A) \[\sqrt{2}\]
B) 1
C) \[\sqrt{3}\]
D) 2
Correct Answer: D
Solution :
The given lines are \[\pm x\pm y=1\] i.e. \[x+y=1,x-y=1,x+y=-1\]and \[x-y=-1\] These lines form a quadrilateral whose vertices are \[A(-1,0),B(0,-1),C(1,0)\]and \[D(0,1)\] Obviously ABCD is a square. Length of each side of this square is \[\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\] Hence area of square is \[\sqrt{2}\times \sqrt{2}=2sq.\]units Trick: Required area = \[\frac{2{{c}^{2}}}{|ab|}=\frac{2\times {{1}^{2}}}{|1\times 1|}=2\].
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