Question

what are the coordinates of the circumcenter of \\( \triangle abc \\)?

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Explanation:

Step1: Identify the type of triangle

Since \(AB\) is horizontal (\(y = - 4\) for points \(A(-5,-4)\) and \(B(3,-4)\)), and the circum - center of an isosceles triangle (where \(AC = BC\)) lies on the perpendicular bisector of the base.

Step2: Find the perpendicular bisector of \(AB\)

The mid - point of \(AB\) is calculated using the mid - point formula \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). For \(A(-5,-4)\) and \(B(3,-4)\), \(x=\frac{-5 + 3}{2}=-1\), \(y=\frac{-4-4}{2}=-4\). The slope of \(AB\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-4+4}{3 + 5}=0\), so the perpendicular bisector of \(AB\) is a vertical line \(x=-1\).

Step3: Check the symmetry

Since the triangle is symmetric about the line \(x = - 1\) (because \(A(-5,-4)\) and \(B(3,-4)\) are symmetric about \(x=-1\) and \(C(0,6)\) also satisfies the symmetry with respect to the perpendicular bisector of \(AB\)), the circum - center lies on the perpendicular bisector of \(AB\).

Answer:

\((-1,-1)\)