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Q.
The area of the region bounded by the locus of a point $P$ satisfying $d(P, A)=4$, where $A$ is $(1,2)$ is
Straight Lines
Solution:
We have, $\max \{|x-1|,|y-2|\}=4$
If $\{|x-1| \geq|y-2|\}$
then $|x-1|=4$,
i.e., if $(x+y-3)(x-y+1) \geq 0$,
then $x=-3$ or 5
If $|y-2| \geq|x-1|$,
then $|y-2|=4$
i.e., $(x+y-3)(x-y+1) \leq 0$,
then $y=-2$ or 6 .
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 $(8)^{2}=64 Sq$. unit.