Question

Locus of midpoint of rod having length \(2 \mathrm{c}\) begins to slide on two perpendicular lines is... (a) \(x^{2}-y^{2}=c^{2}\) (b) \(x^{2}+y^{2}=c^{2}\) (c) \(x^{2}+y^{2}=2 c^{2}\) (d) \(x^{2}-y^{2}=2 c^{2}\)

Short answer

The derived equation for the locus of the midpoint of the rod is \(x^2 + y^2 = 4c^2\). However, none of the given options matches this equation.

Step-by-step solution

Assign Coordinates to the Endpoints of the Rod

First, let's assume the two perpendicular lines as \(x\)-axis and \(y\)-axis. Let one endpoint of the rod be at \((x, 0)\) on the \(x\)-axis, and let the other endpoint be at \((0, y)\) on the \(y\)-axis.

Find the Midpoint of the Rod

To find the midpoint, we need to find the average of the \(x\)-coordinates and the \(y\)-coordinates of the endpoints. The midpoint, M, of the rod will be \[ M \left(\frac{x}{2},\frac{y}{2}\right) \]

Use the Distance Formula to find the Length of the Rod

Now, we will use the distance formula to find the length of the rod, using the coordinates of the two endpoints. \[ \mathrm{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Using the coordinates \((x, 0)\) and \((0, y)\): \[ \mathrm{Length} = \sqrt{(0 - x)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]

Write the Length of the Rod in Terms of c

Since the length of the rod is given as \(2c\), we can write \[ 2c = \sqrt{x^2 + y^2} \]

Square both sides of the equation and simplify

Squaring both sides of the equation, we have \[ (2c)^2 = (x^2 + y^2)^2 \] Simplifying it further, we get \[ 4c^2 = x^2 + y^2 \]

Compare the Derived Equation with the Given Options

Comparing the derived equation with the given options, we can conclude that the locus of the midpoint of the rod is: \[ x^2 + y^2 = 4c^2 \] However, none of the given options matches our derived equation. Thus, the correct answer is not in the options provided.

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using the coordinate system. It links algebra and geometry, allowing us to describe geometric figures using numerical equations. In this system, points are represented by coordinates \((x, y)\) in a plane. These coordinates define a specific location in a 2D space.

In coordinate geometry, lines, shapes, and curves can be expressed algebraically. For example:

• A line in this system can be represented by the equation \(y = mx + b\), where \(m\) is the slope, and \(b\) is the y-intercept.
• Circles are represented by \((x - h)^2 + (y - k)^2 = r^2\) where \((h, k)\) is the center, and \(r\) is the radius.

The beauty of coordinate geometry is in its versatility and ability to solve complex geometric problems using simple algebraic formulas. In the given problem, coordinate geometry helps in assigning locations to the endpoints of the rod, making it easier to compute the midpoint and derive the locus equation.

Midpoint Theorem

The midpoint theorem is a simple, yet powerful tool in geometry. It states that the midpoint of a line segment is the average, or mean, of the coordinates of its endpoints. This can be applied in any coordinate plane to find the center point of a segment. For two endpoints \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint \(M\) is given by:

\[ M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

• This equation shows how averaging the x-coordinates gives the x-point of the midpoint.
• Similarly, averaging the y-coordinates gives the y-point.

In the exercise, this theorem was used to calculate the midpoint of the rod that has endpoints on the x-axis and y-axis, simplifying the problem significantly. Understanding the midpoint theorem can help you solve geometric problems involving bisecting lines and finding centers, which is crucial in many coordinate geometry questions.

Distance Formula

The distance formula is derived from the Pythagorean theorem and allows us to calculate the distance between two points in a plane. Given two points, \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between them is calculated as:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

• This formula gives the straight-line distance, effectively creating a right triangle where the differences in x and y are the two legs, and the distance is the hypotenuse.
• This is vital in scenarios where accuracy is required, such as finding lengths of geometric figures.

In the problem at hand, the distance formula is applied to ensure the rod maintains its constant length of \(2c\) as it slides on the axes. This allows calculation of the correct position of the rod's midpoint.

Equation of Locus

The equation of a locus describes all possible locations of a point that satisfy certain conditions. In essence, it's all about finding a path that fulfills a set of constraints. For a point moving in a plane, its path can often be described by an algebraic equation—a locus.

In the exercise, we needed to determine the locus of the midpoint of a sliding rod. By using the information provided and the midpoint and distance formulas, we derived the equation for the locus as \(x^2 + y^2 = 4c^2\).

• This equation represents a circle centered at the origin with a radius of \(2c\).
• The locus gives us a clear picture of all positions the midpoint can occupy.

Understanding the concept of the equation of a locus is key for problems involving movement and constraints. It explains how and where things are allowed to move in coordinate geometry.