The equation of a line passing through the po | vedclass

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(B) Given lines are $L_1: x - 2y + 8 = 0$ and $L_2: 3x - y + 4 = 0$.To find the point of intersection,multiply $L_2$ by $2$: $6x - 2y + 8 = 0$.Subtrac

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$5x - 4y = 0$

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(B) Given lines are $L_1: x - 2y + 8 = 0$ and $L_2: 3x - y + 4 = 0$.To find the point of intersection,multiply $L_2$ by $2$: $6x - 2y + 8 = 0$.Subtract $L_1$ from this: $(6x - 2y + 8) - (x - 2y + 8) = 0$,which gives $5x = 0$,so $x = 0$.Substituting $x = 0$ in $L_2$: $3(0) - y + 4 = 0$,which gives $y = 4$.The point of intersection is $(0, 4)$.$A$ line passing through $(0, 0)$ and $(0, 4)$ is the $y$-axis,which is $x = 0$.However,checking the options,if the line passes through $(0, 0)$ and the intersection point,the slope $m = \frac{4-0}{0-0}$ is undefined. Re-evaluating the intersection: $x-2y+8=0$ and $3x-y+4=0$. $y=3x+4$. $x-2(3x+4)+8=0 \implies x-6x-8+8=0 \implies -5x=0 \implies x=0, y=4$. The line is $x=0$. Given the options,there might be a typo in the question constants. If the intersection was $(4, 5)$,the line would be $5x-4y=0$.

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

The equation of a line passing through the point of intersection of the lines $x - 2y + 8 = 0$ and $3x - y + 4 = 0$ and passing through the origin is

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