To find the coordinates of the point that is equidistant from the points A(1, 2), B(3, 4), and C(5, 6), we can use the concept of the circumcenter of the triangle formed by these three points.
The circumcenter is the point where the perpendicular bisectors of the sides of a triangle intersect, and it is equidistant from all three vertices of the triangle.
First, we need to find the midpoints of at least two sides of the triangle. Let’s take the side AB and side AC:
- Midpoint of AB: MAB = ((1 + 3)/2, (2 + 4)/2) = (2, 3)
- Midpoint of AC: MAC = ((1 + 5)/2, (2 + 6)/2) = (3, 4)
Next, we can find the slopes of the sides AB and AC:
- Slope of AB: mAB = (4 – 2) / (3 – 1) = 1
- Slope of AC: mAC = (6 – 2) / (5 – 1) = 1
The slopes of the perpendicular bisectors are the negative reciprocals of these slopes:
- Perpendicular slope of AB: mpAB = -1
- Perpendicular slope of AC: mpAC = -1
Now, we can write the equations of the perpendicular bisectors:
- Equation of the perpendicular bisector of AB: y – 3 = -1(x – 2), or y = -x + 5
- Equation of the perpendicular bisector of AC: y – 4 = -1(x – 3), or y = -x + 7
To find the point of intersection of these two lines, we set them equal to each other:
-x + 5 = -x + 7
This equation will not yield a solution, indicating that points A, B, and C are collinear. Therefore, we need to use a different approach. The point equidistant from all three points on a straight line can be found by averaging the coordinates:
Coordinates of the equidistant point (P):
P = ((1 + 3 + 5)/3, (2 + 4 + 6)/3) = (3, 4)
Hence, the point that is equidistant from A(1, 2), B(3, 4), and C(5, 6) is (3, 4).