Circumcenter = (-1,0) The circumcenter of a triangle is the intersection of the perpendicular bisectors of the sides of the triangle. So let's calculate a couple of the bisectors and determine their intersection. Slope AB = (3 - -3)/(2 - -4) = (3+3)/(2+4) = 6/6 = 1 Perpendicular bisector will have a slope of -1 and will pass through point ((2-4)/2,(3-3)/2) = -2/2,0/2) = (-1,0) Equation is of the form y = -x + b Substitute known point 0 = -(-1) + b 0 = 1 + b -1 = b So the equation for the perpendicular bisector of AB is y = -x - 1 Now let's calculate the perpendicular bisector of BC Slope BC = (-3 - -3)/(-4 - 2) = (-3 + 3) / (-6) = 0/-6 = 0. This means that the line is horizontal and that the perpendicular bisector will be a vertical line with infinite slope. A point that line will pass through is ((-4 + 2)/2, (-3 + -3)/2) = (-2/2, 0/2) = (-1,0). So the equation for the line is: x = -1 Now we want the intersection between x = -1 and y = -x - 1 Since we know that x has to be -1, just substitute it into the 2nd equation. y = -x - 1 y = -(-1) - 1 y = 1 - 1 y = 0
So the circumcenter is (-1,0). Let's verify that. The distance from the circumcenter to each vertex of the triangle will be the same. Using the Pythagorean theorem, C^2 = A^2 + B^2. We're not going to bother taking the square root since if the squares are equal, then square roots will also be equal. Distance^2 from (2,3): (2- -1)^2 + (3-0)^2 = 3^2 + 3^2 = 9 + 9 = 18 Distance^2 from (-4,-3): (-4 - -1)^2 + (-3 - 0)^2 = -3^2 + -3^2 = 9 + 9 = 18 Distance^2 from (2,-3): (2 - -1)^2 + (-3 - 0)^2 = 3^2 + -3^2 = 9 + 9 = 18 The distances to all three vertexes is identical, so (-1,0) is verified as the circumcenter.
