<span>(3.5, 3) is the circumcenter of triangle ABC.
The circumcenter of a triangle is the intersection of the perpendicular bisectors of each side. All three of these perpendicular bisectors will intersect at the same point. So you have a nice self check to make sure your math is correct. Now let's calculate the equation for these bisectors.
Line segment AB:
Slope
(4-2)/(1-1) = 2/0 = infinity.
This line segment is perfectly vertical. So the bisector will be perfectly horizontal, and will pass through ((1+1)/2, (4+2)/2) = (2/2, 6/2) = (1,3).
So the equation for this perpendicular bisector is y = 3.
Line segment BC
(2-2)/(6-1) = 0/5 = 0
This line segment is perfectly horizontal. So the bisector will be perfectly vertical, and will pass through ((1+6)/2,(2+2)/2) = (7/2, 4/2) = (3.5, 2)
So the equation for this perpendicular bisector is x=3.5
So those two bisectors will intersect at point (3.5,3) which is the circumcenter of triangle ABC.
Now let's do a cross check to make sure that's correct.
Line segment AC
Slope = (4-2)/(1-6) = 2/-5 = -2/5
The perpendicular will have slope 5/2 = 2.5. So the equation is of the form
y = 2.5*x + b
And will pass through the point
((1+6)/2, (4+2)/2) = (7/2, 6/2) = (3.5, 3)
Plug in those coordinates and calculate b.
y = 2.5x + b
3 = 2.5*3.5 + b
3 = 8.75 + b
-5.75 = b
So the equation for the 3rd bisector is
y = 2.5x - 5.75
Now let's check if the intersection with this line against the other 2 works.
Determining intersection between bisector of AC and AB
y = 2.5x - 5.75
y = 3
3 = 2.5x - 5.75
8.75 = 2.5x
3.5 = x
And we get the correct value. Now to check AC and BC
y = 2.5x - 5.75
x = 3.5
y = 2.5*3.5 - 5.75
y = 8.75 - 5.75
y = 3
And we still get the correct intersection.</span>
<span>arccos (cos pi/2)=? </span>cos pi/2=0, arccos (0)=?, we know that cos pi/2=0, so cos^-1 (o)= Pi/2, but cos^-1 (o)=arccos (0), so arccos (cos pi/2)=Pi/2
