Find the circumcenter of triangleABC with A(1,6), B(1,4), and C (5,4)
To find the circumcenter of triangle ABC, we need to find the intersection point of the perpendicular bisectors of the sides AB, BC, and CA.
Let's start by finding the midpoint and the slope of the line segment AB.
The midpoint of AB is ((1+1)/2, (6+4)/2) = (1, 5).
The slope of AB is (4-6)/(1-1) = -2/0, which is undefined. This means that the line perpendicular to AB is vertical.
Next, let's find the midpoint and the slope of the line segment BC.
The midpoint of BC is ((1+5)/2, (4+4)/2) = (3, 4).
The slope of BC is (4-4)/(5-1) = 0/4 = 0. This means that the line perpendicular to BC is horizontal.
Lastly, let's find the midpoint and the slope of the line segment CA.
The midpoint of CA is ((5+1)/2, (4+6)/2) = (3, 5).
The slope of CA is (6-4)/(1-5) = 2/-4 = -1/2.
Now that we have the midpoints and slopes for each perpendicular bisector, we can find the equations of the lines.
The equation of the perpendicular bisector of AB is x = 1.
The equation of the perpendicular bisector of BC is y = 4.
The equation of the perpendicular bisector of CA can be found using the point-slope form: y - 5 = (-1/2)(x - 3).
Simplifying this equation, we get 2y - 10 = -x + 3.
Rearranging the terms, we get x + 2y = 13.
Now, we need to find the intersection point of these three lines. In this case, the intersection point is (1, 4).
Therefore, the circumcenter of triangle ABC is (1, 4).