For c neq 0, c neq 1,if the straight lines x+ | vedclass

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(A) The given lines are $x+y-1=0$,$2x-y-c=0$,and $bx+2by-c=0$. Since these lines have one common point,they are concurrent. The condition for concurre

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$c < 1 \Rightarrow b \in \left(-3, \frac{3}{4}\right)$

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(A) The given lines are $x+y-1=0$,$2x-y-c=0$,and $bx+2by-c=0$. Since these lines have one common point,they are concurrent. The condition for concurrency is that the determinant of the coefficients must be zero: $\left|\begin{array}{ccc} 1 & 1 & -1 \ 2 & -1 & -c \ b & 2b & -c \end{array}ight| = 0$ Expanding the determinant: $1(c + 2bc) - 1(-2c + bc) - 1(4b + b) = 0$ $c + 2bc + 2c - bc - 5b = 0$ $3c + bc - 5b = 0$ $c(b+3) = 5b$ $c = \frac{5b}{b+3}$ For $c $\frac{5b}{b+3} $\frac{5b - b - 3}{b+3} Using the wavy curve method,the inequality holds for $b \in \left(-3, \frac{3}{4}ight)$.

For $c \neq 0, c \neq 1$,if the straight lines $x+y=1$,$2x-y=c$,and $bx+2by=c$ have one common point,then:

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