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: Find the point of intersection 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 can use methods like elimination or substitution. Let's 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: \[ (a^2 + b^2)x = 0 \] Since \( (a, b) \neq (0, 0) \), we know that \( a^2 + b^2>0 \). Therefore, \( x = 0 \). Now, substitute \( x = 0 \) into equation (1) to find the value of \( y \): \[ a(0) + 2by + 3b = 0 \] \[ 2by = -3b \] If \( b \neq 0 \), we can divide by \( 2b \) to get \( y = -\frac{3b}{2b} = -\frac{3}{2} \). If \( b = 0 \), then from \( a^2 + b^2>0 \), we must have \( a \neq 0 \). Substituting \( b = 0 \) and \( x = 0 \) into equation (2): \[ b(0) - 2ay - 3a = 0 \] \[ -2ay = 3a \] Since \( a \neq 0 \), we can divide by \( -2a \) to get \( y = -\frac{3a}{2a} = -\frac{3}{2} \).
In both cases, the y-coordinate of the intersection point is \( -\frac{3}{2} \). The point of intersection is \( \left(0, -\frac{3}{2}\right) \).
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 equation of the form \( y = c \), where \( c \) is a constant. Since the line passes through the point \( \left(0, -\frac{3}{2}\right) \), the equation of the line is \( y = -\frac{3}{2} \).
Step 3: Determine the position of the line relative to the x-axis.
The equation of the line is \( y = -\frac{3}{2} \). Since the y-coordinate is negative, the line is below the x-axis. The distance of this line from the x-axis (which is \( y = 0 \)) is \( |-\frac{3}{2} - 0| = \left|-\frac{3}{2}\right| = \frac{3}{2} \). Therefore, the line parallel to the x-axis passing through the intersection of the given lines is below the x-axis at a distance of \( \frac{3}{2} \) from it.