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What is the locus of points equidistant from two points A and B that are 8 meters apart?
It is the perpendicular bisector of AB, the line joining the two points.
If a and b are two points in the plane the perpendicular bisector of Ab is the set of all points equidistant from a and b?
The perpendicular bisector of the line segment connecting points ( A ) and ( B ) in the plane is a line that divides the segment into two equal parts at a right angle. Every point on this line is equidistant from points ( A ) and ( B ). This means that if you take any point ( P ) on the perpendicular bisector, the distance from ( P ) to ( A ) will be the same as the distance from ( P ) to ( B ). Thus, the perpendicular bisector is the locus of points satisfying this equidistance condition.
Given two points A and B in the three dimensional space what is the set of points equidistant from A and B?
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.
Where do all the points on a plane equidistant from the endpoints of ab lie?
All points on a plane that are equidistant from the endpoints of a line segment ( ab ) lie on the perpendicular bisector of the segment. This line bisects ( ab ) at a right angle and includes all points that are the same distance from both endpoints ( a ) and ( b ). Therefore, any point on this line is equidistant from ( a ) and ( b ).
If a and b are two points in the plane the perpendicular bisector of a B is the set of all points equidistant from a and b true or false?
True. The perpendicular bisector of the segment connecting points ( a ) and ( b ) is defined as the set of all points that are equidistant from both ( a ) and ( b ). This line is perpendicular to the segment at its midpoint and ensures that any point on this line maintains equal distance to both endpoints.
What is the locus of points equidistant from two points A and B that are 8 meters apart?
It is the perpendicular bisector of AB, the line joining the two points.
If a and b are two points in the plane the perpendicular bisector of Ab is the set of all points equidistant from a and b?
The perpendicular bisector of the line segment connecting points ( A ) and ( B ) in the plane is a line that divides the segment into two equal parts at a right angle. Every point on this line is equidistant from points ( A ) and ( B ). This means that if you take any point ( P ) on the perpendicular bisector, the distance from ( P ) to ( A ) will be the same as the distance from ( P ) to ( B ). Thus, the perpendicular bisector is the locus of points satisfying this equidistance condition.
Given two points A and B in the three dimensional space what is the set of points equidistant from A and B?
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.
Where do all the points on a plane equidistant from the endpoints of ab lie?
All points on a plane that are equidistant from the endpoints of a line segment ( ab ) lie on the perpendicular bisector of the segment. This line bisects ( ab ) at a right angle and includes all points that are the same distance from both endpoints ( a ) and ( b ). Therefore, any point on this line is equidistant from ( a ) and ( b ).
If a and b are two points in the plane the perpendicular bisector of a B is the set of all points equidistant from a and b true or false?
True. The perpendicular bisector of the segment connecting points ( a ) and ( b ) is defined as the set of all points that are equidistant from both ( a ) and ( b ). This line is perpendicular to the segment at its midpoint and ensures that any point on this line maintains equal distance to both endpoints.
If a and b are two points in the plane the perpendicular bisector of ab is the set of all points equidistant from a to b?
The perpendicular bisector of the line segment connecting points ( a ) and ( b ) in a plane is a line that is perpendicular to the segment at its midpoint. This line consists of all points that are equidistant from ( a ) and ( b ). Therefore, if any point lies on the perpendicular bisector, it maintains equal distance from both points. This property is fundamental in geometry and is used in various applications, including triangulation and construction.
How many points are there that are equidistant points a and b and also 2in from the line passing through a and b?
To find the points that are equidistant from points A and B, you would first determine the perpendicular bisector of the line segment AB, which is a line that is equidistant from both points. For the points that are also 2 units from the line passing through A and B, there will be two lines parallel to the perpendicular bisector, each located 2 units away. Therefore, there are exactly two points that satisfy both conditions.
Points A and B are 6 units apart How many points are equidistant fomr A and B and 2 units from A?
None. If a point is 2 units from 'A' and equidistant from 'A' and 'B', then it also has to be2 units from 'B'.But the shortest distance between 'A' and 'B' is 6 units, and the point on that line that's equidistantfrom both of them is the point in the middle, which is 3 units from each.So a point equidistant from 'A' and 'B' must be 3 or more units from each one. 2 units won't do it.
Describe the locus of points that are 9 cm from point B?
a circle 9 cm from point b I was co fused by this but you just do a diagram and write this
In the figure points A B C and D reflect across to coincide with points G J I and H respectively?
In the given scenario, points A, B, C, and D are reflected across a line or point to coincide with points G, J, I, and H, respectively. This reflection implies that each original point and its corresponding reflected point are equidistant from the line of reflection. Therefore, the positions of points A, B, C, and D are symmetrically opposite to points G, J, I, and H concerning the line of reflection. This geometric relationship highlights the properties of reflection in a coordinate plane.
What points are in the first quadrant of the xy-plane?
All ordered pairs, (a,b) such that a > 0 and b > 0
How many planes can be determined by A B and C?
If points A, B, and C are not on the same line, they determine a single plane.
