Problem 77 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The line parallel to the \(x\)-axis and passing through the intersection of the lines \(a x+2 b y+3 b=0\) and \(b x-2 a y\) \(-3 a=0\), where \((a, b) \neq(0,0)\) is (A) below the \(x\)-axis at a distance of \(\frac{3}{2}\) from it (B) below the \(x\)-axis at a distance of \(\frac{2}{3}\) from it (C) above the \(x\)-axis at a distance of \(\frac{3}{2}\) from it (D) above the \(x\)-axis at a distance of \(\frac{2}{3}\) from it

Step-by-Step Solution

Answer

The line is (A) below the x-axis at a distance of \(\frac{3}{2}\).

Step 1: Write the Equation of the Line Parallel to the x-axis

When a line is parallel to the x-axis, it has the form \( y = k \), where \( k \) is a constant. To find \( k \), we need to find the y-coordinate of the intersection point of the given lines.

Step 2: Identify the System of Equations

We have two line equations given: 1. \( ax + 2by + 3b = 0 \)2. \( bx - 2ay - 3a = 0 \).We will solve this system to find the intersection point \((x, y)\).

Step 3: Eliminate One Variable

To eliminate \( x \), we can multiply the first equation by \( b \) and the second by \( a \). This gives us:1. \( abx + 2b^2y + 3b^2 = 0 \)2. \( abx - 2a^2y - 3a^2 = 0 \).Subtract the second modified equation from the first to eliminate \( x \):\((2b^2 + 2a^2)y + (3b^2 + 3a^2) = 0\).

Step 4: Solve for y

From the equation obtained in Step 3, simplify it to find \( y \):\((2b^2 + 2a^2)y = -(3b^2 + 3a^2)\),\[ y = -\frac{3b^2 + 3a^2}{2b^2 + 2a^2} = -\frac{3}{2}. \]

Step 5: Determine the Line's Position Relative to the x-axis

The y-coordinate of the intersection point is \(-\frac{3}{2}\), meaning the line \( y = -\frac{3}{2} \) passes through the intersection point and is parallel to the x-axis. Since \(-\frac{3}{2}\) is negative, the line is below the x-axis.

Key Concepts

Lines and Angles

Understanding lines and angles is fundamental in coordinate geometry, and it involves knowing various types of lines and how they interact with each other. Here, a line parallel to the x-axis is explored. A line is considered parallel to the x-axis when it maintains a constant distance from it, meaning it never intersects with the x-axis.
One can describe this line with the equation form such as \(y = k\), where \(k\) is a constant representing any specific y-coordinate across the entire line length.
If a line is at \(y = 1\), it's one unit above the x-axis, while \(y = -1\) is one unit below, illustrating how negative values signify below the x-axis and positive denote above.
Angles come in when lines intersect; however, for parallel lines, there isn't an angle of intersection with axes because they never meet. This fundamental idea helps set the stage for more complex concepts like systems of equations and intersections.

System of Equations

The system of equations forms the backbone of finding the intersection of two lines in coordinate geometry. In mathematics, a system of equations is a set of equations with multiple variables.
The ultimate goal is to find values of these variables that satisfy all the given equations simultaneously, which is crucial for determining where lines intersect.

• For example, given equations are \(ax + 2by + 3b = 0\) and \(bx - 2ay - 3a = 0\).
• To solve, one can use substitution, elimination, or matrix methods.

The elimination method involves aligning like terms and subtracting equations to "eliminate" one variable. Here, by multiplying each equation by strategic constants \(b\) and \(a\), we form equivalent equations hopefully leading to variable elimination.
Once one variable is eliminated, solving becomes straightforward for the remaining variable.

Intersection of Lines

When two lines have equations, their intersections are determined by the shared solution to the equations of the line, indicating where they cross each other's paths.
This point holds significance in problems involving coordinate geometry as it often has practical applications, like finding a location on a plane.

• Using the earlier system of equations, subtracting one from the other resulted in a simplified form: \((2b^2 + 2a^2)y + (3b^2 + 3a^2) = 0\).
• Solving, we find \(y = -\frac{3}{2}\), clarifying where the y-coordinate of the intersecting point occurs.

This y-value then informs the equation of the new line, \(y = -\frac{3}{2}\), parallel to the x-axis and passing through this specific intersection, spatially situating itself below the x-axis due to the negative value.
Thus, whenever two lines' equations can be solved, not only do we find the intersection but also gain insights on the relative positioning of points and lines within a plane.