Question:
The distance between the lines represented by the equation $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ is
The distance between the lines represented by the equation $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ is
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For parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$, distance is $|c_1-c_2|/\sqrt{a^2+b^2}$.
Updated On: May 14, 2026
- $\frac{1}{\sqrt{5}}$ units
- $\frac{1}{5}$ units
- $\sqrt{5}$ units
- $5$ units
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Concept
The equation represents two parallel lines if the second-degree part is a perfect square. Distance $d = \frac{2\sqrt{g^2 - ac}}{\sqrt{a(a+b)}}$ or $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Step 2: Meaning
$4x^2 + 4xy + y^2 = (2x+y)^2$. The equation is $(2x+y)^2 - 3(2x+y) - 4 = 0$.
Step 3: Analysis
Let $t = 2x+y$. Then $t^2 - 3t - 4 = 0 \implies (t-4)(t+1) = 0$. The lines are $2x+y-4=0$ and $2x+y+1=0$. Distance $d = \frac{|-4 - 1|}{\sqrt{2^2 + 1^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$.
Step 4: Conclusion
The distance between the lines is $\sqrt{5}$ units. Final Answer: (C)
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