The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0

Q: Why can't I just solve for the intersection point by elimination? It feels more direct.

While you *could* solve for $x$ and $y$ using elimination, it leads to a swamp of variables $a$ and $b$. The 'Family of Lines' method is the 'sniper' approach—it hits the target without needing to know the exact coordinates of the intersection point.

Q: How do I know if the line is above or below the x-axis?

Look at the final equation $y = -\frac{3}{2}$. Since the y-coordinate is negative, the line sits below the x-axis. If it were $y = \frac{3}{2}$, it would be above.

Q: Is it always safe to divide by $(a^2 + b^2)$?

Great catch! In general, you must be careful. However, the problem explicitly states $(a, b) \neq (0, 0)$. This guarantees that $a^2 + b^2 > 0$, so we are never dividing by zero. Always check your constraints!

Question

The line parallel to the x-axis and passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$, where $(a, b) 
eq (0, 0)$ is

Options:

(A) below the x-axis at a distance of $\frac{3}{2}$ from it

(B) below the x-axis at a distance of $\frac{2}{3}$ from it

(C) above the x-axis at a distance of $\frac{3}{2}$ from it

(D) above the x-axis at a distance of $\frac{2}{3}$ from it

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at two lines, $L_1: ax + 2by + 3b = 0$ and $L_2: bx - 2ay - 3a = 0$. They cross at a single point.

Calculating the coordinates of that point directly is a trap. It is a path filled with messy algebra that will drain your time.

Instead, we use the 'Family of Lines'—a powerful, elegant concept. Any line passing through the intersection of $L_1$ and $L_2$ can be written as $L_1 + \lambda L_2 = 0$. This is our master key.

### The Algebraic Setup

We take our ingredients and place them into the pot:
$$(ax + 2by + 3b) + \lambda(bx - 2ay - 3a) = 0$$

We haven't changed the geometry; we have just created a flexible equation that represents every possible line passing through that intersection point. Now, we organize by grouping the terms by $x$ and $y$:
$$(a + b\lambda)x + (2b - 2a\lambda)y + (3b - 3a\lambda) = 0$$

This is the standard form of a line, $Ax + By + C = 0$.

### The Condition of Parallelism

The problem demands a line parallel to the $x$-axis. Geometrically, this means the line is horizontal.

Algebraically, a horizontal line has a slope of zero. In the equation $Ax + By + C = 0$, the slope is given by $-\frac{A}{B}$.

For this slope to be zero, $A$ must be zero. Thus, the coefficient of $x$ must vanish:
$$a + b\lambda = 0 \Rightarrow \lambda = -\frac{a}{b}$$

### The Final Dance

We substitute $\lambda = -\frac{a}{b}$ back into our grouped equation. The $x$ term is now zero.

We focus on the $y$ and constant terms. After substitution, the $y$ coefficient becomes:
$$2b - 2a\left(-\frac{a}{b}ight) = \frac{2b^2 + 2a^2}{b} = \frac{2(a^2 + b^2)}{b}$$

The constant term becomes:
$$3b - 3a\left(-\frac{a}{b}ight) = \frac{3b^2 + 3a^2}{b} = \frac{3(a^2 + b^2)}{b}$$

Since $(a, b) 
eq (0, 0)$, the term $(a^2 + b^2)$ is non-zero. We divide the entire equation by $\frac{a^2 + b^2}{b}$, leaving us with:
$$2y + 3 = 0$$

This simplifies to **$y = -\frac{3}{2}$**. This is a horizontal line located below the $x$-axis at a distance of $\frac{3}{2}$. We have arrived at the truth through elegance, not brute force.

Animated Solution for Mathematics - Straight Lines: The line parallel to the x-axis and passing through the intersection of the lines $a x + 2 b y + 3 b = 0$ and $b x - 2 a y - 3 a = 0$ , where $(a, b) \neq = (0, 0)$ is

Select Answer:

Visualized Solution

Visualizing the Intersection

The Family of Lines Concept

Raw Setup: Substitution

Grouping the Terms

Condition for Parallelism

Solving for $λ$

Substituting $λ$ Back

Simplifying the $y$ Coefficient

Simplifying the Constant Term

The Final Equation

Conclusion and Takeaway

The Sigma Insight: Family of Lines

Analyzing the Setup

The Algebraic Setup

The Condition of Parallelism

The Final Dance

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