If `a, b, c` are in `A.P` then the lines represented by `ax + by+ c= 0` are

Options

A. a single line
B. a family of concurrent lines
C. a family of parallel lines
D. none of these

Explanation

To solve the problem, we need to analyze the condition given that \( a, b, c \) are in Arithmetic Progression (A.P.) and how it relates to the line represented by the equation \( ax + by + c = 0 \).

### Step-by-Step Solution:

1. **Understanding A.P.**:
Since \( a, b, c \) are in A.P., we have the relationship:
\[ a + c = 2b \]
This can also be rearranged to express \( c \):
\[ c = 2b - a \]

2. **Substituting \( c \) in the Line Equation**:
We substitute the expression for \( c \) into the line equation \( ax + by + c = 0 \):
\[ ax + by + (2b - a) = 0 \]
Simplifying this gives:
\[ ax + by + 2b - a = 0 \]

3. **Rearranging the Equation**:
We can rearrange the equation:
\[ ax + by = a - 2b \]

4. **Factoring the Equation**:
We can factor the left-hand side:
\[ a(x - 1) + b(y + 2) = 0 \]
This shows that the equation can be expressed as a linear combination of two lines.

5. **Identifying the Lines**:
The equation can be represented as:
\[ L_1: x - 1 = 0 \quad \text{and} \quad L_2: y + 2 = 0 \]
These represent two lines:
- \( L_1 \): A vertical line at \( x = 1 \)
- \( L_2 \): A horizontal line at \( y = -2 \)

6. **Finding the Intersection Point**:
The intersection point of these two lines is:
\[ P(1, -2) \]
This means that any line represented by the equation \( ax + by + c = 0 \) will pass through the point \( P(1, -2) \).

7. **Conclusion**:
Since the lines represented by the equation \( ax + by + c = 0 \) all pass through the fixed point \( P(1, -2) \), we conclude that they form a family of concurrent lines.

### Final Answer:
The lines represented by \( ax + by + c = 0 \) are a family of concurrent lines.

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