Chapter 5.1, Problem 45PPS

To determine

To find: The coordinates of the circum center of the triangle.

Answer to Problem 45PPS

The coordinates of the circum center of the triangle is $\left(5,3\right)$ .

Explanation of Solution

Given information:

The given vertices are $A\left(0,0\right)$ , $B\left(0,6\right)$ , and $C\left(10,0\right)$ .

Calculation:

The perpendicular bisector of the triangle intersects at a point called the circum center that is equidistance from the vertices of the triangle.

The graph for the points $A\left(0,0\right)$ , $B\left(0,6\right)$ , and $C\left(10,0\right)$ is shown in figure (1).

  

Figure (1)

  $\overline{AC}$ is horizontal, and $AC=10$ . So, the perpendicular bisector of $\overline{AC}$ is vertical, and through $\left(0,5\right)$ . The linear equation is $x=5$ .

  $\overline{AB}$ is vertical, and $AB=6$ . So, the perpendicular bisector of $\overline{AB}$ is horizontal, and through $\left(0,3\right)$ . The linear equation is $y=3$ .

Circum center theorem shows the circum center is the intersection point of $x=5$ and $y=3$ .

Therefore, the coordinates of the circum center of the triangle is $\left(5,3\right)$ .