Prove that the lines $3 x + y - 14 = 0$ and $3 x - 8 y + 4 = 0$ are concurrent.

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Solution

Given lines are $3 x + y - 14 = 0, x - 2 y + 0 = 0$ and $3 x - 8 y + 4 = 0$ .
We have,
$∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 & - 14 1 & - 2 & 0 3 & - 8 & 4 \end{matrix} ∣ ∣ ∣ ∣ = 3 (- 8 + 0) - 1 (4 - 0) - 14 (- 8 + 6) = - 24 - 4 + 28 = 0$

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