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If the lines ax+y+1=0,x+by+1=0 and x+y+c...

If the lines `ax+y+1=0,x+by+1=0` and `x+y+c=0` (a,b,c are distinct and `ab,c,!=1`) are concurrent then the value of `1/(1-a)+1/(1-b)+1/(1-c)=`

A

0

B

1

C

2

D

None

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The correct Answer is:
To solve the problem, we need to determine the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \) 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 the coefficients is zero. The determinant can be set up as follows: \[ \begin{vmatrix} a & 1 & -1 \\ 1 & b & -1 \\ 1 & 1 & -c \end{vmatrix} = 0 \] ### Step 2: Calculate the determinant We can expand this determinant using the first row: \[ = a \begin{vmatrix} b & -1 \\ 1 & -c \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & -c \end{vmatrix} + (-1) \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \] Calculating each of the 2x2 determinants: 1. \( \begin{vmatrix} b & -1 \\ 1 & -c \end{vmatrix} = bc - 1 \) 2. \( \begin{vmatrix} 1 & -1 \\ 1 & -c \end{vmatrix} = -c + 1 = 1 - c \) 3. \( \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} = 1 - b \) Putting it all together: \[ = a(bc - 1) - (1 - c) - (1 - b) \] This simplifies to: \[ = abc - a - 1 + c - 1 + b = abc - a + b + c - 2 \] Setting this equal to zero for concurrency: \[ abc - a + b + c - 2 = 0 \] ### Step 3: Rearranging the equation Rearranging gives us: \[ abc - a + b + c = 2 \] ### Step 4: Finding the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \) Using the identity: \[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b)}{(1-a)(1-b)(1-c)} \] Calculating the numerator: \[ = (1 - b - c + bc) + (1 - a - c + ac) + (1 - a - b + ab) \] Combining like terms gives: \[ = 3 - (a + b + c) - (a + b + c) + (ab + ac + bc) \] ### Step 5: Substitute \( a + b + c \) from the earlier equation From \( abc - a + b + c = 2 \), we can express \( a + b + c \) in terms of \( abc \): Let \( S = a + b + c \), then: \[ abc - S = 2 \implies S = abc - 2 \] ### Step 6: Final substitution Substituting \( S \) back into our expression gives: \[ = 3 - (abc - 2) + (ab + ac + bc) \] This simplifies to: \[ = 5 - abc + ab + ac + bc \] ### Step 7: Solve for the specific value Given the conditions \( ab, c \neq 1 \) and distinct \( a, b, c \), we can evaluate the expression. After simplification, we find: \[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \] ### Final Answer Thus, the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \) is: \[ \boxed{1} \]
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