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(B) First,find the vertices of the triangle by solving the equations of the lines pairwise:$1$. Intersection of $y = x$ and $y = 2x$: $x = 2x \Rightar

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$(6, -8)$

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(B) First,find the vertices of the triangle by solving the equations of the lines pairwise:$1$. Intersection of $y = x$ and $y = 2x$: $x = 2x \Rightarrow x = 0, y = 0$. Vertex $A = (0, 0)$.$2$. Intersection of $y = x$ and $y = 3x + 4$: $x = 3x + 4$ $\Rightarrow -2x = 4$ $\Rightarrow x = -2, y = -2$. Vertex $B = (-2, -2)$.$3$. Intersection of $y = 2x$ and $y = 3x + 4$: $2x = 3x + 4 \Rightarrow x = -4, y = -8$. Vertex $C = (-4, -8)$.Let the circumcenter be $O(h, k)$. The distance from $O$ to each vertex must be equal: $OA^2 = OB^2 = OC^2$.$OA^2 = h^2 + k^2$.$OB^2 = (h + 2)^2 + (k + 2)^2 = h^2 + 4h + 4 + k^2 + 4k + 4 = h^2 + k^2 + 4h + 4k + 8$.Equating $OA^2 = OB^2$: $4h + 4k + 8 = 0 \Rightarrow h + k = -2$.$OC^2 = (h + 4)^2 + (k + 8)^2 = h^2 + 8h + 16 + k^2 + 16k + 64 = h^2 + k^2 + 8h + 16k + 80$.Equating $OA^2 = OC^2$: $8h + 16k + 80 = 0 \Rightarrow h + 2k = -10$.Subtracting the first equation from the second: $(h + 2k) - (h + k) = -10 - (-2) \Rightarrow k = -8$.Substituting $k = -8$ into $h + k = -2$: $h - 8 = -2 \Rightarrow h = 6$.Thus,the circumcenter is $(6, -8)$.

The circumcenter of the triangle formed by the lines $y = x$,$y = 2x$,and $y = 3x + 4$ is

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