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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$|x|+|y|=1$

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.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...