The perpendicular bisector  ### Problem:

Where do all points lie that have the same distance from two different given points?

### Consideration:

In the following drawing you can search th positions of C,
where it has the same distance from A and B:
• Drag the point C across the drawing's surface and...
• ...watch the values of the two distance measures:

When you succeed to move the point C in such a way that its distance from point A equals the distance from point B, then you can make the point C draw a trace:

• chose the tool "Record a locus line",
• click the point C,
• then grip it with the pair of pliers....
• ...and drag it over the surface.
Which "curve" does C describe when it always has the same distance from A than it has from B?

After having recognized this, you can construct this "curve": use the tool "Line through 2 points". Now you must assure that C runs always on this line. You can do this by snapping it to the line:
• Click with the right mouse button on the point C.
• In the upcoming "context menu" of the point C you chose the item "Snap point to line";
• Click on the line the point C is to be snapped to!
Now C jumps onto the line and will always stay on this line however you drag it. It's time to check, if now C has always the same distance from Y than it has from B. Maybe, you need to adjust your constructed line a bit...

### Construction:

All points that have equal distance from two points A and B, lie on a straight line. How can you construct this line exactly? Obviously it is sufficient to construct two of its points, because then the line is already determined. But how do you do this with compass and rulers only?

Surely you know that all points with a certain distance from a points A lie on a circle around A that has this distance as radius. So you can draw a circle around A with some fixed radius, und then a second circle around B, with the same radius. This can easily be achieved when you use the length of the segment [AB] as radius for the two circles. After having constructed these two circles, it is necessary to construct their intersection points S and T (use the tool "Intersect two lines"). Afterwardy, you can construct the straight line through S and T (with the tool "Straight line through 2 points").

Now its your turn to carry out the construction:

You can easily check whether your construction is correct:
• Just drag one of the points A or B.
Do all possible points C on the line (ST) have equal distances from A and B?

With the tool "Name object" from the context menu of the constructed straight line you can give this line a name, e.g. "g". When additionally you construct the straight line h through A and B, then you see that g and h are orthogonal. The intersection point M of g and h lies halfway between A and B -- that's clear, because it is a point of g, and all points of g have equal distances from A and B!

### Definition:

The straight line that holds all the points that have equal distances from two given points A and B is called the perpendicular bisector of the segment [AB].

### Result:

Given: two (different) points A and B
Wanted: a straight line g the points C that satisfy the equation: d(A;C) = d(B;C), where d() is the distance function.
Construction text:
• k1 is a circle around A running through B
• k2 is a circle around B running through A
• S and T are the intersection points of k1 and k2
• g is the straight line through S and T

### For further puzzling:

1. You are not forced to use the distance of A to B as radius for the circles in the above constructions. But there is a lower limit for the radius. Can you determine the value of this limit?
Is there an upper limit, too?

2. The point M also lies halfway between S and T, but this seems not yet clear. Can you complete the drawing to become convinced that this is always true?  