If `a + 2b + 3c = 0 " then " a/3+(2b)/3+c=0` and comparing with line ax + by + c, we get `x = 1/3 & y = 2/ 3` so there will be a point `(1/3,2/3)` from where each of the lines of the form ax + by + c = 0 will pass for the given relation between a,b,c . We can say if there exists a linear relation between a,b,c then the family of straight lines of the form of `ax + by +c ` pass through a fixed point .
If a , b,c are in A.P., then the line ax + 2by + c = 0 passes through

Options

A. (0,0)
B. (1,-2)
C. (-1,1)
D. (1,-1)

Answer

To solve the problem, we need to find the point through which the line \( ax + 2by + c = 0 \) passes when \( a, b, c \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Given that \( a + 2b + 3c = 0 \), we can rewrite this as:
\[ \frac{a}{3} + \frac{2b}{3} + c = 0 \]
This implies that the line represented by \( ax + by + c = 0 \) passes through the point \( \left( \frac{1}{3}, \frac{2}{3} \right) \).

**Hint**: Recognize that the coefficients of \( a, b, c \) relate to the coordinates of a point.

2. **Condition for A.P.**:
If \( a, b, c \) are in A.P., then the condition for A.P. is:
\[ 2b = a + c \]
From this, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]

**Hint**: Use the definition of A.P. to relate the terms.

3. **Substituting into the Line Equation**:
Substitute \( c = 2b - a \) into the line equation \( ax + 2by + c = 0 \):
\[ ax + 2by + (2b - a) = 0 \]
Simplifying this gives:
\[ ax + 2by + 2b - a = 0 \]
Rearranging, we have:
\[ ax - a + 2by + 2b = 0 \]
Factoring out \( a \) and \( b \):
\[ a(x - 1) + 2b(y + 1) = 0 \]

**Hint**: Factor the equation to identify the relationships between the variables.

4. **Setting Up the Conditions**:
For the equation \( a(x - 1) + 2b(y + 1) = 0 \) to hold for all \( a \) and \( b \), both coefficients must independently equal zero:
- \( x - 1 = 0 \) implies \( x = 1 \)
- \( 2b(y + 1) = 0 \) implies \( y + 1 = 0 \) or \( y = -1 \)

**Hint**: Set each coefficient to zero to find the fixed point.

5. **Conclusion**:
Thus, the fixed point through which the line \( ax + 2by + c = 0 \) passes when \( a, b, c \) are in A.P. is:
\[ (x, y) = (1, -1) \]

**Final Answer**: The line \( ax + 2by + c = 0 \) passes through the point \( (1, -1) \).