Consider the [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) bound by the x-axis, the y-axis, and the line $x+y=2$. The point [equidistant](https://artofproblemsolving.com/wiki/index.php/Equidistant) from the [vertices](https://artofproblemsolving.com/wiki/index.php/Vertices) of this triangle is the [incenter](https://artofproblemsolving.com/wiki/index.php/Incenter), the [point of intersection](https://artofproblemsolving.com/wiki/index.php/Concurrency) of the [angle bisectors](https://artofproblemsolving.com/wiki/index.php/Angle_bisector) and the center of the inscribed [circle](https://artofproblemsolving.com/wiki/index.php/Circle). Now, remove the coordinate system. Let the origin be $O$, the y-intercept of the line be $A$, the x-intercept of the line be $B$, and the point be $P$.

Notice that $x$ in the diagram is what we are looking for: the [distance](https://artofproblemsolving.com/wiki/index.php/Distance) from the point to the x-axis ($OB$). Also, $OP, BP,$ and $AP$ are angle bisectors since $P$ is the incenter. $OPC\cong OPD$ by $AAS$, and $PD=OC$, since $PC||OD$, so $OC=CP=PD=DO=x$. Therefore, since $OA=OB=2$, we have $DA=CB=2-x$. Also, $CPB\cong FPB$ and $ADP\cong AFP$ by $AAS$, so $AF=FB=DA=CB=2-x$, and $AB=4-2x$. However, we know from the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) that $AB=2\sqrt{2}$. Therefore, $4-2x=2\sqrt{2}\implies x=2-\sqrt{2}, \boxed{\text{C}}$.
A point $(x, y)$ is to be chosen in the [coordinate plane](https://artofproblemsolving.com/wiki/index.php/Coordinate_plane) so that it is equally distant from the [x-axis](https://artofproblemsolving.com/wiki/index.php/X-axis), the [y-axis](https://artofproblemsolving.com/wiki/index.php/Y-axis), and the [line](https://artofproblemsolving.com/wiki/index.php/Line) $x+y=2$. Then $x$ is
$\mathrm{(A) \ }\sqrt{2}-1 \qquad \mathrm{(B) \ }\frac{1}{2} \qquad \mathrm{(C) \ } 2-\sqrt{2} \qquad \mathrm{(D) \ }1 \qquad \mathrm{(E) \ } \text{Not uniquely determined}$