If the lines ax + y + 1 = 0, x + by + 1 = 0 a | vedclass

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(B) If the given lines are concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vma

Vedclass · Accepted Answer

$1$

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(B) If the given lines are concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\begin{vmatrix} a & 1-a & 1-a \ 1 & b-1 & 0 \ 1 & 0 & c-1 \end{vmatrix} = 0$Expanding the determinant:$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$Dividing the entire equation by $(1-a)(1-b)(1-c)$ (since $a, b, c 
eq 1$):$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$$\frac{-a}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = 0$Adding $1$ to both sides:$1 - \frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$\frac{1-a+a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

If the lines $ax + y + 1 = 0, x + by + 1 = 0$ and $x + y + c = 0$ ($a, b, c$ being distinct and different from $1$) are concurrent,then $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = $

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