The line parallel to the x- axis and passing | vedclass

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(C) Given lines are:$ax + 2by + 3b = 0$  $(1)$$bx - 2ay - 3a = 0$  $(2)$To find the intersection point,multiply $(1)$ by $a$ and $(2)$ by $b$:$a^2x +

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below $x-$ axis at a distance $3/2$ from it

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(C) Given lines are:$ax + 2by + 3b = 0$  $(1)$$bx - 2ay - 3a = 0$  $(2)$To find the intersection point,multiply $(1)$ by $a$ and $(2)$ by $b$:$a^2x + 2aby + 3ab = 0$$b^2x - 2aby - 3ab = 0$Adding these equations: $(a^2 + b^2)x = 0$. Since $(a, b) 
eq (0, 0)$,$a^2 + b^2 
eq 0$,so $x = 0$.Substituting $x = 0$ into $(1)$: $2by = -3b$. Since $b$ cannot be $0$ (otherwise the lines would not intersect at a unique point or would be coincident),we get $y = -3/2$.The point of intersection is $(0, -3/2)$.$A$ line parallel to the $x-$ axis has the equation $y = k$. Since it passes through $(0, -3/2)$,the equation is $y = -3/2$.This represents a line below the $x-$ axis at a distance of $3/2$ from it.

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 line parallel to the $x-$ axis and passing through the point of intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$,where $(a, b) \neq (0, 0)$,is:

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