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

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(B) For three lines to be concurrent,the determinant of their coefficients must be zero:$\left| \begin{array}{ccc} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c

Vedclass · Accepted Answer

$1$

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(B) For three lines to be concurrent,the determinant of their coefficients must be zero:$\left| \begin{array}{ccc} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{array} ight| = 0$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\left| \begin{array}{ccc} a & 1-a & 1-a \ 1 & b-1 & 0 \ 1 & 0 & c-1 \end{array} ight| = 0$Expanding along the first row:$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$Divide the entire equation by $(1-a)(1-b)(1-c)$:$\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$$-\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$Since $-\frac{a}{1-a} = \frac{1-a-1}{1-a} = \frac{1}{1-a} - 1$,we substitute this into the equation:$(\frac{1}{1-a} - 1) + \frac{1}{1-b} + \frac{1}{1-c} = 0$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

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

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