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If $3 x+6 y+2=0, x+y+1=0,2 x-y+3=0$ are three given lines then the point $\left(\frac{-4}{3}, \frac{1}{3}\right)$ is
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Verified Answer
The correct answer is:
the point of concurrence of the lines
Here we are given that
$$
\begin{aligned}
& 3 x+6 y+2=0 \\
& x+y+1=0 \\
& 2 x-y+3=0
\end{aligned}
$$
$\because \quad$ Here $\left(\frac{-4}{3}, \frac{1}{3}\right)$ satisfied all of these three lines
$\Rightarrow\left(\frac{-4}{3}, \frac{1}{3}\right)$ is the concurrent point of given three lines.
$$
\begin{aligned}
& 3 x+6 y+2=0 \\
& x+y+1=0 \\
& 2 x-y+3=0
\end{aligned}
$$
$\because \quad$ Here $\left(\frac{-4}{3}, \frac{1}{3}\right)$ satisfied all of these three lines
$\Rightarrow\left(\frac{-4}{3}, \frac{1}{3}\right)$ is the concurrent point of given three lines.
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