Question

The distance between the lines represented by the equation $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ is

• A. $\frac{1}{\sqrt{5}}$ units
• B. $\frac{1}{5}$ units
• C. $\sqrt{5}$ units
• D. $5$ units

Hint:

For parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$, distance is $|c_1-c_2|/\sqrt{a^2+b^2}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given equation is a second-degree general equation representing a pair of straight lines.
Since the second-degree terms form a perfect square ($4x^2 + 4xy + y^2 = (2x+y)^2$), the lines are parallel.
We need to factorize the equation to find the individual equations of the lines and then use the formula for distance between parallel lines.
Step 2: Key Formula or Approach:
Factorize the equation into the form $(ax + by + c_1)(ax + by + c_2) = 0$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by:
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]
Step 3: Detailed Explanation:
Given equation: $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$
Observe the second-degree part: $4x^2 + 4xy + y^2 = (2x + y)^2$.
We can rewrite the given equation as:
\[ (2x + y)^2 - 3(2x + y) - 4 = 0 \]
This is a quadratic equation in terms of $(2x + y)$. Let $u = 2x + y$.
\[ u^2 - 3u - 4 = 0 \]
Factorizing the quadratic equation:
\[ u^2 - 4u + u - 4 = 0 \]
\[ u(u - 4) + 1(u - 4) = 0 \]
\[ (u - 4)(u + 1) = 0 \]
So, $u = 4$ or $u = -1$.
Substituting back $u = 2x + y$:
Line 1: $2x + y - 4 = 0$ (here $c_1 = -4$)
Line 2: $2x + y + 1 = 0$ (here $c_2 = 1$)
Both lines have $a = 2$ and $b = 1$.
The distance between them is:
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]
\[ d = \frac{|-4 - 1|}{\sqrt{2^2 + 1^2}} \]
\[ d = \frac{|-5|}{\sqrt{4 + 1}} = \frac{5}{\sqrt{5}} \]
Rationalizing the denominator:
\[ d = \frac{5}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5} \]
Step 4: Final Answer:
The distance is $\sqrt{5}$ units.