Find the point in the xy plane which is equidistant from (1,1), (7,1), and (1,9).
Please and thanks.
To find the point in the xy-plane that is equidistant from three given points, we can use the concept of circumcenter. The circumcenter is the point equidistant from all three points.
Let's solve this step by step:
Step 1: Plot the three given points on the xy-plane: (1,1), (7,1), and (1,9).
Step 2: To find the circumcenter, we need to find the perpendicular bisectors for at least two pairs of line segments connecting the given points.
A perpendicular bisector is a line that cuts a line segment into two equal halves and is perpendicular to that line segment.
In this case, let's find the perpendicular bisectors for line segments connecting (1,1) and (7,1), and (1,1) and (1,9).
- Perpendicular bisector 1: (1,1) and (7,1)
The midpoint of this line segment is ((1+7)/2, (1+1)/2) = (4,1).
The slope of the line segment is (1-1)/(7-1) = 0.
The slope of the perpendicular bisector is the negative reciprocal of the slope of the line segment, which is undefined (since the slope of the line segment is 0).
Therefore, the equation of the perpendicular bisector 1 passing through (4,1) is x = 4.
- Perpendicular bisector 2: (1,1) and (1,9)
The midpoint of this line segment is ((1+1)/2, (1+9)/2) = (1, 5).
The slope of the line segment is (9-1)/(1-1) = undefined (since the denominator is 0).
The slope of the perpendicular bisector is 0.
Therefore, the equation of perpendicular bisector 2 passing through (1,5) is y = 5.
Step 3: Now we intersect the two perpendicular bisectors to find the circumcenter.
The intersection of equations x = 4 and y = 5 is the point (4,5).
Hence, the point (4,5) is the circumcenter, which is equidistant from (1,1), (7,1), and (1,9).
Note: You can also verify this by calculating the distance from the circumcenter (4,5) to each of the given points. The distance should be the same for all three points to confirm that it is equidistant.