Question:
The straight line which is parallel to X-axis and passing through the intersection of the lines \( ax+2by+3b=0 \) and \( bx-2ay-3a=0 \), \( (a,b)\neq(0,0) \) is
The straight line which is parallel to X-axis and passing through the intersection of the lines \( ax+2by+3b=0 \) and \( bx-2ay-3a=0 \), \( (a,b)\neq(0,0) \) is
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When a question specifies that a relation holds for any non-zero parameter pair \( (a,b) \), pick easy numbers like \( a = 1, b = 0 \). The lines become \( x = 0 \) and \( -2y - 3 = 0 \implies y = -1.5 \). The answer is found instantly!
Updated On: Jun 8, 2026
- above the X-axis at a distance of \( \frac{3}{2} \) units from it
- above the X-axis at a distance of \( \frac{2}{3} \) units from it
- below the X-axis at a distance of \( \frac{3}{2} \) units from it
- below the X-axis at a distance of \( \frac{2}{3} \) units from it
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The Correct Option is C
Solution and Explanation
Concept:
Any line parallel to the X-axis is of the form \( y = c \), where \( c \) is the y-coordinate of any point it passes through. Therefore, we only need to extract the \( y \)-coordinate of the intersection point of the two given lines.
Step 1: Eliminating variable \( x \) from the two equations.
1) \( ax + 2by = -3b \)
2) \( bx - 2ay = 3a \)
To eliminate \( x \), multiply equation (1) by \( b \) and equation (2) by \( a \): \[ abx + 2b^2y = -3b^2 \quad \cdots (3) \] \[ abx - 2a^2y = 3a^2 \quad \cdots (4) \]
Step 2: Subtracting the equations to isolate \( y \).
Subtracting equation (4) from equation (3): \[ (2b^2y) - (-2a^2y) = -3b^2 - 3a^2 \] \[ 2(a^2 + b^2)y = -3(a^2 + b^2) \] Since \( (a,b) \neq (0,0) \), \( a^2 + b^2 \neq 0 \), so we can divide both sides by \( (a^2 + b^2) \): \[ 2y = -3 \implies y = -\frac{3}{2} \]
Step 3: Interpreting the geometric location.
Since \( y = -\frac{3}{2} \), the line lies strictly below the X-axis (due to the negative sign) at a perpendicular distance of \( \frac{3}{2} \) units.
Step 1: Eliminating variable \( x \) from the two equations.
1) \( ax + 2by = -3b \)
2) \( bx - 2ay = 3a \)
To eliminate \( x \), multiply equation (1) by \( b \) and equation (2) by \( a \): \[ abx + 2b^2y = -3b^2 \quad \cdots (3) \] \[ abx - 2a^2y = 3a^2 \quad \cdots (4) \]
Step 2: Subtracting the equations to isolate \( y \).
Subtracting equation (4) from equation (3): \[ (2b^2y) - (-2a^2y) = -3b^2 - 3a^2 \] \[ 2(a^2 + b^2)y = -3(a^2 + b^2) \] Since \( (a,b) \neq (0,0) \), \( a^2 + b^2 \neq 0 \), so we can divide both sides by \( (a^2 + b^2) \): \[ 2y = -3 \implies y = -\frac{3}{2} \]
Step 3: Interpreting the geometric location.
Since \( y = -\frac{3}{2} \), the line lies strictly below the X-axis (due to the negative sign) at a perpendicular distance of \( \frac{3}{2} \) units.
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