The area enclosed within the curve |x| + |y| | vedclass

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Question

(d) 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 vertic

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(d) 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 $\sqrt 2 	imes \sqrt 2 = 2sq.$units

Trick: Requird area = $\frac{{2{c^2}}}{{|ab|}} = \frac{{2 	imes {1^2}}}{{|1 	imes 1|}} = 2$.

The area enclosed within the curve $|x| + |y| = 1$ is

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