The vertices A, B and C of a triangle are (2, 1), (5, 2) and (3, 4), respectively. What are the coordinates of circumcentre?
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(3.25, 2.25)
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(2, 6)
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(6.5, 4.5)
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(13, 9)
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None of the above
Circumcentre (x, y) is equidistant from (2, 1), (5, 2), (3, 4). (x-2)^2 + (y-1)^2 = (x-5)^2 + (y-2)^2 => -4x + 4 - 2y + 1 = -10x + 25 - 4y + 4 => 6x + 2y = 24 => 3x + y = 12. (x-2)^2 + (y-1)^2 = (x-3)^2 + (y-4)^2 => -4x + 4 - 2y + 1 = -6x + 9 - 8y + 16 => 2x + 6y = 20 => x + 3y = 10. Solving 3x+y=12 and x+3y=10: x = 3.25, y = 2.25.
The circumcentre is the intersection of the perpendicular bisectors of the triangle's sides. The midpoint of segment AB is ((2+5)/2, (1+2)/2), which is (3.5, 1.5), and the slope of AB is (2-1)/(5-2), which is 1/3. The equation of the perpendicular bisector of AB is y - 1.5 = -3(x - 3.5). The midpoint of segment BC is (4, 3) and the slope of BC is (4-2)/(3-5), which is -1. The perpendicular bisector of BC is y - 3 = 1(x - 4), simplifying to y = x - 1. Substituting x - 1 into the first equation gives x - 2.5 = -3x + 10.5, yielding x = 3.25, and substituting this into the second equation gives y = 2.25. The coordinates are (3.25, 2.25).