A rod AB of length 30 cm moves such that its ends always touching the co-ordinate axes. Determine the locus of a point C of the rod if AC : BC = 2 : 1

Answer

To solve the problem, we need to find the locus of point C on the rod AB, given that the rod has a fixed length of 30 cm and that it touches the coordinate axes at points A and B. The ratio of AC to BC is given as 2:1.

### Step-by-Step Solution:

1. **Define the Points**:
Let point A be at (0, y_A) and point B be at (x_B, 0) on the coordinate axes. The length of the rod AB is given as 30 cm, so we have:
\[ OA + OB = 30 \quad \text{or} \quad y_A + x_B = 30 \]

2. **Set Up the Ratios**:
Since AC : BC = 2 : 1, we can express the lengths AC and BC in terms of a variable k:
\[ AC = 2k \quad \text{and} \quad BC = k \]
Therefore, the total length of the rod can also be expressed as:
\[ AC + BC = 2k + k = 3k \]
Setting this equal to the length of the rod:
\[ 3k = 30 \quad \Rightarrow \quad k = 10 \]
Hence, we have:
\[ AC = 20 \quad \text{and} \quad BC = 10 \]

3. **Find Coordinates of Point C**:
The coordinates of point C can be found using the section formula. Since C divides AB in the ratio 2:1, we can find the coordinates of C (x_C, y_C) using:
\[ x_C = \frac{2x_B + 1 \cdot 0}{2 + 1} = \frac{2x_B}{3} \]
\[ y_C = \frac{2 \cdot 0 + 1y_A}{2 + 1} = \frac{y_A}{3} \]

4. **Express y_A in terms of x_B**:
From the equation \(y_A + x_B = 30\), we can express \(y_A\):
\[ y_A = 30 - x_B \]

5. **Substituting into Coordinates of C**:
Substitute \(y_A\) into the equation for \(y_C\):
\[ y_C = \frac{30 - x_B}{3} = 10 - \frac{x_B}{3} \]

6. **Express x_B in terms of x_C**:
From the equation for \(x_C\):
\[ x_C = \frac{2x_B}{3} \quad \Rightarrow \quad x_B = \frac{3x_C}{2} \]

7. **Substitute x_B into y_C**:
Now substitute \(x_B\) into the equation for \(y_C\):
\[ y_C = 10 - \frac{1}{3} \left(\frac{3x_C}{2}\right) = 10 - \frac{x_C}{2} \]

8. **Formulate the Locus Equation**:
Now we have:
\[ y_C = 10 - \frac{x_C}{2} \]
Rearranging gives:
\[ y_C + \frac{x_C}{2} = 10 \]

9. **Convert to Standard Form**:
Rearranging further, we can express it as:
\[ \frac{x_C}{20} + \frac{y_C}{10} = 1 \]
This represents the equation of an ellipse.

### Final Locus Equation:
The locus of point C is given by:
\[ \frac{x^2}{20^2} + \frac{y^2}{10^2} = 1 \]