \((x - 2y)(x - 2y + k) = 0\).
The lines are \(x - 2y = 0\) and \(x - 2y + k = 0\), which are parallel.
\[ d = \frac{|k|}{\sqrt{1^2 + 2^2}} = \frac{|k|}{\sqrt{5}} = 3 \]
So \(|k| = 3\sqrt{5}\).
Option A would give a distance of \(1\), option B would give \(\frac{3\sqrt{3}}{\sqrt5}\) and option C would give \(\sqrt{\frac35}\), none of which equals \(3\).

Final Answer:
\(k = 3\sqrt{5}\) (taking the positive value), option (D). \[ \boxed{3\sqrt{5}} \]