Let `A(2, 0) and B(-2, 0)` are two fixed vertices of `DeltaABC`. If the vertex C moves in the first quadrant in such a way that `cotA+cotB=2,` then the locus of the point C is

Options

A. `y=2`
B. `x=4`
C. `x=2`
D. `y=1`

Answer

To find the locus of point C in triangle ABC, where A(2, 0) and B(-2, 0) are fixed vertices and C moves in the first quadrant such that cot A + cot B = 2, we can follow these steps:

### Step 1: Define the coordinates of point C
Let the coordinates of point C be (h, k), where h > 0 and k > 0 since C is in the first quadrant.

### Step 2: Calculate cot A and cot B
To find cot A and cot B, we need to drop perpendiculars from point C to line AB (the x-axis). Let D be the foot of the perpendicular from C to the x-axis.

- The distance AD (base for cot A) = 2 - h
- The height CD (perpendicular) = k
- Therefore, \( \cot A = \frac{AD}{CD} = \frac{2 - h}{k} \)

For cot B:
- The distance BD (base for cot B) = h + 2
- The height CD (perpendicular) = k
- Therefore, \( \cot B = \frac{BD}{CD} = \frac{h + 2}{k} \)

### Step 3: Set up the equation based on the given condition
According to the problem, we have:
\[ \cot A + \cot B = 2 \]
Substituting the expressions for cot A and cot B:
\[ \frac{2 - h}{k} + \frac{h + 2}{k} = 2 \]

### Step 4: Combine the fractions
Combining the fractions gives:
\[ \frac{(2 - h) + (h + 2)}{k} = 2 \]
This simplifies to:
\[ \frac{4}{k} = 2 \]

### Step 5: Solve for k
Cross-multiplying gives:
\[ 4 = 2k \]
Dividing both sides by 2 results in:
\[ k = 2 \]

### Step 6: Write the locus of point C
Since k represents the y-coordinate of point C, we have:
\[ y = 2 \]
Thus, the locus of point C is the line \( y = 2 \).

### Final Answer
The locus of point C is the line \( y = 2 \).

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