If `Delta` be the determinant given in last question and `Delta=0` then system of lines given by the equation `ax+by+c=0` pass through the point

Options

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

Answer

To solve the problem, we need to determine the point through which the system of lines represented by the equation \( ax + by + c = 0 \) passes when the determinant \( \Delta = 0 \).

### Step-by-Step Solution:

1. **Given Condition**: We know that \( \Delta = 0 \). This implies that the lines represented by the equations are concurrent, meaning they intersect at a common point.

2. **Understanding the Variables**: The coefficients \( a \), \( b \), and \( c \) are given to be in Arithmetic Progression (AP). If \( a \), \( b \), and \( c \) are in AP, then:
\[ 2b = a + c \]
From this, we can express \( b \) as:
\[ b = \frac{a + c}{2} \]

3. **Substituting \( b \) into the Line Equation**: The line equation can be rewritten by substituting the value of \( b \):
\[ ax + \left(\frac{a + c}{2}\right)y + c = 0 \]

4. **Clearing the Fraction**: Multiply the entire equation by 2 to eliminate the fraction:
\[ 2ax + (a + c)y + 2c = 0 \]

5. **Rearranging the Equation**: Rearranging gives:
\[ 2ax + ay + cy + 2c = 0 \]

6. **Factoring the Equation**: We can factor the equation:
\[ A(2x + y) + C(y + 2) = 0 \]
Here, \( A = a \) and \( C = c \).

7. **Setting Each Factor to Zero**: For the equation to hold, either:
\[ 2x + y = 0 \quad \text{(Equation 1)} \]
or
\[ y + 2 = 0 \quad \text{(Equation 2)} \]

8. **Solving Equation 2**: From Equation 2:
\[ y + 2 = 0 \implies y = -2 \]

9. **Substituting \( y \) into Equation 1**: Substitute \( y = -2 \) into Equation 1:
\[ 2x - 2 = 0 \implies 2x = 2 \implies x = 1 \]

10. **Finding the Point**: Thus, the point through which the lines pass is:
\[ (x, y) = (1, -2) \]

### Final Answer:
The system of lines given by the equation \( ax + by + c = 0 \) passes through the point \( (1, -2) \).