A and B are two fixed points. The vertex C of a `DeltaBC` moves such that `cotA+cotB=` constant. Locus of C is a straight line

Options

A. `_|_` to AB
B. parallel to AB
C. Inclined at an angle of `A-B` to AB
D. none

Answer

To solve the problem, we need to find the locus of point C in triangle ABC, where A and B are fixed points, and the condition given is that \( \cot A + \cot B = \text{constant} \).

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Let A and B be two fixed points in the Cartesian plane. Let C be a point that moves such that the sum of the cotangents of angles A and B is a constant.

2. **Define the Angles**:
Let \( \angle ACB = C \), \( \angle ABC = B \), and \( \angle BAC = A \).

3. **Using Cotangent Definitions**:
The cotangent of an angle in a triangle can be expressed in terms of the lengths of the sides:
- \( \cot A = \frac{AD}{CD} \)
- \( \cot B = \frac{BD}{CD} \)
where D is the foot of the perpendicular from C to line AB.

4. **Setting Up the Equation**:
Given that \( \cot A + \cot B = \text{constant} \), we can write:
\[ \cot A + \cot B = \frac{AD}{CD} + \frac{BD}{CD} = \frac{AD + BD}{CD} \]
This implies:
\[ \frac{AD + BD}{CD} = \text{constant} \]

5. **Finding the Locus**:
Since \( AD + BD = AB \) (the sum of the segments from D to A and B is equal to the length of AB), we can substitute:
\[ \frac{AB}{CD} = \text{constant} \]
Let’s denote \( AB = K \), where K is a constant.

6. **Conclusion About the Locus**:
Rearranging gives us:
\[ CD = \frac{K}{\text{constant}} \]
Since K is constant and the denominator is also constant, CD must be a fixed value. This means that as C moves, the height from C to line AB (which is represented by CD) remains constant.

7. **Final Result**:
Therefore, the locus of point C is a straight line parallel to line AB.