11.2.4 Circles in the Coordinate Plane (continued) Example 3 Meteorology Application
The point equidistant from three points in the coordinate plane is found in this application example. The coordinates of the three points are given. To find the point that is equidistant from all three points, the circumcenter, draw perpendicular bisectors of two of the sides of the triangle. The point of intersection of the two perpendicular bisectors with each other is the circumcenter of the triangle. To find the midpoint of two of the line segments, use the Midpoint Formula. The midpoint for SM is ⎛− + − + . The slope of SM is (2 − ( − 5))/(2 − ( − 5)) = 7/7 = 1. The perpendicular line to SM has slope − 1. Draw the line using the slope and intersection point. Repeat these calculations for MG . The point of intersection of the two lines is ( − 1, − 2). ⎝ ⎜ ⎞ ⎠ ⎟ = − − ⎛ ⎝ ⎜ M 3 2 , ⎞ ⎠ ⎟ M 5 2 2 , 5 2 2 3 2 Notice that the circumcenter of the triangle is also the center point of a circle with radius 5 that passes through all three vertices of the triangle.
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