The lines $x - y - 2 = 0, x + y - 4 = 0, x + 3 y = 6$ meet at the point

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(3, 1)

(1, 2)

(1, 1)

(2, 2)

answer is C.

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Detailed Solution

L_{1} = x - y - 2 = 0 \to (1)

$L_{2} = x + y - 4 = 0 \to (2)$

$L_{3} = x + 3 y - 6 = 0 \to (3)$

$(1) + (2)$

$\begin{matrix} x - y - 2 = 0 \\ \underset{________________}{x + y - 4 = 0} \\ 2 x - 6 = 0 \end{matrix}$

$x = 3$

$\Rightarrow 3 - y - 2 = 0$

$\Rightarrow y = 1$

$L_{1} & L_{2}$ are meet at point $(3, 1)$

$(3) - (2)$

$\begin{matrix} x + 3 y - 6 = 0 \\ \underset{_____________________}{{}_{-}x - {}_{+}y - {}_{+}2 = 0} \\ 4 y - 4 = 0 \end{matrix}$

$y = 1$

$x + 3 (1) - 6 = 0$

$x = 3$

$∴$ $L_{2} & L_{3}$ are meet at point $(3, 1)$

$∴$ $L_{1}, L_{2}, L_{3}$ are meet at point $(3, 1)$

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