Point point

If $\triangle ABC$ is an equilateral triangle, it satisfies $AB = AC$ and $BA = BC$ . The set of points $C$ that satisfy the first condition is a circumference with center $A$ and radius $AB$ , while the set of points $C$ that satisfy the second condition is a circumference with center $B$ and radius $BA$ . Both circumferences intersect at two points, which are the only points $C$ such that $\triangle ABC$ is equilateral.

Only two if you can imagine how it's work, just for example if AB is in line y (x=0) so equilateral triangle that can happen is where C is to right line x (+) and to left line x (-).

Ex. Consider two points in the x axis of the plane would be A=2, B=6. In an equilateral triangle all sides should have same angle. according to this, the point C would be in 4 of x axis. the point can be placed above and below the line A and B So the answer is 2

so an equilateral triangle sides are equal so c=b=a c=2

The point C can be on either side of line joining A to B . So number of points =2. On one side of the line segment there will be only one point since two rays will intersect once only.

The pts. lie on the perpendicular bisector of line AB at a distance = AB from A and B . you can try it manually also

That's easy, we have 1 especific C that we can made a equilateral but is one above and one below, so 2

If there are two points in a plane that are 1 inch apart, then the other point must be 1 inch apart from the two points. It can be to the left and then reflected to the right. This results in two possible points.