$(a)\;2\qquad(b)\;4\qquad(c)\;5\qquad(d)\;7$

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- Distance between two parallel lines $ax+by+c_1=0 $ and $ax+by+c_2=0$ is $\bigg|\large\frac{c_1-c_2}{\sqrt {a^2+b^2}}\bigg|$

The given lines are

$x+y+1=0$--------(1)

$x+y-1=0$--------(2)

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

$x-y-1=0$--------(4)

Distance between the parallel lines (1) & (2) say $P_1$ is

$P_1=\large\frac{\mid -1-1\mid}{\sqrt{(1)^2+(1)^2}}$

$P_1=\sqrt 2$

Similarly distance between (3) & (4) is also $P_2=\sqrt 2$

Here $P_1=P_2$

$\therefore$ The height of the rhombus$=\sqrt 2$

Solving (1) and (3) we get the point $A(-1,0) $ and

Solving (1) and (4) we get the point $B(0,-1)$

Distance $AB=\sqrt {(1)^2+(-1)^2}=\sqrt 2$

$i.e.,$Base of the rhombus$=\sqrt 2$

Area of the rhombus=$base\times height$$=\sqrt 2\times\sqrt 2=2$

Hence (a) is the correct answer.

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