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The circumcenter of a triangle formed by (4,2), (3,-3) and (2,2) is located at (3, -0.4).
The circumcenter of a triangle is the point of intersection of the three perpendicular bisectors of the sides of a triangle.
Let A = vertex at (4,2)
B = vertex at (3,-3)
C = vertex at (2,2)
Get the midpoint of each side.
midpoint = [(x₂ + x₁)/2 , (y₂ + y₁)/2]
side AB : midpoint = [(3 + 4)/2 , (-3 + 2)/2] = (3.5, -0.5)
side AC : midpoint = [(4 + 2)/2 , (2 + 2)/2] = (3, 2)
side BC : midpoint = [(3 + 2)/2 , (-3 + 2)/2] = (2.5, -0.5)
Get the slope of each side.
slope = (y₂ - y₁) / (x₂ - x₁)
side AB : slope = (-3 - 2) / (3 - 4) = 5
side AC : slope = (2 - 2) / (2 - 4) = 0/-2 = 0
side BC : slope = (2 - -3) / (2 - 3) = -5
Determine the equation of the line of each of the three perpendicular bisectors of the sides of a triangle, that passes through the midpoint and with a slope equal to the negative reciprocal of the slope of the side.
y = mx + b
perpendicular bisector of AB :
-0.5 = -1/5(3.5) + b ; b = 0.2 ; y = -1/5x + 0.2
perpendicular bisector of AC :
since AC is a horizontal line, x = 3
perpendicular bisector of BC :
-0.5 = 1/5(2.5) + b ; b = -1 ; y = 1/5x - 1
Since x is already equal to 3, substitute the value of x in the two other equation and solve for y.
y = -1/5x + 0.2
y = -1/5(3) + 0.2
y = -0.4
y = 1/5x - 1
y = 1/5(3) - 1
y = -0.4
Hence, the circumcenter of the triangle is located at (3, -0.4).
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