Find the circumcenter of ΔABC
with A(1, 6), B(1, 4), and C(5, 4).(1 point)
Responses
(1, 4)
(1, 4)
(5, 4)
(5, 4)
(3, 5)
(3, 5)
(1, 6)
The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect.
To find the perpendicular bisectors, we first need to find the slopes of the sides.
The slope of side AB can be found using the formula:
m = (y2 - y1)/(x2 - x1)
mAB = (4 - 6)/(1 - 1) = -2/0
Since the denominator is 0, the slope is undefined. This means that side AB is a vertical line.
The midpoint of AB can be found by averaging the x-coordinates and y-coordinates of A and B respectively:
x = (x1 + x2)/2 = (1 + 1)/2 = 1
y = (y1 + y2)/2 = (6 + 4)/2 = 5
So the midpoint of AB is (1, 5).
The perpendicular bisector of AB will be a horizontal line passing through the midpoint (1, 5).
Since the line is horizontal, the slope of the perpendicular bisector will be 0.
So the equation of the perpendicular bisector of AB is:
y - 5 = 0
y = 5
Now let's find the perpendicular bisector of side BC.
The slope of side BC can be found using the formula:
mBC = (4 - 4)/(5 - 1) = 0
Since the slope is 0, side BC is a horizontal line.
The midpoint of BC is (3, 4).
The equation of the perpendicular bisector of BC passing through the midpoint (3, 4) is:
x - 3 = 0
x = 3
Finally, let's find the perpendicular bisector of side AC.
The slope of side AC can be found using the formula:
mAC = (4 - 6)/(5 - 1) = -2/4 = -1/2
The midpoint of AC can be found by averaging the x-coordinates and y-coordinates of A and C respectively:
x = (x1 + x2)/2 = (1 + 5)/2 = 6/2 = 3
y = (y1 + y2)/2 = (6 + 4)/2 = 10/2 = 5
So the midpoint of AC is (3, 5).
The slope of the perpendicular bisector is the negative reciprocal of the slope of AC:
m_perpendicular = -1/mAC = -1/(-1/2) = 2
Using the point-slope form, the equation of the perpendicular bisector of AC passing through the midpoint (3, 5) is:
y - 5 = 2(x - 3)
Simplifying the equation, we get:
2x - 6 - y + 5 = 0
2x - y - 1 = 0
-y = -2x + 1
y = 2x - 1
To find the circumcenter, we need to find the intersection point of the perpendicular bisectors.
Since the equations of the perpendicular bisectors of AB and BC are both horizontal lines (y = 5 and x = 3 respectively), they intersect at (3, 5).
Therefore, the circumcenter of triangle ABC is (3, 5).
The correct answer is (3, 5).