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(C) Let the vertices of the square be $A(a, 0)$,$B(1, y_B)$,$C(b, 0)$,and $D(1, 2)$.Since $AC$ is a diagonal on the $x$-axis,the other diagonal $BD$ m

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$x - y + 1 = 0, x + y - 3 = 0$

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(C) Let the vertices of the square be $A(a, 0)$,$B(1, y_B)$,$C(b, 0)$,and $D(1, 2)$.Since $AC$ is a diagonal on the $x$-axis,the other diagonal $BD$ must be a vertical line $x = 1$ because the diagonals of a square are perpendicular bisectors.The intersection of diagonals is $E(1, 0)$.Since $E$ is the midpoint of $AC$,$\frac{a+b}{2} = 1 \implies a+b = 2$.Also,the distance $AE = EC = ED = EB$. $AE = |1-a|$,$EC = |b-1|$,$ED = 2$.Since $AE = ED$,$|1-a| = 2 \implies 1-a = 2$ or $1-a = -2$.If $1-a = 2$,$a = -1$,then $b = 3$. If $1-a = -2$,$a = 3$,then $b = -1$.Taking $A(-1, 0)$ and $C(3, 0)$,the sides passing through $D(1, 2)$ are $AD$ and $CD$.Slope of $AD = \frac{2-0}{1-(-1)} = \frac{2}{2} = 1$. Equation: $y - 2 = 1(x - 1) \implies x - y + 1 = 0$.Slope of $CD = \frac{2-0}{1-3} = \frac{2}{-2} = -1$. Equation: $y - 2 = -1(x - 1) \implies x + y - 3 = 0$.

One diagonal of a square is the $x$-axis. If one vertex of the square is $(1, 2)$,find the equations of the sides passing through this vertex.

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