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If the lines $3 \mathrm{y}+4 \mathrm{x}=1, \mathrm{y}=\mathrm{x}+5$ and $5 \mathrm{y}+\mathrm{bx}=3$ are concurrent, then what is the value of b?
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The correct answer is:
6
Given lines, $3 \mathrm{y}+4 \mathrm{x}=1 \Rightarrow 4 \mathrm{x}+3 \mathrm{y}-1=0$
$y=x+5 \Rightarrow x-y+5=0$
$5 y+b x=3 \Rightarrow b x+5 y-3=0$
Since, these lines are concurrent,
$\left|\begin{array}{ccc}4 & 3 & -1 \\ 1 & -1 & 5 \\ b & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 b)-1(5+b)=0$
$\Rightarrow 4(-22)+9+15 b-5-b=0$
$\Rightarrow-88+4+14 b=0$
$\Rightarrow-84+14 b=0$
$\Rightarrow \mathrm{b}=6$
$y=x+5 \Rightarrow x-y+5=0$
$5 y+b x=3 \Rightarrow b x+5 y-3=0$
Since, these lines are concurrent,
$\left|\begin{array}{ccc}4 & 3 & -1 \\ 1 & -1 & 5 \\ b & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 b)-1(5+b)=0$
$\Rightarrow 4(-22)+9+15 b-5-b=0$
$\Rightarrow-88+4+14 b=0$
$\Rightarrow-84+14 b=0$
$\Rightarrow \mathrm{b}=6$
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