If the lines ax+y+1=0, x+by+1=0 and x+y+c=0 (a,b and c being distinct and different from 1) are concurrent the value of `(a)/(a-1)+(b)/(b-1)+(c)/(c-1)` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the value of \(\frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1}\) given that the lines \(ax + y + 1 = 0\), \(x + by + 1 = 0\), and \(x + y + c = 0\) are concurrent.

### Step 1: Set up the determinant for concurrency
The lines are concurrent if the determinant of their coefficients is equal to zero. The coefficients of the lines can be arranged in a determinant as follows:

\[ \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
We will calculate the determinant step by step:

\[ D = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} \]

Using the determinant expansion method, we can expand along the first row:

\[ D = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} = bc - 1\)

2. \(\begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} = c - 1\)

3. \(\begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} = 1 - b\)

Putting these back into the determinant:

\[ D = a(bc - 1) - (c - 1) + (1 - b) \]

This simplifies to:

\[ D = abc - a - c + 1 + 1 - b \]
\[ D = abc - a - b - c + 2 \]

### Step 3: Set the determinant to zero
Since the lines are concurrent, we set the determinant to zero:

\[ abc - a - b - c + 2 = 0 \]

Rearranging gives us:

\[ abc = a + b + c - 2 \]

### Step 4: Express \(\frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1}\)
Now we need to find the value of:

\[ \frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1} \]

This can be rewritten using the identity:

\[ \frac{x}{x-1} = 1 + \frac{1}{x-1} \]

Thus,

\[ \frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1} = 3 + \left(\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1}\right) \]

### Step 5: Find a common denominator
Let’s find a common denominator for \(\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1}\):

\[ \frac{(b-1)(c-1) + (a-1)(c-1) + (a-1)(b-1)}{(a-1)(b-1)(c-1)} \]

### Step 6: Simplify the numerator
The numerator expands to:

\[ (bc - b - c + 1) + (ac - a - c + 1) + (ab - a - b + 1) \]

Combining like terms gives:

\[ abc - (a + b + c) + 3 \]

### Step 7: Substitute \(abc\)
Using \(abc = a + b + c - 2\):

\[ abc - (a + b + c) + 3 = (a + b + c - 2) - (a + b + c) + 3 = 1 \]

### Final Result
Thus, we find:

\[ \frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} = \frac{1}{(a-1)(b-1)(c-1)} \]

So, we have:

\[ \frac{a}{a-1} + \frac{b}{b-1} + \frac{c}{c-1} = 3 + 1 = 4 \]

Therefore, the final answer is:

\[ \boxed{4} \]