Consider the lines given by L_1: x+3y-5=0,L_2 | vedclass

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(B) The lines are $L_1: x+3y-5=0$,$L_2: 3x-ky-1=0$,and $L_3: 5x+2y-12=0$.$(A)$ For concurrency,the lines must intersect at a single point. Solving $L_

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$(A) \rightarrow (s); (B) \rightarrow (p, q); (C) \rightarrow (r); (D) \rightarrow (p, q, s)$

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(B) The lines are $L_1: x+3y-5=0$,$L_2: 3x-ky-1=0$,and $L_3: 5x+2y-12=0$.$(A)$ For concurrency,the lines must intersect at a single point. Solving $L_1$ and $L_3$: $x+3y=5$ and $5x+2y=12$. Multiplying $L_1$ by $5$: $5x+15y=25$. Subtracting $L_3$: $13y=13 \Rightarrow y=1$. Then $x=2$. The point of intersection is $(2, 1)$. Substituting into $L_2$: $3(2)-k(1)-1=0 \Rightarrow 6-k-1=0 \Rightarrow k=5$. Thus,$(A) ightarrow (s)$.$(B)$ For lines to be parallel: $L_1 \parallel L_2: \frac{1}{3} = \frac{3}{-k} \Rightarrow k=-9$.$L_2 \parallel L_3: \frac{3}{5} = \frac{-k}{2} \Rightarrow k=-\frac{6}{5}$.$L_1 \parallel L_3$ is not possible as slopes are $-1/3$ and $-5/2$. Thus,$(B) ightarrow (p, q)$.$(C)$ Lines form a triangle if they are not concurrent and no two lines are parallel. This happens when $k 
eq 5, -9, -\frac{6}{5}$. Among the options,only $k=\frac{5}{6}$ satisfies this. Thus,$(C) ightarrow (r)$.$(D)$ Lines do not form a triangle if they are concurrent or parallel. This happens when $k=5, -9, -\frac{6}{5}$. Thus,$(D) ightarrow (p, q, s)$.

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	$(s)$ $k=5$

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