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 determine the point through which the line \( ax + 2by + c = 0 \) passes, given that \( a, b, c \) are in Arithmetic Progression (A.P.).

### Step 1: Understand the condition for A.P.
The condition for three numbers \( a, b, c \) to be in A.P. is given by:
\[ 2b = a + c \]
This means that the middle term \( b \) is the average of \( a \) and \( c \).

### Step 2: Express \( c \) in terms of \( a \) and \( b \)
From the A.P. condition, we can rearrange the equation to express \( c \):
\[ c = 2b - a \]

### Step 3: Substitute \( c \) into the line equation
Now, we substitute \( c \) into the line equation \( ax + 2by + c = 0 \):
\[ ax + 2by + (2b - a) = 0 \]
This simplifies to:
\[ ax + 2by + 2b - a = 0 \]
Rearranging gives:
\[ ax + 2by = a - 2b \]

### Step 4: Factor out common terms
We can factor out \( a \) and \( b \) from the left-hand side:
\[ a(x - 1) + 2b(y + 1) = 0 \]

### Step 5: Set each factor to zero
For the equation to hold true for all values of \( a \) and \( b \), we set each factor to zero:
1. \( x - 1 = 0 \) implies \( x = 1 \)
2. \( 2b(y + 1) = 0 \) implies \( y + 1 = 0 \) (since \( b \neq 0 \)), thus \( y = -1 \)

### Step 6: Conclusion
The point through which the line \( ax + 2by + c = 0 \) passes is:
\[ (1, -1) \]

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

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