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

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

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IM Commentary

This job is part of a collection presenting vital foundational geometricresults and constructions which are fundamental for more elaborate arguments.They room presented without a actual worldcontext so as to see the important hypotheses and logical steps associated as clearly as possible. Further snucongo.org, showing exactly how theseresults have the right to be used in context, will additionally be developed. Teachers should choosean proper mixture that abstract and also contextual snucongo.org which best fostersthe advance of your students" geometric intuition and understanding.

This task provides the essential characterization that the perpendicular bisectorof a heat segment as the collection of points equidistant from the endpoints that the segment. In the first part of the task, the instructor may need to suggestthat there space two situations to consider:

the instance where $C$ is top top the line segment $overlineAB$ (this is a special case andthe students might forget to take into consideration this)the case where $C$ is not on the line segment $overlineAB$.

The same two cases occur in component (b) because that the allude $P$.

The instructor may also wish to walk over the logic of component (b) that the task and make sure that the students know what needs to it is in shown and how that is various from part (a). Both (a) and also (b) continue via triangle congruence for this reason the students have to be familiar with and also confident in implementing these criteria. Due to the fact that the task is reasonably long and also detailed the is recommended greatly for instructional purposes.

This task contains an experimental GeoGebra worksheet, through the intentthat instructors might use that to much more interactively demonstrate therelevant content material. The file should be considered a draftversion, and also feedback on it in the comment section is highlyencouraged, both in regards to suggestions for advancement and for ideason using it effectively. The paper can be run via the complimentary onlineapplication GeoGebra, or runlocally if GeoGebra has actually been set up on a computer.

This file can be supplied individually to walk v the given set of problems. It can likewise be used in a presentation layout for one audience. Each step has actually solutions which have the right to be presented by clicking the check boxes the appear. The buttons are exactly how you switch in between steps and at any type of time you are able to traction the red suggest on the perpendicular bisector up and also down.


Solution

We consider an initial the instance where $C$ is no on segment $overlineAB$ together in the picturebelow. The suggest of intersection of $L$ and also the segment $overlineAB$ is labelled $O$:

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By hypothesis, angle $COA$ and also $COB$ are both best angles and also so arecongruent. Side $overlineCO$ is congruent come itself and also side $overlineOA$ is congruent toside $overlineOB$ since $L$ bisects segment $overlineAB$. By SAS we conclude thattriangle $COA$ is congruent come triangle $COB$. For this reason segment $overlineCA$ is congruentto segment $overlineCB$ and so $C$ is equidistant native $A$ and also $B$ together desired.

Next we think about the case where $C$ is top top segment $overlineAB$ together pictured below:

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This situation is easier than the ahead one due to the fact that we understand by hypothesis that$L$ bisects segment $overlineAB$ and also so $overlineCA$ is congruent to $overlineCB$ and also hence$C$ is equidistant native $A$ and $B$ right here too.

Here we assume that $P$ is equidistant indigenous $A$ and $B$. There room again twocases to consider. First, $P$ might be top top segment $overlineAB$ and also then $P$ mustbe the midpoint of segment $overlineAB$ and so $P$ is top top $L$. Following suppose $P$ isnot on line $L$. In this case, we let $D$ signify the midpoint of $overlineAB$ and $M$the line joining $P$ and $D$ together pictured below:

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We know that $overlinePA$ is congruent to $overlinePB$ since $P$ is equidistant from $A$ and$B$ by assumption. We know that $overlineDA$ is congruent come $overlineDB$ because $D$is the midpoint that $overlineAB$. Lastly $overlinePD$ is congruent come $overlinePD$. Therefore bySSS, triangle $PDA$ is congruent to triangle $PDB$. Since angles $PDA$ and$PDB$ space congruent and include up come $180$ degrees, they should both beright angles. Because $overlineDA$ is congruent come $overlineDB$ us conclude the line $M$ isthe perpendicular bisector of segment $overlineAB$, the is $M=L$.In part (a) we observed that every allude on the perpendicular bisector the segment$overlineAB$ is equidistant native points $A$ and also $B$. In component (b) we experienced that any kind of pointequidistant indigenous points $A$ and $B$ is on the perpendicular bisector the segment $overlineAB$. Thus the perpendicular bisector of $overlineAB$ is consisted of exactly the thosepoints $P$ which are equidistant native $A$ and $B$.

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Suppose $A$ and $B$ space two distinct points in the airplane and $L$ isthe perpendicular bisector of segment $overlineAB$ as pictured below:

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If $C$ is a point on $L$, display that $C$ is equidistant native $A$ and also $B$,that is present 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 ~ above $L$.Conclude that the perpendicular bisector of $overlineAB$ is precisely the collection of pointswhich space equidistant indigenous $A$ and $B$.