If the lines $x+3 y-9=0,4 x+b y-2=0$ and $2 x-y-4=0$ are concurrent, then the equation of the line passing…

If the lines $x+3 y-9=0,4 x+b y-2=0$ and $2 x-y-4=0$ are concurrent, then the equation of the line passing through the point $(b, 0)$ and concurrent with the given lines, is
  1. $2x + y + 10 = 0$
  2. $4x - 7y + 20 = 0$
  3. $x - y + 5 = 0$
  4. $x - 4y + 5 = 0$

Solution

Given that, $x+3 y-9=0, \quad 4 x+b y-2=0$ $2 x-y-4=0$ are concurrent. $\begin{aligned} & \therefore \quad\left|\begin{array}{ccc} 1 & 3 & -9 \\ 4 & b & -2 \\ 2 & -1 & -4 \end{array}\right|=0 \\ & 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0 \\ & -4 b-2+36+36+18 b=0 \\ & 14 b=-70 \quad \Rightarrow b=-5 \end{aligned}$ From equations $x+3 y-9=0$ and $2 x-y-4=0$ $\Rightarrow \quad x=3 \text { and } y=2$ $\therefore \quad$ Concurrency point is $(3,2)$ Equation of line passing through $(-5,0)$ and $(3,2)$ is $\begin{aligned} & y-0=\frac{2-0}{3+5}(x+5) \\ & y=\frac{1}{4}(x+5) \quad \Rightarrow 4 y=x+5 \\ & \Rightarrow x-4 y+5=0 \end{aligned}$

Asked in: AP EAMCET 2016

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