A rod of length / slides with its ends on two perpendicular lines. Find the locus of its mid-point.

Answer

To find the locus of the midpoint of a rod of length \( L \) sliding with its ends on two perpendicular lines, we can follow these steps:

### Step 1: Set Up the Coordinate System
Let the two perpendicular lines be the x-axis and the y-axis. We can denote the endpoints of the rod as \( A(a, 0) \) on the x-axis and \( B(0, b) \) on the y-axis.

**Hint:** Visualize the problem by sketching the axes and the rod positioned between them.

### Step 2: Define the Midpoint
The midpoint \( P(h, k) \) of the rod can be calculated using the midpoint formula:
\[ h = \frac{a + 0}{2} = \frac{a}{2} \]
\[ k = \frac{0 + b}{2} = \frac{b}{2} \]

**Hint:** Remember that the midpoint is the average of the coordinates of the endpoints.

### Step 3: Apply the Pythagorean Theorem
According to the problem, the length of the rod \( AB \) is given by:
\[ AB = L \]
Using the distance formula, we can express this as:
\[ AB^2 = OA^2 + OB^2 \]
where \( OA = a \) and \( OB = b \). Therefore:
\[ L^2 = a^2 + b^2 \]

**Hint:** This step involves recognizing that the rod forms a right triangle with the axes.

### Step 4: Substitute for \( a \) and \( b \)
From the midpoint coordinates, we can express \( a \) and \( b \) in terms of \( h \) and \( k \):
\[ a = 2h \quad \text{and} \quad b = 2k \]
Substituting these into the equation from Step 3 gives:
\[ L^2 = (2h)^2 + (2k)^2 \]
This simplifies to:
\[ L^2 = 4h^2 + 4k^2 \]

**Hint:** Be careful with the algebra when substituting and simplifying.

### Step 5: Rearrange the Equation
Dividing the entire equation by \( L^2 \) gives:
\[ 1 = \frac{4h^2}{L^2} + \frac{4k^2}{L^2} \]
This can be rearranged to:
\[ \frac{h^2}{\left(\frac{L}{2}\right)^2} + \frac{k^2}{\left(\frac{L}{2}\right)^2} = 1 \]

**Hint:** Recognize that this is the standard form of the equation of a circle.

### Step 6: Identify the Locus
The equation derived represents a circle centered at the origin with a radius of \( \frac{L}{2} \).

**Final Result:**
The locus of the midpoint \( P(h, k) \) of the rod is given by:
\[ h^2 + k^2 = \left(\frac{L}{2}\right)^2 \]