If the lines L_1 equiv x-2y+3=0,L_2 equiv 2x+ | vedclass

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(A) The lines $L_1: x-2y+3=0$ and $L_2: 2x+y+1=0$ intersect at a point. Solving these equations:$x-2y = -3$$2x+y = -1 \implies y = -1-2x$Substituting

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$7$

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(A) The lines $L_1: x-2y+3=0$ and $L_2: 2x+y+1=0$ intersect at a point. Solving these equations:$x-2y = -3$$2x+y = -1 \implies y = -1-2x$Substituting $y$ in the first equation: $x-2(-1-2x) = -3 \implies x+2+4x = -3 \implies 5x = -5 \implies x = -1$.Then $y = -1-2(-1) = 1$. The point of intersection is $(-1, 1)$.Since the lines are concurrent,$(-1, 1)$ must satisfy $L_3: 3x+y+c=0$.$3(-1)+1+c = 0 \implies -3+1+c = 0 \implies c = 2$.Now,the slopes of $L_1$ and $L_3$ are $m_1 = 1/2$ and $m_3 = -3$.The acute angle $	heta$ between them is given by $	an 	heta = \left| \frac{m_1-m_3}{1+m_1m_3} ight|$.$	an 	heta = \left| \frac{1/2 - (-3)}{1 + (1/2)(-3)} ight| = \left| \frac{7/2}{1 - 3/2} ight| = \left| \frac{7/2}{-1/2} ight| = |-7| = 7$.

If the lines $L_1 \equiv x-2y+3=0$,$L_2 \equiv 2x+y+1=0$,and $L_3 \equiv 3x+y+c=0$ are concurrent and $\theta$ is the acute angle between the lines $L_1=0$ and $L_3=0$,then $\tan \theta=$

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