Given two straight lines $x + y - 7 = 0$ and $x - y + 3 = 0$ . Find point of intersection of them.

(2, 5)

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(- 2, - 5)

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(2, - 5)

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(- 2, 5)

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Solution

The correct option is A $(2, 5)$
$x + y - 7 = 0....... e q 1 x - y + 3 = 0...... e q 2$
Add both the equations
Thus we get, $2 x - 4 = 0 2 x = 4 x = \frac{4}{2} x = 2 2 + y - 7 = 0 y = 7 - 2 = 5 (x, y) = (2, 5)$

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Given two straight lines x+y−7=0 and x−y+3=0 . Find point of intersection of them.

The correct option is A (2,5)x+y−7=0.......eq1x−y+3=0......eq2Add both the equationsThus we get, 2x−4=02x=4x=42x=22+y−7=0y=7−2=5(x,y)=(2,5)

Given two straight lines $x + y - 7 = 0$ and $x - y + 3 = 0$ . Find point of intersection of them.

The correct option is A $(2, 5)$
$x + y - 7 = 0....... e q 1 x - y + 3 = 0...... e q 2$
Add both the equations
Thus we get, $2 x - 4 = 0 2 x = 4 x = \frac{4}{2} x = 2 2 + y - 7 = 0 y = 7 - 2 = 5 (x, y) = (2, 5)$