JEE Main & Advanced Mathematics Straight Line Question Bank Critical Thinking

  • question_answer The area enclosed within the curve \[|x|+|y|=1\]is   [RPET 1990, 1997; IIT 1981; UPSEAT 2003]

    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 = \[\frac{2{{c}^{2}}}{|ab|}=\frac{2\times {{1}^{2}}}{|1\times 1|}=2\].


You need to login to perform this action.
You will be redirected in 3 sec spinner