Answer:
Therefore, the circumcenter of triangle ABC is the point (−3/2,−5).
Explanation:
To find the circumcenter of triangle ABC, we need to find the point where the perpendicular bisectors of the sides intersect.
First, let's find the equations of the perpendicular bisectors of AB and BC.
The midpoint of AB is:
((−7−4)/2,(−8+4)/2) = (−11/2,−2)
The slope of AB is:
(4−(−8))/(−4−(−7))=12/3=4
So the slope of a line perpendicular to AB is:
−1/4
Therefore, the equation of the perpendicular bisector of AB is:
y−(−2)=−1/4(x−(−11/2))
y+2=1/4(x+11/2)
y=1/4x−(23/4)
Now let's find the midpoint and slope of BC.
The midpoint of BC is:
(−(4−7)/2,(4+4)/2) = (−3/2,4)
The slope of BC is:
(4−4)/(−4−(−7))=0/3=0
So the slope of a line perpendicular to BC is undefined.
Therefore, the equation of the perpendicular bisector of BC is simply:
x=−3/2
Now we need to find where these two lines intersect. Substituting the equation of the second line into the first, we get:
y=1/4x−(23/4)
x=−3/2
So:
y=1/4(−3/2)−(23/4)
y=−5
Therefore, the circumcenter of triangle ABC is the point (−3/2,−5).