Question

If the distance between the lines represented by \((x-2y)^2+k(x-2y) = 0\) is \(3\) units and \(k > 0\) then the equations of the lines are

• A. \(x-2y = 0\) and \(x-2y+\sqrt{3} = 0\).
• B. \(x-2y = 0\) and \(x-2y+\sqrt{5} = 0\).
• C. \(x-2y = 0\) and \(x-2y+3\sqrt{3} = 0\).
• D. \(x-2y = 0\) and \(x-2y+3\sqrt{5} = 0\).

Hint:

Factor the equation into x-2y=0 and x-2y+k=0, then use the parallel line distance formula.

Answer 1

The Correct Option is D

Step 1: Factor the equation:
\((x-2y)^2+k(x-2y)=(x-2y)(x-2y+k)=0\). The two lines are \(x-2y=0\) and \(x-2y+k=0\).

Step 2: Find the distance between them:
These lines are parallel. The distance is \(\dfrac{|k-0|}{\sqrt{1^2+(-2)^2}}=\dfrac{k}{\sqrt5}\) for \(k>0\).

Step 3: Solve for k:
\(\dfrac{k}{\sqrt5}=3\), so \(k=3\sqrt5\). The lines are \(x-2y=0\) and \(x-2y+3\sqrt5=0\). Option D.

Step 4: Why the other options are wrong.
A, B, C have \(k=\sqrt3,\ \sqrt5,\ 3\sqrt3\), which give distances \(\sqrt3/\sqrt5\), 1 and \(3\sqrt3/\sqrt5\), none equal to 3.

Final Answer:
The lines are x - 2y = 0 and x - 2y + 3 sqrt 5 = 0. \[ \boxed{\text{(D) }x-2y=0\ \text{and}\ x-2y+3\sqrt5=0} \]