L_1 and L_2 are two lines having slopes 2 and | vedclass

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(B) First,find the point of intersection of the lines $x-y+2=0$ and $2x+y+3=0$. Adding the two equations: $(x-y+2) + (2x+y+3) = 0 \implies 3x+5=0 \imp

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$7$

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(B) First,find the point of intersection of the lines $x-y+2=0$ and $2x+y+3=0$. Adding the two equations: $(x-y+2) + (2x+y+3) = 0 \implies 3x+5=0 \implies x = -\frac{5}{3}$. Substituting $x$ into $x-y+2=0$: $-\frac{5}{3} - y + 2 = 0 \implies y = 2 - \frac{5}{3} = \frac{1}{3}$. The point of concurrency is $P(-\frac{5}{3}, \frac{1}{3})$.For line $L_1$ with slope $m_1 = 2$: $y - \frac{1}{3} = 2(x + \frac{5}{3}) \implies y = 2x + \frac{10}{3} + \frac{1}{3} \implies y = 2x + \frac{11}{3} \implies 2x - y + \frac{11}{3} = 0$. The intercepts are $x = -\frac{11}{6}$ and $y = \frac{11}{3}$. Sum of absolute values: $|-\frac{11}{6}| + |\frac{11}{3}| = \frac{11}{6} + \frac{22}{6} = \frac{33}{6} = 5.5$.For line $L_2$ with slope $m_2 = -\frac{1}{2}$: $y - \frac{1}{3} = -\frac{1}{2}(x + \frac{5}{3}) \implies y = -\frac{1}{2}x - \frac{5}{6} + \frac{1}{3} \implies y = -\frac{1}{2}x - \frac{1}{2} \implies x + 2y + 1 = 0$. The intercepts are $x = -1$ and $y = -\frac{1}{2}$. Sum of absolute values: $|-1| + |-\frac{1}{2}| = 1 + 0.5 = 1.5$.The total sum is $5.5 + 1.5 = 7$.

$L_1$ and $L_2$ are two lines having slopes $2$ and $-\frac{1}{2}$ respectively. If both $L_1$ and $L_2$ are concurrent with the lines $x-y+2=0$ and $2x+y+3=0$,then the sum of the absolute values of the intercepts made by the lines $L_1$ and $L_2$ on the coordinate axes is:

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