If the lines x+3y-9=0,4x+by-2=0,and 2x-y-4=0 | vedclass

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(D) Given that the lines $x+3y-9=0$,$4x+by-2=0$,and $2x-y-4=0$ are concurrent. The condition for concurrency is that the determinant of the coefficien

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$x - 4y + 5 = 0$

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(D) Given that the lines $x+3y-9=0$,$4x+by-2=0$,and $2x-y-4=0$ are concurrent. The condition for concurrency is that the determinant of the coefficients must be zero: $\begin{vmatrix} 1 & 3 & -9 \ 4 & b & -2 \ 2 & -1 & -4 \end{vmatrix} = 0$ Expanding the determinant: $1(-4b - 2) - 3(-16 + 4) - 9(-4 - 2b) = 0$ $-4b - 2 + 36 + 36 + 18b = 0$ $14b + 70 = 0$ $\Rightarrow 14b = -70$ $\Rightarrow b = -5$ Now,find the point of concurrency by solving $x+3y-9=0$ and $2x-y-4=0$: From $2x-y-4=0$,we get $y = 2x-4$. Substituting into $x+3y-9=0$: $x + 3(2x-4) - 9 = 0$ $\Rightarrow x + 6x - 12 - 9 = 0$ $\Rightarrow 7x = 21$ $\Rightarrow x = 3$. Then $y = 2(3) - 4 = 2$. The point of concurrency is $(3, 2)$. The required line passes through $(b, 0) = (-5, 0)$ and $(3, 2)$. The equation is $y - 0 = \frac{2-0}{3 - (-5)}(x - (-5))$ $y = \frac{2}{8}(x+5)$ $\Rightarrow y = \frac{1}{4}(x+5)$ $\Rightarrow 4y = x+5$ $\Rightarrow x - 4y + 5 = 0$.

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

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