(a, b) is the point of concurrency of the lin | vedclass

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(B) The lines $x-3y+3=0$ and $2x+y-8=0$ intersect at the point of concurrency $(a, b)$.Solving these two equations:$x-3y = -3$$2x+y = 8 \Rightarrow y

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$p$

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(B) The lines $x-3y+3=0$ and $2x+y-8=0$ intersect at the point of concurrency $(a, b)$.Solving these two equations:$x-3y = -3$$2x+y = 8 \Rightarrow y = 8-2x$Substituting $y$ in the first equation: $x-3(8-2x) = -3$ $\Rightarrow x-24+6x = -3$ $\Rightarrow 7x = 21$ $\Rightarrow x=3$.Then $y = 8-2(3) = 2$.So,$(a, b) = (3, 2)$.Since $(3, 2)$ lies on $kx+y+k=0$,we have $3k+2+k=0$ $\Rightarrow 4k = -2$ $\Rightarrow k = -\frac{1}{2}$.The line $L$ is $ax-by+2k=0$,which becomes $3x-2y+2(-\frac{1}{2}) = 0 \Rightarrow 3x-2y-1=0$.The perpendicular distance $p$ from the origin $(0, 0)$ to $3x-2y-1=0$ is $p = \frac{|3(0)-2(0)-1|}{\sqrt{3^2+(-2)^2}} = \frac{1}{\sqrt{13}}$.The perpendicular distance from $(2, 3)$ to $3x-2y-1=0$ is $d = \frac{|3(2)-2(3)-1|}{\sqrt{3^2+(-2)^2}} = \frac{|6-6-1|}{\sqrt{13}} = \frac{1}{\sqrt{13}}$.Since $d = p$,the correct option is $p$.

$(a, b)$ is the point of concurrency of the lines $x-3y+3=0$,$kx+y+k=0$,and $2x+y-8=0$. If the perpendicular distance from the origin to the line $L \equiv ax-by+2k=0$ is $p$,then the perpendicular distance from the point $(2, 3)$ to $L=0$ is

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