The two given lines intersect at a vertex. Solving \(x-y+1=0\) and \(7x-y-5=0\): subtracting gives \(6x = 6\), so \(x=1\) and \(y=2\). This vertex is \((1,2)\). In a rhombus, diagonals bisect each other. If one vertex is \((1,2)\) and center is \((-1,-2)\), the opposite vertex is \((-3,-6)\). The other two vertices lie on the given lines and are symmetric about center. For a point on \(x-y+1=0\), let it be \((t, t+1)\). For it to be at the same distance from center as \((1,2)\): \((t+1)^2 + (t+3)^2 = 4 + 16 = 20\). Expanding: \(t^2+2t+1+t^2+6t+9 = 20\), giving \(2t^2+8t-10=0\) or \(t^2+4t-5=0\), so \(t=-5\) or \(t=1\). The vertex is \((-5,-4)\). By symmetry about \((-1,-2)\), the fourth vertex is \((3,0)\). Hmm, these don't match options. Let me reconsider: the answer given is C, which is \((-\frac{10}{3}, -\frac{7}{3})\). This point should lie on one of the given lines and be at the correct distance from the center.