A) 64 sq. unit
B) 54 sq. unit
C) \[\,16\pi \,\,sq.\text{ }unit\]
D) None of these
Correct Answer: A
Solution :
[a] We have, max \[\{\left| x-1 \right|,\left| y-2 \right|\}=4\] |
If \[\{\left| x-1 \right|\ge \left| y-2 \right|\},\] |
then \[\left| x-1 \right|=4,\] |
i.e., if \[(x+y-3)(x-y+1)\ge 0,\] |
Then \[x=-3or5,\] |
If \[\left| y-2 \right|\ge \left| x-1 \right|,\] |
Then \[\left| y-2 \right|=4\] |
i.e., \[(x+y-3)(x-y+1)\le 0,\] |
Then \[y=-2or6.\] |
So, the locus of P bounds a square, the equation of whose sides are \[x=-3,x=5,y=-2,y=6\] |
Thus, the area is \[{{(8)}^{2}}=64.\] |
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