Question 837930: Coordinate geometry proof:
Prove- All 3 perpendicular bisectors of a triangle bisect at the same point.
I drew a picture and labeled it A, B, C. I did midpoints for AB(a, b), BC(2a, b) , and AC(a, 0). What else would I have to do? What about a conclusion statement?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! You should first specify the coordinates of A, B, C are.
Without loss of generality, let A (0,0), B (1,0), and C (m,n) (it is safe to make these assumptions because we can shift/scale/rotate the triangle). Let us find the intersection of the perpendicular bisectors of AB and AC.
The equation for the perpendicular bisector of AB is given by x = 1/2, and for AC it is (x - \frac{m}{2})) , which may be written as x + \frac{m^2 + n^2}{2n}) (this can be found by using point-slope form). Substitute x = 1/2 to get
 + \frac{m^2 + n^2}{2n} = \frac{m^2 + n^2 - m}{2n})
So these two perpendicular bisectors intersect at ) . Now all we need to do is show that this point is on the line formed by the perpendicular bisector of AC. The slope of AC is  so we can show that the slope of the line connecting O and the midpoint of BC is  .
The midpoint of BC is ) so the corresponding slope is
Hence we have shown that the line connecting O and the midpoint of BC is perpendicular to BC, so the perpendicular bisector of BC must contain O. Therefore all three perpendicular bisectors are concurrent. They meet at the circumcenter of the triangle (the center of the circumscribed circle).
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