Question

what are the coordinates of the circumcenter of this triangle? enter your answer by filling in the boxes. (\boxed{ }, \boxed{ })

Explanation:

Step1: Identify triangle vertices

From the graph, vertices are \( A(-2, 5) \), \( B(-2, -1) \), \( C(8, -1) \).

Step2: Find perpendicular bisectors

• For \( AB \) (vertical line \( x = -2 \)): midpoint is \( (-2, 2) \), perpendicular bisector is horizontal line (since \( AB \) is vertical), so \( y = 2 \)? Wait, no—wait, \( AB \) is vertical (same \( x \)-coordinate), so its perpendicular bisector is horizontal, passing through midpoint \( (-2, \frac{5 + (-1)}{2}) = (-2, 2) \)? Wait, no, wait \( B \) is at \( (-2, -1) \), \( A \) at \( (-2, 5) \). So midpoint of \( AB \) is \( (-2, \frac{5 + (-1)}{2}) = (-2, 2) \). The perpendicular bisector of \( AB \) (vertical segment) is horizontal line \( y = 2 \)? Wait, no, wait \( AB \) is vertical, so perpendicular is horizontal, so slope 0. So equation \( y = 2 \)? Wait, no, wait \( BC \): \( B(-2, -1) \), \( C(8, -1) \) is horizontal (same \( y \)-coordinate). Midpoint of \( BC \) is \( (\frac{-2 + 8}{2}, -1) = (3, -1) \). Perpendicular bisector of \( BC \) (horizontal segment) is vertical line \( x = 3 \)? Wait, no, wait \( BC \) is horizontal, so perpendicular is vertical, so equation \( x = 3 \)? Wait, no, wait the circumcenter is the intersection of perpendicular bisectors. Wait, \( AB \) is vertical (x=-2), \( BC \) is horizontal (y=-1). Wait, no, \( A(-2,5) \), \( B(-2,-1) \): vertical line. \( B(-2,-1) \), \( C(8,-1) \): horizontal line. So triangle is right-angled at \( B \)? Wait, \( AB \) vertical, \( BC \) horizontal, so angle at \( B \) is right angle. Then circumcenter of right triangle is midpoint of hypotenuse \( AC \).

Step3: Midpoint of hypotenuse \( AC \)

\( A(-2, 5) \), \( C(8, -1) \). Midpoint formula: \( (\frac{-2 + 8}{2}, \frac{5 + (-1)}{2}) = (\frac{6}{2}, \frac{4}{2}) = (3, 2) \). Wait, but earlier mistake: \( AB \) is vertical (x=-2), \( BC \) is horizontal (y=-1), so right angle at \( B \), so hypotenuse is \( AC \), circumcenter is midpoint of \( AC \).

Wait, let's check: perpendicular bisector of \( AB \): \( AB \) is vertical, so perpendicular bisector is horizontal line through midpoint of \( AB \): midpoint of \( AB \) is (-2, 2), so horizontal line \( y = 2 \). Perpendicular bisector of \( BC \): \( BC \) is horizontal, midpoint (3, -1), so vertical line \( x = 3 \). Intersection of \( x = 3 \) and \( y = 2 \) is (3, 2). Wait, but midpoint of \( AC \) is (3, 2), which matches. So that's the circumcenter.

Answer:

(3, 2)