If 3x + 6y + 2 = 0,x + y + 1 = 0,and 2x - y + | vedclass

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(B) Given the lines: $L_1: 3x + 6y + 2 = 0$ $L_2: x + y + 1 = 0$ $L_3: 2x - y + 3 = 0$ Check if the point $P\left(\frac{-4}{3}, \frac{1}{3}\right)$ sa

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the point of concurrence of the lines

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(B) Given the lines: $L_1: 3x + 6y + 2 = 0$ $L_2: x + y + 1 = 0$ $L_3: 2x - y + 3 = 0$ Check if the point $P\left(\frac{-4}{3}, \frac{1}{3}\right)$ satisfies all three equations: For $L_1$: $3\left(\frac{-4}{3}\right) + 6\left(\frac{1}{3}\right) + 2 = -4 + 2 + 2 = 0$. For $L_2$: $\left(\frac{-4}{3}\right) + \left(\frac{1}{3}\right) + 1 = -1 + 1 = 0$. For $L_3$: $2\left(\frac{-4}{3}\right) - \left(\frac{1}{3}\right) + 3 = \frac{-8-1+9}{3} = 0$. Since the point satisfies all three equations,it is the point of concurrence of the three lines.

If $3x + 6y + 2 = 0$,$x + y + 1 = 0$,and $2x - y + 3 = 0$ are three given lines,then the point $\left(\frac{-4}{3}, \frac{1}{3}\right)$ is

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