Question

Triangle XYZ has vertices at X(2, 5), Y(8, 5) and Z(8, 1). Find the coordinates of the circumcenter(1 point)

Answer 1

To find the circumcenter of a triangle, we need to find the intersection of the perpendicular bisectors of its sides.

First, let's find the midpoint of each side:

Midpoint of XY:
x = (2 + 8)/2 = 5
y = (5 + 5)/2 = 5

Midpoint of YZ:
x = (8 + 8)/2 = 8
y = (5 + 1)/2 = 3

Midpoint of ZX:
x = (2 + 8)/2 = 5
y = (5 + 1)/2 = 3

Now, let's find the equations of the perpendicular bisectors using the midpoints:

Perpendicular bisector of XY:
The slope of XY is (5-5)/(8-2) = 0/6 = 0.

The negative reciprocal of the slope is undefined, so the perpendicular bisector is a vertical line passing through the midpoint (5,5), which is the equation x = 5.

Perpendicular bisector of YZ:
The slope of YZ is (1-5)/(8-8) = 4/0, which is undefined.

Since the line is vertical, the perpendicular bisector is a horizontal line passing through the midpoint (8,3), which is the equation y = 3.

Perpendicular bisector of ZX:
The slope of ZX is (3-5)/(5-2) = -2/3.

The negative reciprocal of the slope is 3/2, so the perpendicular bisector can be written as y = (3/2)x + b.

To find b, we can substitute the coordinates of the midpoint (5,3):

3 = (3/2)(5) + b
3 = 15/2 + b
b = 3 - 15/2
b = 6/2 - 15/2
b = -9/2

So, the equation of the perpendicular bisector of ZX is y = (3/2)x - 9/2.

Now let's find the coordinates of the intersection of the perpendicular bisectors:

Solving the system of equations x = 5 and y = (3/2)x - 9/2, we find:

x = 5
y = (3/2)(5) - 9/2
y = 15/2 - 9/2
y = 6/2
y = 3

Therefore, the circumcenter of triangle XYZ is (5, 3).