First, it needs to be a line segment: a line is infinitely long and so has no midpoint.Suppose the line to be bisected is AB. Place the point of a pair of compasses at A with an arc which is greater than half AB. Draw arcs above and below the line segment. Then, move the compass point to B and without changing the arc width, draw fresh arcs to intercept the previous ones at points X and Y. The intersection of the line segment XY and AB is the midpoint of AB.
perpendicular
segment AB
If you mean end point A is (3, 5) and midpoint of line AB is (-2, 8) then end point B is (-7, 11)
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.
The perpendicular bisector of a line segment AB is the straight line perpendicular to AB through the midpoint of AB.
If M is the midpoint of segment AB, then AMis congruent to MB.
Find the midpoint of the two diagonals
(5/2,11/2)
First, it needs to be a line segment: a line is infinitely long and so has no midpoint.Suppose the line to be bisected is AB. Place the point of a pair of compasses at A with an arc which is greater than half AB. Draw arcs above and below the line segment. Then, move the compass point to B and without changing the arc width, draw fresh arcs to intercept the previous ones at points X and Y. The intersection of the line segment XY and AB is the midpoint of AB.
B is (-5, 9).
Median of a trapezoid is a line segment found on the midpoint of the legs of a trapezoid. It is also known as mid-line or mid-segment. Its basic formula is AB + CD divided by 2.
If the coordinate of A is x, and that of the midpoint of AB, M, is m then the distance AM is m-x so the distance AB = 2*(m-x) So the coordinate of B is x + 2*(m-x) = 2m-x For coordinates in more than one dimension, apply the above rule separately for each dimension.
perpendicular
segment AB
If you mean points of (-1, 5) and (6, -3) then the midpoint is (2.5, 1)
I cannot complete the problem because no preliminary measurements were given.