If the quadratic equation `ax^2+bx+c=0` has `-2` as one of its roots then `ax + by + c = 0` represents

Options

A. A family of concurrent lines
B. A family of parallel lines
C. A single line
D. A line perpendicular to x - axis

Answer

To solve the problem, we need to analyze the given quadratic equation and the linear equation provided.

### Step-by-Step Solution:

1. **Given Information**: We have a quadratic equation of the form \( ax^2 + bx + c = 0 \) with one of its roots as \( -2 \).

2. **Using the Root**: Since \( -2 \) is a root, we can substitute \( x = -2 \) into the quadratic equation:
\[ a(-2)^2 + b(-2) + c = 0 \]
This simplifies to:
\[ 4a - 2b + c = 0 \quad \text{(Equation 1)} \]

3. **Linear Equation**: We are also given the linear equation \( ax + by + c = 0 \).

4. **Rearranging the Linear Equation**: We can rearrange the linear equation as follows:
\[ ax + by = -c \quad \text{(Equation 2)} \]

5. **Finding Points**: Now, we need to find specific points that satisfy both equations. From Equation 1, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 4a - 2b \]

6. **Substituting \( c \) in Equation 2**: Substitute \( c \) from Equation 1 into Equation 2:
\[ ax + by = -(4a - 2b) \]
This simplifies to:
\[ ax + by = -4a + 2b \]

7. **Finding Specific Values**: To find specific values of \( x \) and \( y \), let's set \( x = 4 \) (from \( 4a \)) and \( y = -2 \) (from \( -2b \)):
\[ a(4) + b(-2) = -c \]
Since \( c = 4a - 2b \), substituting gives:
\[ 4a - 2b = - (4a - 2b) \]
This confirms that \( (4, -2) \) is a point on the line represented by \( ax + by + c = 0 \).

8. **Conclusion**: Since the linear equation \( ax + by + c = 0 \) can be rearranged to show that it passes through the point \( (4, -2) \) for any value of \( c \), we conclude that it represents a family of concurrent lines that all pass through the point \( (4, -2) \).

### Final Answer:
The equation \( ax + by + c = 0 \) represents a family of concurrent lines passing through the point \( (4, -2) \).