Problem 19 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A line of fixed length 2 units moves so that its ends are on the positive \(x\)-axis and that part of the line \(x+y=\) 0 which lies in the second quadrant. The locus of the mid-point of the line has the equation (A) \((x+2 y)^{2}+y^{2}=1\) (B) \((x-2 y)^{2}+y^{2}=1\) (C) \((x+2 y)^{2}-y^{2}=1\) (D) none of these

Step-by-Step Solution

Answer

Option (D) is correct.

Step 1: Define the ends of the moving line

Let's consider a line segment of fixed length 2 units. One end is on the positive x-axis, represented as the point \((x_1, 0)\), and the other end is on the line \(x + y = 0\) in the second quadrant, represented as \((-y_2, y_2)\). Thus, these can be seen as coordinates \((0, y_2)\) transformed to \((-y_2, y_2)\).

Step 2: Use the distance formula

The length of the line segment is constant at 2 units, so we apply the distance formula between \((x_1, 0)\) and \((-y_2, y_2)\): \[sqrt{(x_1 + y_2)^2 + (0 - y_2)^2} = 2. \]Simplifying the equation gives us:\[(x_1 + y_2)^2 + y_2^2 = 4.\]

Step 3: Find coordinates of the midpoint

The midpoint \((x, y)\) of the line segment is calculated as:\[x = \frac{x_1 - y_2}{2}, \quad y = \frac{y_2}{2}.\]

Step 4: Express in terms of midpoint coordinates

Substitute \(x_1 = 2x + y\) and \(y_2 = 2y\) in the previous segment's length equation:\[((2x + y) + 2y)^2 + (2y)^2 = 4.\] Simplify to:\[(2x + 3y)^2 + 4y^2 = 4.\] This reduces to:\[ (x + \frac{3}{2}y)^2 + y^2 = 1. \]

Step 5: Compare with given options

Simplify and rewrite the above equation:\[(2(x + \frac{3}{2}y))^2 + (2y)^2 = 4,\]giving:\[(x + 3y)^2 + y^2 = 1.\]This is the standard form, and none of the provided options (A)-(C) match this equation.

Key Concepts

Distance Formula

The distance formula is a fundamental concept in coordinate geometry used to calculate the length between two points in a plane. It is derived from the Pythagorean theorem and is essential for determining distances in a coordinate system. The formula is:

• Distance = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points you are evaluating.
In this exercise, the distance formula is key to ensuring that the line segment remains at a fixed length of 2 units while moving between specified endpoints on the positive \(x\)-axis and a line in the second quadrant.
This setup allows understanding of how relationships between coordinates of endpoints maintain constant distances, which is critical when formulating equations of loci.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics where algebraic equations describe geometric objects. It provides a way to analyze geometric shapes and relations using a coordinate system.
The basic concepts involve:

• Points, represented as coordinates \((x, y)\) on a Cartesian plane.
• Lines and their equations, such as the familiar linear equation \(y = mx + c\) that describes a line's slope \(m\) and y-intercept \(c\).
• The intersection and relationships between different geometric entities.

In this example, coordinate geometry aids in locating the points where the endpoints of the line segment land, namely on the \(x\)-axis and the line \(x+y=0\).
Furthermore, it's instrumental in defining and computing the coordinates of the midpoint, which reveal the line's balance point as it moves across the coordinate plane.

Locus Equation

A locus equation describes a set of points that satisfy certain geometric conditions, essentially forming a geometric shape. In mathematical terms, the locus is the path traced by a point that moves according to a rule or set of conditions.

• Each point on the locus satisfies specific equations or inequalities.
• Locating this area or path involves setting the rule for the point’s journey.

For the described problem, the locus is the path traced by the midpoint of the moving line segment.
To find the locus of the midpoint, we substitute the determined midpoint coordinates into the length equation derived using the distance formula:

• \((x = \frac{x_1 - y_2}{2}, \, y = \frac{y_2}{2})\)

This substitution and simplification yield the locus equation \((x + \frac{3}{2}y)^2 + y^2 = 1\), which can further be analyzed to understand the nature of the path followed by the midpoint.
By understanding how the locus equation is derived, students can solve for the paths traced by specific conditions in coordinate geometry.