A and B are fixed points such that AB=2a. The vertex C of `DeltaABC` such that `cotA+cotB`=constant. Then locus of C is
A
straight line perpendicular to AB
B
straight line parallel to AB
C
circle
D
none of these
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to find the locus of point C such that \( \cot A + \cot B = \text{constant} \) given that points A and B are fixed and the distance \( AB = 2a \).
### Step-by-Step Solution:
1. **Understanding the Setup**:
Let points A and B be fixed points on the coordinate plane. We can place A at \( (-a, 0) \) and B at \( (a, 0) \) so that the distance \( AB = 2a \).
2. **Defining the Triangle**:
Let C be a point \( (x, y) \) in the plane. We need to analyze the angles A and B in triangle ABC.
3. **Using Cotangent Formula**:
The cotangent of an angle in a triangle can be expressed in terms of the lengths of the sides. Specifically, for triangle ABC:
\[
\cot A = \frac{AD}{CD} \quad \text{and} \quad \cot B = \frac{BD}{CD}
\]
where D is the foot of the perpendicular from C to line AB.
4. **Finding Lengths**:
The lengths can be expressed as follows:
- \( AD = \sqrt{(x + a)^2 + y^2} \)
- \( BD = \sqrt{(x - a)^2 + y^2} \)
- \( CD = y \) (since D lies on the x-axis)
5. **Setting Up the Equation**:
We can substitute these lengths into the cotangent expressions:
\[
\cot A + \cot B = \frac{\sqrt{(x + a)^2 + y^2}}{y} + \frac{\sqrt{(x - a)^2 + y^2}}{y} = \text{constant}
\]
Simplifying gives:
\[
\frac{\sqrt{(x + a)^2 + y^2} + \sqrt{(x - a)^2 + y^2}}{y} = \text{constant}
\]
6. **Rearranging the Equation**:
Let \( k \) be the constant. Then we have:
\[
\sqrt{(x + a)^2 + y^2} + \sqrt{(x - a)^2 + y^2} = k \cdot y
\]
7. **Analyzing the Locus**:
The equation derived indicates that as C moves such that the sum of the distances from A and B remains proportional to the vertical distance (y-coordinate), the locus of C will be a line parallel to the line segment AB.
8. **Conclusion**:
Therefore, the locus of point C is a straight line parallel to line AB.
### Final Answer:
The locus of C is a straight line parallel to AB.
To solve the problem, we need to find the locus of point C such that \( \cot A + \cot B = \text{constant} \) given that points A and B are fixed and the distance \( AB = 2a \).
### Step-by-Step Solution:
1. **Understanding the Setup**:
Let points A and B be fixed points on the coordinate plane. We can place A at \( (-a, 0) \) and B at \( (a, 0) \) so that the distance \( AB = 2a \).
2. **Defining the Triangle**:
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