If a,b,c are in A.P then the straight lines `ax+by+c=0` will always pass through the point

Options

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

Answer

To solve the problem, we need to show that if \( a, b, c \) are in Arithmetic Progression (A.P.), then the straight line given by the equation \( ax + by + c = 0 \) will always pass through the point \( (1, -2) \).

### Step-by-Step Solution:

1. **Understanding A.P.**:
If \( a, b, c \) are in A.P., then by definition, we have:
\[ 2b = a + c \]

2. **Rearranging the Line Equation**:
We start with the line equation:
\[ ax + by + c = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ by = -ax - c \]
\[ y = -\frac{a}{b}x - \frac{c}{b} \]

3. **Substituting the A.P. Condition**:
From the A.P. condition \( 2b = a + c \), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]

4. **Substituting \( c \) into the Line Equation**:
Substitute \( c \) back into the line equation:
\[ ax + by + (2b - a) = 0 \]
Simplifying this gives:
\[ ax + by + 2b - a = 0 \]
Rearranging:
\[ ax + by = a - 2b \]

5. **Finding the Point (1, -2)**:
Now, we need to check if the point \( (1, -2) \) lies on this line. Substitute \( x = 1 \) and \( y = -2 \):
\[ a(1) + b(-2) + c = 0 \]
This simplifies to:
\[ a - 2b + c = 0 \]

6. **Using the A.P. Relation**:
Substitute \( c = 2b - a \) into the equation:
\[ a - 2b + (2b - a) = 0 \]
This simplifies to:
\[ a - 2b + 2b - a = 0 \]
Thus, we have:
\[ 0 = 0 \]
This confirms that the equation holds true.

### Conclusion:
Since the equation is satisfied, we conclude that the line \( ax + by + c = 0 \) will always pass through the point \( (1, -2) \) when \( a, b, c \) are in A.P.

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