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!
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 isC
Solution and Explanation
Step 1: Understand the goal. A line parallel to the X-axis looks like $y=c$, where $c$ is a fixed number. So we only need the $y$-value of the point where the two given lines meet. Step 2: Write the two equations clearly. \[ ax+2by=-3b \quad\cdots(1) \] \[ bx-2ay=3a \quad\cdots(2) \] Step 3: Make the $x$ terms match so we can remove them. 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 4: Subtract to keep only $y$. Subtract (4) from (3). The $abx$ terms cancel: \[ 2b^2y+2a^2y=-3b^2-3a^2 \] \[ 2(a^2+b^2)\,y=-3(a^2+b^2) \] Step 5: Solve for $y$. Since $(a,b)\neq(0,0)$, the factor $a^2+b^2$ is not zero, so we can divide it out. \[ 2y=-3 \implies y=-\tfrac{3}{2} \] Step 6: Read off the position. The line is $y=-\tfrac{3}{2}$. The minus sign means it sits below the X-axis, and its distance from the axis is $\tfrac{3}{2}$ units. \[ \boxed{\text{below the X-axis at a distance of }\tfrac{3}{2}\text{ units}} \]