If `a^2+b^2-c^2-2ab = 0 ` , then the family of straight lines ax + by + c = 0 is concurrent at the points

Options

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

Answer

To solve the problem, we start with the given equation and the family of straight lines. We need to determine the point at which the family of lines is concurrent.

### Step-by-Step Solution:

1. **Given Equation**:
We start with the equation:
\[ a^2 + b^2 - c^2 - 2ab = 0 \]

2. **Rearranging the Equation**:
Rearranging the equation, we can group the terms:
\[ a^2 - 2ab + b^2 - c^2 = 0 \]
Notice that \(a^2 - 2ab + b^2\) can be factored:
\[ (a - b)^2 - c^2 = 0 \]

3. **Using the Difference of Squares**:
We can apply the difference of squares formula:
\[ (a - b + c)(a - b - c) = 0 \]
This gives us two equations:
\[ a - b + c = 0 \quad \text{or} \quad a - b - c = 0 \]

4. **Solving for 'a'**:
From the first equation, we can express \(a\):
\[ a = b - c \]
From the second equation, we can express \(a\) as:
\[ a = b + c \]

5. **Substituting 'a' into the Family of Lines**:
The family of straight lines is given by:
\[ ax + by + c = 0 \]
We will substitute both values of \(a\) into this equation.

**For \(a = b - c\)**:
\[ (b - c)x + by + c = 0 \]
Simplifying this:
\[ bx - cx + by + c = 0 \quad \Rightarrow \quad b(x + y) + c(1 - x) = 0 \]

**For \(a = b + c\)**:
\[ (b + c)x + by + c = 0 \]
Simplifying this:
\[ bx + cx + by + c = 0 \quad \Rightarrow \quad b(x + y) + c(x + 1) = 0 \]

6. **Finding the Concurrent Point**:
From the equations derived, we can find the points of concurrency.

**From \(a = b - c\)**:
Setting \(x + y = 0\) gives:
\[ y = -x \]

**From \(a = b + c\)**:
Setting \(x + 1 = 0\) gives:
\[ x = -1 \]
Substituting \(x = -1\) into \(y = -x\):
\[ y = 1 \]

7. **Final Result**:
Therefore, the point of concurrency is:
\[ (x, y) = (-1, 1) \]

### Conclusion:
The family of straight lines \(ax + by + c = 0\) is concurrent at the point \((-1, 1)\).