Answer:
⊥ to AB : x +y = 6
⊥ to AC : x -3y = 0
⊥ to BC : x +3y = 9
see the attachment for a graph
Explanation:
You want the equations and graphs of the perpendicular bisectors of AB, BC, and AC where the points are A(1, 2), B(4, 5), and C(2, -1). You want to relate this to finding the center of a circle through three non-collinear points.
Perpendicular bisector
The perpendicular bisector of a segment PQ with P(x1, y1) and Q(x2, y2) can be written as ...
(x2-x1)(x -(x1+x2)/2) +(y2-y1)(y -(y1+y2)/2) = 0
(a) Equations
Applying this to the given segments, we have ...
⊥ to AB : (4 -1)(x -(4+1)/2) +(5 -2)(y -(5+2)/2) = 0 ⇒ x +y = 6
⊥ to AC : (2 -1)(x -(2+1)/2) +(-1 -2)(y -(-1+2)/2) = 0 ⇒ x -3y = 0
⊥ to BC : (2 -4)(x -(2+4)/2) +(-1 -5)(y -(-1+5)/2) = 0 ⇒ x +3y = 9
(b) Graph
The attachment shows a graph of the lines and their single point of intersection.
Each perpendicular bisector is the locus of the centers of circles through the end points of the respective segments. Then the intersection of those bisectors is the center of a circle through all of the points.
Solving any pair of these equations simultaneously gives the circle center as (4.5, 1.5). This is point D on the graph.
(c) Discussion
The given points are non-collinear, and the intersection of the perpendicular bisectors is the center of a circle through them. The method described in this problem is a suitable method for finding the circle center. (If the three points are considered a triangle, this center is called the "circumcenter.")
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Additional comment
We have written the equations for the lines in standard form, with a positive leading coefficient and mutually prime integers. You will notice the graphing program wrote the same equations for those lines.