What are the coordinates of the circumcenter of triangle ABC with vertices A (0,0), B (3,0), and C (3,2)
To find the circumcenter of a triangle, we need to find the intersection point of the perpendicular bisectors of the triangle's sides.
The midpoint of side AB is the average of the coordinates of A and B: ((0+3)/2, (0+0)/2) = (1.5, 0).
The slope of side AB is (change in y) / (change in x) = (0-0) / (3-0) = 0 / 3 = 0. Therefore, the slope of the perpendicular bisector of AB is undefined (since the slope of a line perpendicular to a line with slope 0 is undefined).
The midpoint of side AC is the average of the coordinates of A and C: ((0+3)/2, (0+2)/2) = (1.5, 1).
The slope of side AC is (change in y) / (change in x) = (2-0) / (3-0) = 2 / 3. Therefore, the slope of the perpendicular bisector of AC is -3 / 2.
Using the point-slope form of a line, we have the equations of the perpendicular bisectors:
For AB: y - 0 = undefined(x - 1.5)
=> x - 1.5 = 0
=> x = 1.5
For AC: y - 1 = (-3/2)(x - 1.5)
=> y - 1 = (-3/2)x + (9/4)
=> y = (-3/2)x + (9/4) + 1
=> y = (-3/2)x + (13/4)
To find the intersection point, we can solve the two equations simultaneously:
x = 1.5
y = (-3/2)(1.5) + (13/4)
= -4.5/2 + 13/4
= -9/4 + 13/4
= 4/4
= 1
Therefore, the circumcenter of triangle ABC is at coordinates (1.5, 1).