Suppose $A$ and also $B$ space two distinctive points in the plane and $L$ isthe perpendicular bisector the segment $overlineAB$ as pictured below:


If $C$ is a suggest on $L$, present that $C$ is equidistant from $A$ and also $B$,that is show that $overlineAC$ and $overlineBC$ space congruent.Conversely, show that if $P$ is a suggest which is equidistant from $A$and $B$, climate $P$ is ~ above $L$.Conclude that the perpendicular bisector the $overlineAB$ is specifically the set of pointswhich room equidistant native $A$ and also $B$.

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This task is component of a collection presenting essential foundational geometricresults and constructions i beg your pardon are an essential for much more elaborate arguments.They room presented without a genuine worldcontext so regarding see the vital hypotheses and also logical steps affiliated as clearly as possible. Additional, showing just how theseresults have the right to be offered in context, will also be developed. Teachers have to choosean ideal mixture of abstract and contextual jobs which finest fostersthe development of your students" geometric intuition and understanding.

This task gives the crucial characterization that the perpendicular bisectorof a heat segment together the collection of clues equidistant indigenous the endpoints the the segment. In the an initial part that the task, the instructor might need to suggestthat there space two cases to consider:

the case where $C$ is on the heat segment $overlineAB$ (this is a special case andthe students might forget to think about this)the case where $C$ is no on the line segment $overlineAB$.

The exact same two cases occur in part (b) for the allude $P$.

The instructor may additionally wish to walk over the logic of component (b) that the task and also make certain that the students understand what needs to be shown and how the is various from part (a). Both (a) and (b) continue via triangle congruence so the students must be familiar with and also confident in implementing this criteria. Since the task is fairly long and also detailed it is recommended greatly for instructional purposes.

This task contains an speculative GeoGebra worksheet, v the intentthat instructors could use it to much more interactively show therelevant content material. The record should be taken into consideration a draftversion, and feedback on it in the comment section is highlyencouraged, both in regards to suggestions for development and for ideason using it effectively. The record can be run via the totally free onlineapplication GeoGebra, or runlocally if GeoGebra has actually been set up on a computer.

This paper can be provided individually come walk through the given collection of problems. It can additionally be provided in a presentation format for an audience. Every step has actually solutions which can be shown by clicking the check boxes that appear. The buttons are how you switch in between steps and also at any kind of time you space able to drag the red point on the perpendicular bisector up and also down.


We consider very first the case where $C$ is not on segment $overlineAB$ as in the picturebelow. The suggest of intersection that $L$ and also the segment $overlineAB$ is labelled $O$:


By hypothesis, angle $COA$ and $COB$ space both right angles and so arecongruent. Next $overlineCO$ is congruent to itself and also side $overlineOA$ is congruent toside $overlineOB$ because $L$ bisects segment $overlineAB$. By SAS us conclude thattriangle $COA$ is congruent to triangle $COB$. Hence segment $overlineCA$ is congruentto segment $overlineCB$ and also so $C$ is equidistant native $A$ and $B$ together desired.

Next we take into consideration the instance where $C$ is on segment $overlineAB$ together pictured below:


This case is simpler than the ahead one due to the fact that we understand by hypothesis that$L$ bisects segment $overlineAB$ and so $overlineCA$ is congruent to $overlineCB$ and also hence$C$ is equidistant from $A$ and $B$ below too.

Here us assume the $P$ is equidistant from $A$ and $B$. There room again twocases to consider. First, $P$ could be ~ above segment $overlineAB$ and also then $P$ mustbe the midpoint of segment $overlineAB$ and also so $P$ is top top $L$. Following suppose $P$ isnot on line $L$. In this case, us let $D$ signify the midpoint the $overlineAB$ and $M$the heat joining $P$ and $D$ as pictured below:


We recognize that $overlinePA$ is congruent to $overlinePB$ since $P$ is equidistant indigenous $A$ and$B$ through assumption. We know that $overlineDA$ is congruent come $overlineDB$ due to the fact that $D$is the midpoint of $overlineAB$. Finally $overlinePD$ is congruent to $overlinePD$. Therefore bySSS, triangle $PDA$ is congruent come triangle $PDB$. Because angles $PDA$ and$PDB$ room congruent and include up to $180$ degrees, they should both beright angles. Due to the fact that $overlineDA$ is congruent come $overlineDB$ we conclude the line $M$ isthe perpendicular bisector the segment $overlineAB$, that is $M=L$.In component (a) we witnessed that every suggest on the perpendicular bisector that segment$overlineAB$ is equidistant from points $A$ and $B$. In part (b) we experienced that any type of pointequidistant indigenous points $A$ and also $B$ is on the perpendicular bisector the segment $overlineAB$. Thus the perpendicular bisector the $overlineAB$ is comprised exactly that thosepoints $P$ which are equidistant indigenous $A$ and $B$.

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Suppose $A$ and $B$ room two distinctive points in the plane and $L$ isthe perpendicular bisector of segment $overlineAB$ together pictured below:


If $C$ is a point on $L$, present that $C$ is equidistant indigenous $A$ and also $B$,that is show that $overlineAC$ and also $overlineBC$ room congruent.Conversely, present that if $P$ is a suggest which is equidistant indigenous $A$and $B$, climate $P$ is top top $L$.Conclude that the perpendicular bisector the $overlineAB$ is specifically the set of pointswhich space equidistant native $A$ and also $B$.