Question

For l ∈ R, the equation (2l - 3) x² + 2lxy - y² = 0 represents a pair of distinct lines 

• A. only when l = 0
• B. for all values of l ∈ (-3, 1)
• C. for all values of l ∈ R - (0, 1)
• D. for all values of l ∈ R - (-3, 1)

Answer 1

The Correct Option is D

To determine the values of $l$ for which the equation $(2l-3)x^2 + 2lxy - y^2 = 0$ represents a pair of distinct lines, we proceed:

1. Recognizing the Equation Type:
The given equation $(2l-3)x^2 + 2lxy - y^2 = 0$ is a homogeneous equation of degree 2, suggesting it may represent a pair of lines passing through the origin.

2. Condition for Distinct Lines:
For a homogeneous quadratic equation $ax^2 + 2hxy + by^2 = 0$ to represent a pair of distinct lines, the discriminant condition must be satisfied: $h^2 - ab > 0$. Identify the coefficients by comparing with the standard form:

$ a = 2l - 3, \quad h = l, \quad b = -1 $

3. Computing the Discriminant:
Calculate the discriminant:

$ h^2 - ab = l^2 - (2l - 3)(-1) = l^2 + (2l - 3) = l^2 + 2l - 3 $
For distinct lines, we need:

$ l^2 + 2l - 3 > 0 $

4. Solving the Inequality:
Solve the quadratic inequality $l^2 + 2l - 3 > 0$. First, find the roots of the equation $l^2 + 2l - 3 = 0$:

$ (l + 3)(l - 1) = 0 $
Roots are $l = -3$ and $l = 1$. Since the parabola $l^2 + 2l - 3$ opens upward, test the sign in the intervals $(-\infty, -3)$, $(-3, 1)$, and $(1, \infty)$:

- For $l = -4$: $(-4 + 3)(-4 - 1) = (-1)(-5) = 5 > 0$
- For $l = 0$: $(0 + 3)(0 - 1) = 3 \cdot (-1) = -3 < 0$
- For $l = 2$: $(2 + 3)(2 - 1) = 5 \cdot 1 = 5 > 0$
Thus, $(l + 3)(l - 1) > 0$ when $l < -3$ or $l > 1$.

5. Expressing the Solution:
The solution to the inequality is:

$ l \in (-\infty, -3) \cup (1, \infty) $
This can be written as:

$ l \in \mathbb{R} - [-3, 1] $

Final Answer:
The equation represents a pair of distinct lines for all $l \in \mathbb{R} - [-3, 1]$.