Question

The line parallel to the \(\mathrm{x}\) -axis and passing through the intersection of 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) above the \(\mathrm{x}\) -axis at a distance of \((2 / 3)\) from it (b) above the \(\mathrm{x}\) -axis at a distance \((3 / 2)\) from it. (c) below that \(\mathrm{x}\) -axis at a distance \((2 / 3)\) from it. (d) below the \(\mathrm{x}\) -axis at a distance \((3 / 2)\) from it.

Short answer

The line parallel to the x-axis and passing through the intersection of lines \(a x+2 b y+3 b=0\) and \(b x-2 a y-3 a=0\) is (b) above the x-axis at a distance \(\frac{3}{2}\) from it.

Step-by-step solution

Find the intersection point of the given lines

To find the intersection point of the two lines, we need to solve the system of linear equations: \(a x + 2 b y + 3 b = 0 \newline b x - 2 a y - 3 a = 0\) To solve the system, let's express x in terms of y from the first equation and substitute it in the second equation. \( x = \frac{-2by - 3b}{a}\) Now substitute this into the second equation: \(b\left(\frac{-2by - 3b}{a}\right)- 2ay - 3a = 0\) Now, simplify the equation and solve for y.

Simplify the equation and solve for y

Multiplying through by a, we get: \(-2b^2y - 3b^2 - 2a^2y - 3a^2 = 0\) Now, combine the like terms: \((-2b^2 - 2a^2)y = 3a^2 + 3b^2\) Now, divide both sides by \(-2b^2 - 2a^2\), we get: \(y = \frac{3a^2 + 3b^2}{2a^2 + 2b^2}\) Now, substitute the value of y back into the expression for x:

Calculate x from y

Substitute the value of y into the expression for x: \(x = \frac{-2b\left(\frac{3a^2 + 3b^2}{2a^2 + 2b^2}\right) - 3b}{a}\) Simplify the expression: \(x = \frac{-6ab^2 - 6b^3 - 6ab^2}{2a^3 + 2a^2b}\) \(x = \frac{-6b(2a^2 + b^2)}{2a(a^2 + ab)}\) \(x = \frac{-2b}{a}\) Now, our intersection point is: Intersection Point: \((x, y) = \left(-\frac{2b}{a}, \frac{3(a^2 + b^2)}{2(a^2 + b^2)}\right)\)

Determine the equation of the line parallel to the x-axis

A line parallel to the x-axis has the equation of the form: \(y = n\) Where n is a constant. As this line passes through the intersection point, the constant n will be equal to the y-coordinate of the intersection point. \(y = \frac{3(a^2 + b^2)}{2(a^2 + b^2)}\)

Find the distance of this line from the x-axis

The distance from the line to the x-axis is the absolute value of the constant n: Distance = \(|\frac{3(a^2 + b^2)}{2(a^2 + b^2)}|\)

Identify which of the given options match our result

The distance calculated above is equal to \(\frac{3}{2}\). Since a line parallel to the x-axis will never intersect the x-axis, the intersection point will be either above or below the x-axis. In this case, since the value of y is positive, the line is above the x-axis. Therefore, the answer is: (b) above the x-axis at a distance \(\frac{3}{2}\) from it.

Key concepts

Parallel Lines

Imagine two train tracks running alongside each other, stretching off into the distance without ever meeting. In mathematics, we refer to these as parallel lines—straight lines in a plane that, no matter how far they extend, never intersect. This is because they have the same slope or gradient.

When working with a coordinate plane, if two lines are parallel, they can be represented by equations that, though distinct, share the same coefficient ratios for their x and y terms (except vertical lines, which are parallel to the y-axis). In an algebraic sense, if you have two linear equations, say, Ax + By = C and Ax + By = D where C and D are different constants, the lines will be parallel because their slopes, given by -A/B, are identical.

In the context of the given exercise, the line parallel to the x-axis is a horizontal line, which means it will have a slope of zero. Such a line can be represented by the equation y = k, where k is the y-coordinate through which the line passes, and it provides a visual understanding of the concept of parallelism within a coordinate geometry framework.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is like a meeting point for algebra and geometry where we use numbers to represent geometric shapes on a grid called a coordinate plane. This plane is divided into four quadrants by the x-axis (horizontal) and the y-axis (vertical).

The location of any point on this plane can be described with a pair of numbers known as coordinates, written as (x,y). Here, x tells us how far along the point is and y tells us how far up or down. By plotting these coordinates, we can draw shapes, lines, and curves and then use algebraic equations to analyze their properties.

For example, in the exercise we're considering, we turned the problem of finding where two lines cross into an algebra problem by setting up equations based on the given line formulas. Then, by manipulating these equations, we found a specific point (x,y), which represents the precise spot where the two lines meet on the graph.

Intersection Point of Lines

In the world of lines on a graph, an intersection point is like a shared secret—where two lines cross paths and say 'Hello!' To find this point, algebra swoops in to save the day. You set up equations for each line and solve them as a system, which tells you the x and y values where they meet.

If you think about two roads crossing, the intersection point is like the street corner—it has a specific address, which, in math, is expressed as coordinates. In situations where lines are parallel, there's a bit of a twist—there is no street corner, no 'Hello!' because they are equidistant at all times and never meet.

In the exercise we scrutinised, we found the intersection of two lines by solving their equations to discover that special point where they cross. Once we had this point, its y-coordinate told us the 'height' of a new line parallel to the x-axis going through the intersection, helping us determine whether it's above or below the x-axis—an important piece of information to answer the question correctly.