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 consecutive odd integers then the line ax + by + c = 0 will pass through

Options

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

Answer

To solve the problem step by step, we need to analyze the given equation and the conditions provided.

### Step 1: Understand the given equation
We start with the equation:
\[ a + 2b + 3c = 0 \]

### Step 2: Rewrite the equation
We can rewrite this equation in terms of \( c \):
\[ 3c = - (a + 2b) \]
\[ c = -\frac{1}{3}(a + 2b) \]

### Step 3: Substitute into the line equation
We need to compare this with the line equation of the form:
\[ ax + by + c = 0 \]
Substituting \( c \) into the line equation gives:
\[ ax + by - \frac{1}{3}(a + 2b) = 0 \]

### Step 4: Rearranging the equation
Multiplying through by 3 to eliminate the fraction:
\[ 3ax + 3by - (a + 2b) = 0 \]
This simplifies to:
\[ 3ax + 3by - a - 2b = 0 \]

### Step 5: Grouping terms
Rearranging gives:
\[ a(3x - 1) + b(3y - 2) = 0 \]

### Step 6: Analyzing the coefficients
For this equation to hold for all values of \( a \) and \( b \), we must have:
\[ 3x - 1 = 0 \quad \text{and} \quad 3y - 2 = 0 \]

### Step 7: Solving for \( x \) and \( y \)
From \( 3x - 1 = 0 \):
\[ 3x = 1 \]
\[ x = \frac{1}{3} \]

From \( 3y - 2 = 0 \):
\[ 3y = 2 \]
\[ y = \frac{2}{3} \]

### Step 8: Conclusion about the point
Thus, we find that the point through which all lines of the form \( ax + by + c = 0 \) pass is:
\[ \left( \frac{1}{3}, \frac{2}{3} \right) \]

### Step 9: Considering consecutive odd integers
If \( a, b, c \) are consecutive odd integers, we can represent them as:
- Let \( a = n \) (first odd integer)
- Let \( b = n + 2 \) (second odd integer)
- Let \( c = n + 4 \) (third odd integer)

### Step 10: Substitute into the line equation
Substituting these into the line equation \( ax + by + c = 0 \):
\[ n x + (n + 2) y + (n + 4) = 0 \]

### Step 11: Rearranging
This can be rearranged to:
\[ n(x + 1) + 2y + 4 = 0 \]

### Step 12: Finding fixed point
To find the fixed point, we can set \( n = 0 \) (which is valid since we are looking for a point independent of \( n \)):
\[ 2y + 4 = 0 \]
\[ y = -2 \]

Thus, the line will pass through the point:
\[ (1, -2) \]

### Final Answer
The line \( ax + by + c = 0 \) will pass through the point:
\[ (1, -2) \]

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