what are the coordinates of the circumcenter of \\( \\triangle abc \\)?\nenter your answer in the boxes.\n(…

Explanation:

Step1: Recall the property of circum - center for a triangle

The circum - center of a triangle is the intersection of the perpendicular bisectors of the sides of the triangle. For a triangle with a horizontal side (like side (AB) in (\triangle ABC)), the perpendicular bisector of a horizontal line segment (AB) is a vertical line passing through the mid - point of (AB).

Step2: Find the mid - point of side (AB)

The formula for the mid - point (M) of two points (A(x_1,y_1)) and (B(x_2,y_2)) is (M=\left(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2}\right)). If (A(-5,-4)) and (B(2,-4)) (from the graph, by counting the grid units), then (x_1=-5,x_2 = 2,y_1=-4,y_2=-4). The (x) - coordinate of the mid - point of (AB) is (\frac{-5 + 2}{2}=\frac{-3}{2}=-1.5), and the (y) - coordinate of the mid - point of (AB) is (\frac{-4+( - 4)}{2}=-4). Since (AB) is horizontal ((y=-4) for both (A) and (B)), the perpendicular bisector of (AB) is the vertical line (x=-1.5). Since the circum - center lies on the perpendicular bisector of (AB), and by symmetry (or by checking the perpendicular bisectors of other sides, but for an isosceles triangle with (AB) as the base and (C) as the vertex, the circum - center lies on the perpendicular bisector of (AB)), the (y) - coordinate of the circum - center is the same as the (y) - coordinate of the mid - point of (AB) (because of the symmetry of the triangle with respect to the perpendicular bisector of (AB)).

Answer:

((-1.5,-4))