Key Concepts and Formulas

• Family of Lines: The equation of a line passing through the intersection of two lines $L_1 = 0$ and $L_2 = 0$ can be written as $L_1 + \lambda L_2 = 0$, where $\lambda$ is a real number.
• Line Parallel to the x-axis: A line parallel to the x-axis has the form $y = k$, where $k$ is a constant. In the general form $Ax + By + C = 0$, this implies $A = 0$.

Step-by-Step Solution

Step 1: Form the equation of the family of lines.

We are given the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$. We want to find a line that passes through the intersection of these two lines. Using the concept of the family of lines, we can write the equation of any line passing through the intersection as: $(ax + 2by + 3b) + \lambda(bx - 2ay - 3a) = 0$ This equation represents all possible lines passing through the intersection of the two given lines, parameterized by $\lambda$.

Step 2: Rearrange the equation into the form $Ax + By + C = 0$.

To easily apply the condition for a line parallel to the x-axis, we need to rewrite the equation by grouping the $x$, $y$, and constant terms: $ax + 2by + 3b + \lambda bx - 2\lambda ay - 3\lambda a = 0$ $(a + b\lambda)x + (2b - 2a\lambda)y + (3b - 3a\lambda) = 0$ Now we have the equation in the form $Ax + By + C = 0$, where $A = a + b\lambda$, $B = 2b - 2a\lambda$, and $C = 3b - 3a\lambda$.

Step 3: Apply the condition for the line to be parallel to the x-axis.

Since the line is parallel to the x-axis, the coefficient of $x$ must be zero. Therefore, we set $A = 0$: $a + b\lambda = 0$ Solving for $\lambda$, we get: $b\lambda = -a$ $\lambda = -\frac{a}{b}$ This gives us the specific value of $\lambda$ that corresponds to the line parallel to the x-axis. Note that since $(a,b) \ne (0,0)$, if $b=0$, then $a \ne 0$. However, if $b=0$, then $a+b\lambda = a$, which cannot be zero since $a \ne 0$. Thus, we assume $b \ne 0$ to proceed with $\lambda = -a/b$.

Step 4: Substitute the value of $\lambda$ back into the equation of the family of lines.

Substitute $\lambda = -\frac{a}{b}$ into the equation from Step 1: $(ax + 2by + 3b) - \frac{a}{b}(bx - 2ay - 3a) = 0$

Step 5: Simplify the equation to find the value of $y$.

Expand and simplify the equation: $ax + 2by + 3b - \frac{a}{b}bx + \frac{a}{b}2ay + \frac{a}{b}3a = 0$ $ax + 2by + 3b - ax + \frac{2a^2}{b}y + \frac{3a^2}{b} = 0$ $2by + \frac{2a^2}{b}y + 3b + \frac{3a^2}{b} = 0$ $y\left(2b + \frac{2a^2}{b}\right) = -3b - \frac{3a^2}{b}$ $y\left(\frac{2b^2 + 2a^2}{b}\right) = -\frac{3b^2 + 3a^2}{b}$ $y\left(\frac{2(a^2 + b^2)}{b}\right) = -\frac{3(a^2 + b^2)}{b}$ $y = -\frac{3(a^2 + b^2)}{b} \cdot \frac{b}{2(a^2 + b^2)}$ Since $(a,b) \ne (0,0)$, $a^2 + b^2 \ne 0$, so we can safely divide by $a^2 + b^2$. Also, we assumed $b \ne 0$, so we can divide by $b$. Thus: $y = -\frac{3}{2}$

Step 6: Interpret the result.

The equation of the line is $y = -\frac{3}{2}$. Since the y-value is negative, the line is below the x-axis. The distance from the x-axis is the absolute value of y, which is $|-\frac{3}{2}| = \frac{3}{2}$.

Therefore, the line is below the x-axis at a distance of $\frac{3}{2}$ from it.

Common Mistakes & Tips

• Careless Algebra: Double-check your algebra, especially when substituting and simplifying. A small sign error can lead to a wrong answer.
• Understanding the Family of Lines: Make sure you understand why $L_1 + \lambda L_2 = 0$ represents all lines passing through the intersection of $L_1 = 0$ and $L_2 = 0$.
• Special Cases: Be mindful of cases where $b=0$. In such cases, the condition $a + b\lambda = 0$ can't be used to find $\lambda$. We assumed $b\ne 0$ here, which is generally implied by the problem setup.

Summary

We found the equation of the line parallel to the x-axis and passing through the intersection of the two given lines. We used the family of lines concept and the condition for a line to be parallel to the x-axis to determine the equation $y = -\frac{3}{2}$. This indicates that the line is below the x-axis at a distance of $\frac{3}{2}$ from it.

The final answer is \boxed{\text{below the } x \text{ - axis at a distance of } {3 \over 2} \text{ from it}}, which corresponds to option (A).