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Bisect two arcs above and below the given points or line and the perpendicular of these arcs cuts through the midpoint.

Q: How do you find a point which is equidistant from a given pair of points?

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It is the circumcentre of the triangle formed by the three points. Draw the perpendicular bisectors of two of the lines joining the three points. They will meet at the point that is equidistant from the three points.

The equidistant point of a straight line is the middle. Measure the distance from one end to the other and half it.

A plane is the set of all points in 3-D space equidistant from two points, A and B. If it will help to see it, the set of all points in a plane that are equidistant from points A and B in the plane will be a line. Extend that thinking off the plane and you'll have another plane perpendicular to the original plane, the one with A and B in it. And the question specified that A and B were in 3-D space. Another way to look at is to look at a line segment between A and B. Find the midpoint of that line segment, and then draw a plane perpendicular to the line segment, specifying that that plane also includes the midpoint of the line segment AB. Same thing. The set of all points that make up that plane will be equidistant from A and B. At the risk of running it into the ground, given a line segment AB, if the line segment is bisected by a plane perpendicular to the line segment, it (the plane) will contain the set of all points equidistant from A and B.

-- Draw a line segment from one point to the other.-- Construct the perpendicular bisector of the line segment..-- Every point on the perpendicular bisector of the line segmentis equidistant from the two original points.==========================================================================Whereupon the first contributor observed:Yes, that works, no matter how you set the compass, as long as it's more than 1/2 the distancebetween the two points. Every setting of the compass will give you a pair of points that areequal distances from the original two. As you find more and more of them ... with differentsettings of the compass ... you'll see that all the equal-distance points you're finding all lieon the same straight line. That line is the perpendicular bisector of the line between the twooriginal points, just as we described up above.

Infinitely many. There are infinitely many points in the plane and although any pair of points define a line, no matter how many lines you are given, it is always possible to find a point that is not on any of them - that is, a point that is not collinear.

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In spherical geometry we look at the globe as the sphere S^2. Any plane intersecting the sphere will create a great circle. Now if you take any point on the globe and reflect it across that plane, you have another point that is equidistant from the plane. The sets of all these points will be equidistant from the great circle.

To find a point equidistant from three other points, construct perpendicular bisectors for two of the segments formed from three points. Note: this will be the center of the circle that has all three points on it's circumference. Three points, not in a straight line, form three pairs of points with each pair defining a different line. Take any pair of points and draw the perpendicular bisector of the line joining them. Repeat for one of the other pairs. These two perpendicular bisectors will meet at the point which is equidistant from all three points - the circumcenter of the triangle formed by the three points.

It is the circumcentre of the triangle formed by the three points. Draw the perpendicular bisectors of two of the lines joining the three points. They will meet at the point that is equidistant from the three points.

The equidistant point of a straight line is the middle. Measure the distance from one end to the other and half it.

It takes 3 non collinear points to define one specific circle. With only two points an infinite number of circles can be drawn. Proof: Given two points A, B draw the line between them. Then find the perpendicular bisector of the line AB. Any point on the perpendicular bisector is equidistant from the two original points, A and B. A circle with center C and radius AC will then pass through points A and B. There are infinite point C's on the perpendicular bisector so there are infinite circles. Given three points A, B and D you can find the perpendicular bisector for line segements AB and then the perpendicular bisector fof line segment BC. The two perpedicular bisectors will not be parallel because the points A, B and D are non collinear. This means the two perpeniducar bisectors will intercept at only one point C(like any two intercepting lines). This point C is equidistant from points A, B, and D. A circle with center C and radius AC will then pass through all three of the points. Since there is only one point C that lies on both perpendicular bisectors, there is only one circle possible.

A plane is the set of all points in 3-D space equidistant from two points, A and B. If it will help to see it, the set of all points in a plane that are equidistant from points A and B in the plane will be a line. Extend that thinking off the plane and you'll have another plane perpendicular to the original plane, the one with A and B in it. And the question specified that A and B were in 3-D space. Another way to look at is to look at a line segment between A and B. Find the midpoint of that line segment, and then draw a plane perpendicular to the line segment, specifying that that plane also includes the midpoint of the line segment AB. Same thing. The set of all points that make up that plane will be equidistant from A and B. At the risk of running it into the ground, given a line segment AB, if the line segment is bisected by a plane perpendicular to the line segment, it (the plane) will contain the set of all points equidistant from A and B.

To find the slope (steepness, not height) of a line when given two points, do the following: Slope = (y2-y1)/(x2-x1), where (x1, y1) is one point, and (x2,y2) is the second point.

The points are Dependent. Just pot the points and put two arrows at the end of the lines.

since you know of one points and the halfway point between the other point. just multiply the halfway point by 2 and this is the total distance between the two points.

-- Draw a line segment from one point to the other.-- Construct the perpendicular bisector of the line segment..-- Every point on the perpendicular bisector of the line segmentis equidistant from the two original points.==========================================================================Whereupon the first contributor observed:Yes, that works, no matter how you set the compass, as long as it's more than 1/2 the distancebetween the two points. Every setting of the compass will give you a pair of points that areequal distances from the original two. As you find more and more of them ... with differentsettings of the compass ... you'll see that all the equal-distance points you're finding all lieon the same straight line. That line is the perpendicular bisector of the line between the twooriginal points, just as we described up above.

Unfortunately, no. Unless the user messages you and tells you that they've given you a trust point.

Infinitely many. There are infinitely many points in the plane and although any pair of points define a line, no matter how many lines you are given, it is always possible to find a point that is not on any of them - that is, a point that is not collinear.