Browse Questions

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

$\begin{array}{1 1}(A)\;2\;sq.units\\(B)\;3\;sq.units\\(C)\;4\;sq.units\\(D)\;5\;sq.units\end{array}$

$|x|+|y|=1$
The curve represent four lines.
$x+y=1,\quad x-y=1, \quad -x+y=1, \quad -x-y=1$
Which enclose a square of side = Distance between opposite sides.
$x+y=1 , \quad x-y=-1$
$=> \large\frac{1+1}{\sqrt {1+1}}=\frac{2}{\sqrt 2}$$=\sqrt 2 Perpendicular distance between two parallel lines, we have, ax+by+c_1=0 ax+by+c_2=0 is given by \bigg| \large\frac{c_1-c_2}{\sqrt {a^2+b^2}}\bigg|$$=P$
Required area $=(side)^2=2\;sq.units$
Hence A is the correct answer.