... in the same plane.
if a function is increasing, the average change of rate between any two points must be positive.
No, because Of any three points on a line there exists no more than one that lies between the other two.
no,three points can be non collinear
if i put three points on the common arm ,then they are common points for both the two angles
No. Any pair of the three will describe a plane, so the three possible pairs describe three planes.
No. A trinagle does not require four points, three are sufficient. And any three points, if they are not colinear, must be coplanar.
A unique plane is defined by three non-collinear points. This means that the points must not all lie on the same straight line. If the three points are collinear or if only two points are given, they do not suffice to define a unique plane. Thus, the key restriction is that the three points must be non-collinear.
They must be collinear.
There are no planes containing any number of given points. Two points not the same define a line. Three points not in a line define a plane. For four or more points to lie in the same plane, three can be arbitrary but not on the same line, but the fourth (and so on) points must lie in that same plane.
In 3-dimensional space, yes, any three points are coplanar.
Any 3 points
Two points (which must lie on a line) and the third point NOT on that line.
Any three points anywhere in space can be the vertices of a triangle, as long as all three are not colinear.
no
Yes. In fact any three points that are not collinear define a plane and therefore MUST lie on a plane.
No, given any three points, it is possible for one of the points not to be on the line defined by the other two points. Only two points on a line are needed to identify the exact position of the line. The positions of any three points gives you the exact position of the plane that includes those three points.No, it is not true. If it were true, all triangles would be straight lines !?!
Since any two point must be collinear and must, therefore, define a line, the answer is 5C2, the number of combinations of two [points] out of five. This is 5*4/(2*1) = 10