The coordinates of the circumcenter of triangle ABC with the given vertices are (-2.5, -4.5).
In order to determine the coordinates of the circumcenter of triangle ABC with the given vertices, let its circumcenter be P(x, y). Thus, PA = PB = PC.
By squaring all sides of triangle ABC, we have:
PA² = PB² = PC²
From PA² = PB², we have:
(x - 3)² + (y - (-1))² = (x - (-3))² + (y - (-11))²
(x - 3)² + (y + 1)² = (x + 3)² + (y + 11)²
x² - 6x + 9 + y² + 2y + 1 = x² + 6x + 9 + y² + 22y + 121
-12x - 20y = 120
-3x - 5y = 30 ..........equation 1.
From PB² = PC², we have:
(x - (-3))² + (y - (-11))² = (x - 3)² + (y - (-8))²
(x + 3)² + (y + 11)² = (x - 3)² + (y + 8)²
x² + 6x + 9 + y² + 22y + 121 = x² - 6x + 9 + y² + 16y + 64
12x + 6y = -57 ..........equation 2.
Solving eqn. 1 and eqn. 2 simultaneously, we have:
x = -2.5 and y = -4.5.
Therefore, the coordinates of the circumcenter of triangle ABC with the given vertices are (-2.5, -4.5).
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