If the lines `a x+y+1=0,x+b y+1=0` and `x+y+c=0(a ,b ,c` being distinct and different from `1)` are concurrent, then prove that `1/(1-a)+1/(1-b)+1/(1-c)=1.`

Options

A. 0
B. 1
C. 1/(a+b+c)
D. none of these

Explanation

To prove that \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \) given that the lines \( ax + y + 1 = 0 \), \( x + by + 1 = 0 \), and \( x + y + c = 0 \) are concurrent, we will follow these steps:

### Step 1: Write the equations in standard form
The equations of the lines can be rewritten as:
1. \( ax + y + 1 = 0 \)
2. \( x + by + 1 = 0 \)
3. \( x + y + c = 0 \)

### Step 2: Form the coefficient matrix
The coefficient matrix for the lines is:
\[ \begin{bmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{bmatrix} \]

### Step 3: Set the determinant to zero
For the lines to be concurrent, the determinant of the coefficient matrix must be zero:
\[ \text{Det} = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Using the determinant formula, we expand:
\[ \text{Det} = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ = a(bc - 1) - (c - 1) + (1 - b) \]
\[ = abc - a - c + 1 + 1 - b \]
\[ = abc - a - b - c + 2 \]

### Step 5: Set the determinant to zero
Setting the determinant to zero gives:
\[ abc - a - b - c + 2 = 0 \]
Thus,
\[ abc = a + b + c - 2 \]

### Step 6: Rearranging the equation
Now, we can rearrange the equation:
\[ abc - a - b - c + 2 = 0 \implies 1 = a + b + c - abc \]

### Step 7: Multiply by \( \frac{1}{(1-a)(1-b)(1-c)} \)
We multiply the entire equation by \( \frac{1}{(1-a)(1-b)(1-c)} \):
\[ \frac{1}{(1-a)(1-b)(1-c)} = \frac{a + b + c - abc}{(1-a)(1-b)(1-c)} \]

### Step 8: Simplifying the left-hand side
Now, we can express \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \):
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b)}{(1-a)(1-b)(1-c)} \]

### Step 9: Expanding the numerator
Expanding the numerator:
\[ = (1 - b - c + bc) + (1 - a - c + ac) + (1 - a - b + ab) \]
\[ = 3 - (a + b + c) + (ab + ac + bc) \]

### Step 10: Using the previous result
Using the result from Step 6, we can substitute \( a + b + c \) and \( abc \):
\[ = 3 - (a + b + c) + (ab + ac + bc) = 1 \]

### Conclusion
Thus, we have shown that:
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \]