If the points `a+b,a-b and a+kb` be collinear, then k is equal to

Options

A. A. 0
B. B. 2
C. C. `-2`
D. D. any real number

Explanation

To solve the problem, we need to determine the value of \( k \) such that the points \( a + b \), \( a - b \), and \( a + kb \) are collinear. Here’s a step-by-step solution:

### Step 1: Define the Points
Let:
- Point A = \( a + b \)
- Point B = \( a - b \)
- Point C = \( a + kb \)

### Step 2: Find the Vectors
We need to find the vectors \( \overrightarrow{AB} \), \( \overrightarrow{BC} \), and \( \overrightarrow{AC} \).

1. **Vector \( \overrightarrow{AB} \)**:
\[ \overrightarrow{AB} = B - A = (a - b) - (a + b) = a - b - a - b = -2b \]

2. **Vector \( \overrightarrow{BC} \)**:
\[ \overrightarrow{BC} = C - B = (a + kb) - (a - b) = a + kb - a + b = (k + 1)b \]

3. **Vector \( \overrightarrow{AC} \)**:
\[ \overrightarrow{AC} = C - A = (a + kb) - (a + b) = a + kb - a - b = (k - 1)b \]

### Step 3: Use the Collinearity Condition
For the points to be collinear, the vectors \( \overrightarrow{AB} \), \( \overrightarrow{BC} \), and \( \overrightarrow{AC} \) must satisfy the condition that one vector can be expressed as a linear combination of the others.

We can express:
\[ \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} \]

Substituting the vectors we found:
\[ -2b + (k + 1)b = (k - 1)b \]

### Step 4: Simplify the Equation
Combine the terms:
\[ (-2 + k + 1)b = (k - 1)b \]

This simplifies to:
\[ (k - 1 - 2 + k + 1)b = 0 \]
\[ (2k - 2)b = 0 \]

### Step 5: Solve for \( k \)
Since \( b \neq 0 \) (as it is a vector), we can set the coefficient to zero:
\[ 2k - 2 = 0 \]
\[ 2k = 2 \implies k = 1 \]

### Conclusion
The value of \( k \) that makes the points collinear is \( k = 1 \).