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The point of intersection of the lines $x + y - 1 = 0$ , $x - y - 1 = 0$ is

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(1, 2)

(0, 1)

(0, 0)

(1, 0)

answer is B.

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Detailed Solution

L_{1} = x + y - 1, L_{2} = x - y - 1 = 0

$a_{1} = 1, b_{1} = 1, c_{1} = - 1$

$a_{2} = 1, b_{2} = - 1, c_{2} = - 1$

The point intersection of $L_{1} & L_{2}$ is

$P = [\frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}, \frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}}]$

$= [\frac{1 (- 1) - (- 1) (- 1)}{1 (- 1) - (1) (1)}, \frac{- 1 (1) - (- 1) (1)}{1 (- 1) - (1 \times 1)}]$

$P = [\frac{- 1 - 1}{- 1 - 1}, \frac{- 1 + 1}{- 1 - 1}]$

$= (1, 0)$

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