Question

The line parallel to the x-axis passing through the intersection of the lines \[ ax + 2by + 3b = 0 \quad \text{and} \quad bx - 2ay - 3a = 0 \] 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:

For horizontal lines: \begin{itemize} \item Find y-coordinate of intersection. \item Sign determines above/below x-axis. \end{itemize}

Answer 1

The Correct Option is C

Concept: Line parallel to x-axis ⇒ equation \( y = \text{constant} \). So we find y-coordinate of intersection point. Step 1: {\color{red}Solve equations.} \[ ax + 2by = -3b \quad (1) \] \[ bx - 2ay = 3a \quad (2) \] Step 2: {\color{red}Eliminate \( x \).} Multiply (1) by \( b \), (2) by \( a \): \[ abx + 2b^2y = -3b^2 \] \[ abx - 2a^2y = 3a^2 \] Subtract: \[ 2b^2y + 2a^2y = -3b^2 - 3a^2 \] \[ 2(a^2 + b^2)y = -3(a^2 + b^2) \] Step 3: {\color{red}Solve for \( y \).} Since \( a^2 + b^2 \neq 0 \): \[ y = -\frac{3}{2} \] Step 4: {\color{red}Interpret result.} Line is: \[ y = -\frac{3}{2} \] So it lies below x-axis at distance \( \frac{3}{2} \).