SOLUTION: It is true in general, that the perpendicular bisectors of the sides of a triangle are concurrent and that the point of intersection is the centre of the circle through all three v

Question 826407: It is true in general, that the perpendicular bisectors of the sides of a triangle are concurrent and that the point of intersection is the centre of the circle through all three vertices. Prove this result by placing the vertices at A(2a,0), B(-2a,0) and C(2b,2c) and proceeding as follows:
a) Find the gradients of AB,BC and CA, and hence find the equations of the three perpendicular bisectors.
While I do know how to do the rest of the steps (that's why I didn't write it down), I seem to be having trouble with the first step (written above). Would someone be able to help me with this particular question?
Thanks!!
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
For each line, the gradient (we call it slope) is the differences in y-coordinates divided by the difference in x-coordinates.

Segment AB is part of the horizontal line , the x-axis.
Its midpoint is (0,0) , the origin.
The perpendicular bisector to AB is the line perpendicular to the x-axis through the origin. That is the line , the y-axis.
All that is so by design (clever choice of coordinates, placing the triangle so it would be so).

The other two perpendicular bisectors are not so easy to find.

The gradient/slope of AC is
.
A perpendicular line would have a gradient/slope of

The midpoint of AC has

With that we can write the equation for the perpendicular bisector to AC in point-slope form as

The intersection of that line and the line has (of course) and

Similarly,
the midpoint of BC has

The gradient of BC is
.
A perpendicular line would have a gradient/slope of

With that we can write the equation for the perpendicular bisector to AC in point-slope form as

So the intersection of that line and has (of course) and

So all 3 perpendicular bisectors pass through the point with
and .
Since all points on the perpendicular bisector of a segment are equidistant from the ends of the segment, the point above should be at the same distance from A, B, and C, so a circle centered at that point, with that equal distance as its radius would pass through A, B, and C.
Its radius can be calculated as the distance from that point to A(2a,0)

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