Answer:
the circumcenter of triangle ABC is the point (3, 7/3).
Explanation:
To find the circumcenter of triangle ABC, we need to find the intersection point of the perpendicular bisectors of its sides.
First, let's find the midpoint and slope of each side of the triangle:
Side AB: midpoint = (0,1), slope = undefined (vertical line)
Side AC: midpoint = (3,2), slope = -1/3
Side BC: midpoint = (3,-1/2), slope = 3/2
Next, we need to find the equations of the perpendicular bisectors of each side. The perpendicular bisector of a segment is the line that passes through its midpoint and is perpendicular to the segment.
Perpendicular bisector of AB: x = 0 (it is a vertical line passing through the midpoint of AB)
Perpendicular bisector of AC: passes through the midpoint (3,2) and has a slope of the negative reciprocal of AC's slope, which is 3
Therefore, the equation of the perpendicular bisector of AC is y - 2 = -1/3 (x - 3), which simplifies to y = -x/3 + 8/3
Perpendicular bisector of BC: passes through the midpoint (3,-1/2) and has a slope of the negative reciprocal of BC's slope, which is -2/3
Therefore, the equation of the perpendicular bisector of BC is y + 1/2 = -2/3 (x - 3), which simplifies to y = -2x/3 + 7/2
Now we need to find the intersection point of any two of these perpendicular bisectors. We can choose any two, but it is usually easier to choose the ones that have equations in slope-intercept form, which are the perpendicular bisectors of AC and BC.
Solving the system of equations y = -x/3 + 8/3 and y = -2x/3 + 7/2, we get x = 3 and y = 7/3.
Therefore, the circumcenter of triangle ABC is the point (3, 7/3).