Find the point in the xy plane which is equidistant from (1,1), (7,1), and (1,9).
Please and thanks.
the desired point will be the center of the circle passing through the three points. So, if that is (h,k), we have
(x-1)^2 + (y-1)^2 = r^2
(x-7)^2 + (y-1)^2 = r^2
(x-1)^2 + (y-9)^2 = r^2
Now, you can solve those to find (h,k) and r, or you can note that the three points form a right triangle, whose hypotenuse is the line joining (1,9) and (7,1).
A right triangle's hypotenuse is the diameter of the circumscribed circle. So, that means that the center of the circle is the midpoint of that segment: (4,5).
Now all we have to do is find the distance from (4,5) to any of the points. The distance from (4,5) to (1,1) is 5, so our circle is
(x-4)^2 + (y-5)^2 = 25
thank you Steve
To find the point in the xy-plane that is equidistant from three given points, we can use the concept of circumcenter.
First, let's label the given points:
A(1, 1), B(7, 1), and C(1, 9).
1. Calculate the midpoints of any two sides of the triangle formed by the three points. In this case, let's find the midpoints of AB and BC.
The midpoint of AB, M1, can be found by averaging the x-coordinates and y-coordinates of points A and B:
M1 = ((1+7)/2, (1+1)/2) = (4, 1).
The midpoint of BC, M2, can be found by averaging the x-coordinates and y-coordinates of points B and C:
M2 = ((7+1)/2, (1+9)/2) = (4, 5).
2. Compute the slopes of lines AB and BC. The slope of a line passing through two points (x1, y1) and (x2, y2) can be calculated using the formula:
slope = (y2 - y1) / (x2 - x1).
The slope of line AB, m1, can be calculated using points A(1, 1) and B(7, 1):
m1 = (1 - 1) / (7 - 1) = 0 / 6 = 0.
The slope of line BC, m2, can be calculated using points B(7, 1) and C(1, 9):
m2 = (9 - 1) / (1 - 7) = 8 / -6 = -4/3.
3. Find the equations of the perpendicular bisectors of AB and BC.
The perpendicular bisector of AB passes through the midpoint M1(4, 1) with a slope perpendicular to m1.
The equation of the perpendicular bisector of AB can be written as:
y - y1 = -(1/m1)(x - x1).
Since m1 = 0, the equation simplifies to: x = 4.
The perpendicular bisector of BC passes through the midpoint M2(4, 5) with a slope perpendicular to m2.
The equation of the perpendicular bisector of BC can be written as:
y - y2 = -(1/m2)(x - x2).
Since m2 = -4/3, the equation simplifies to: y - 5 = (3/4)(x - 4).
4. Calculate the intersection point of the two perpendicular bisectors.
To find the point equidistant from A, B, and C, we need to find the intersection point of the equations x = 4 and y - 5 = (3/4)(x - 4).
Substitute the x-coordinate of 4 into the equation y - 5 = (3/4)(x - 4):
y - 5 = (3/4)(4 - 4) = 0.
Therefore, the intersection point is (4, 5).
Hence, the point (4, 5) in the xy-plane is equidistant from the points (1, 1), (7, 1), and (1, 9).