Question

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

• A. above x-axis at a distance \( \frac{3}{2} \) from it.
• B. above x-axis at a distance \( \frac{2}{3} \) from it.
• C. below x-axis at a distance \( \frac{3}{2} \) from it.
• D. below x-axis at a distance \( \frac{2}{3} \) from it.

Hint:

When solving for the intersection of two lines, be careful about potential cases where coefficients might be zero. However, in this problem, the condition \( (a, b) \neq (0, 0) \) ensures that we don't have trivial cases where both \( a \) and \( b \) are zero simultaneously.

Answer 1

The Correct Option is C

Step 1: Determine the intersection point of the two lines.
We have the system of linear equations:
\begin{align} ax + 2by + 3b &= 0 \quad \cdots (1)
bx - 2ay - 3a &= 0 \quad \cdots (2) \end{align} To find the intersection point \( (x, y) \), we'll use elimination. Multiply equation (1) by \( a \) and equation (2) by \( b \):
\begin{align} a^2x + 2aby + 3ab &= 0 \quad \cdots (3)
b^2x - 2aby - 3ab &= 0 \quad \cdots (4) \end{align} Adding equations (3) and (4) eliminates the \( y \) term:\n\[\n(a^2 + b^2)x = 0\n\]\nSince \( (a, b) \neq (0, 0) \), then \( a^2 + b^2>0 \). Therefore, \( x = 0 \).\n\nSubstitute \( x = 0 \) into equation (1) to find \( y \):\n\[\na(0) + 2by + 3b = 0\n\]\n\[\n2by = -3b\n\]\nIf \( b \neq 0 \), divide by \( 2b \) to get \( y = -\frac{3b}{2b} = -\frac{3}{2} \).\n\nIf \( b = 0 \), then from \( a^2 + b^2>0 \), we have \( a \neq 0 \). Substituting \( b = 0 \) and \( x = 0 \) into equation (2):\n\[\nb(0) - 2ay - 3a = 0\n\]\n\[\n-2ay = 3a\n\]\nSince \( a \neq 0 \), divide by \( -2a \) to get \( y = -\frac{3a}{2a} = -\frac{3}{2} \).
In both cases, the y-coordinate is \( -\frac{3}{2} \). The intersection point is \( \left(0, -\frac{3}{2}\right) \).\n\n
Step 2: Find the equation of the line parallel to the x-axis passing through the intersection point.
A line parallel to the x-axis has the form \( y = c \). Since it passes through \( \left(0, -\frac{3}{2}\right) \), the equation is \( y = -\frac{3}{2} \).\n\n
Step 3: Determine the line's position relative to the x-axis.
The equation is \( y = -\frac{3}{2} \). The y-coordinate is negative, so the line is below the x-axis. The distance from the x-axis (where \( y = 0 \)) is \( |-\frac{3}{2} - 0| = \left|-\frac{3}{2}\right| = \frac{3}{2} \).\n\nTherefore, the line is below the x-axis at a distance of \( \frac{3}{2} \).