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Chapter 14: Problem 1325

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

Expert verified
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

01

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.
02

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 \[ M \left(\frac{x}{2},\frac{y}{2}\right) \]
03

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 \((0, y)\): \[ \mathrm{Length} = \sqrt{(0 - x)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]
04

Write the Length of the Rod in Terms of c

Since the length of the rod is given as \(2c\), we can write \[ 2c = \sqrt{x^2 + y^2} \]
05

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 \[ 4c^2 = x^2 + y^2 \]
06

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.

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