the circumcentre lies on the intersection of the right-bisectors of any two chords.
I noticed that AB forms a horizontal line, so that one will be easy.
midpoint is (1/2 , -1) , so the equation of the right-bisector will be x = 1/2
I also noticed that AC is a vertical line, so the equation of the right-bisector of AC is y = -9/2
circumcentre is (1/2 , -9/2)
That was simple since Triangle ABC was right-angled with vertical and horizontal sides
Had I seen that right away, I could have simply taken the midpoint of the hypotenuse, which would have been
( ((-2+3)/2 , (-1-8)/2 )
= (1/2 , -9/2)
find the coordinates of the circumcenter of tringleABC
A(3,-1),B(-2,-1),C(3,-8)
1 answer
1 answer