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OpenSessame
 one year ago
Prove using coordinate geometry: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
You are given a line segment (AB) and a perpendicular line segment (CD).
OpenSessame
 one year ago
Prove using coordinate geometry: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. You are given a line segment (AB) and a perpendicular line segment (CD).

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1With coordinate geometry proofs, you need the coordinate plane.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1dw:1377488818630:dw

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1That is what you are proving right?

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1In the triangle ADC and BDC, AD and BD are congruent because D is the midpoint of the segment AB. Also the side DC is common in both the triangles. This mean that angle CDA and angle CDB are both 90 degrees meaning they are congruent. This means the triangles are congruent because of the postulate SideAngleSide. So since the triangles are congruent, the distance from the two endpoints have to be equal. ^ that is what i said first but its wrong...

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1You need to label your drawing. Where are the points you are referring to?

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1dw:1377488962191:dw

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1dw:1377489047397:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1dw:1377489011250:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1The way you labeled it, D is not the midpoint of AB because D is not even on segment AB.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Okay so it would be a new point, C.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Okay...Well I cant draw a picture to answer it has to be in words only.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1dw:1377489315487:dw

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Okay...Im a bit lost now...

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1dw:1377489369615:dw

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Why the new F point?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Given: Segment CD is the perpendicular bisector of segment AB. Prove: F, a point on the perpendicular bisector of AB, is equidistant from A and B.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Yes, but this needs to use the coordinates. That is what a coordinate proof is.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1So how would I do that, thats the part im not getting...

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1AE = a  0 = a = a BE = a  0 = a = a AE = BE = a

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1FE = d  0 = d = d

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhh...Okay makes sense!!!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1We want to show that AF = BF. AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Right, So use the Pythagorean theorem with the coordinates?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Exactly. Pythagoras with coordinates.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\((AF)^2 = (AE)^2 + (FE)^2\) \( AF = \sqrt{ (AE)^2 + (FE)^2 } \) \( AF = \sqrt{ a^2 + d^2 } \)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Now we do the same for FB.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1\[(AF)^{2}=(BE)^{2}+(FE)^{2}\]

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Yes, but we are doing BF now, not AF. \( (BF)^2=(BE)^2+(FE)^2 \) \( BF=\sqrt{(BE)^2+(FE)^2 } \) \( BF=\sqrt{a^2+d^2} \)

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Im a little confused on how you got from the 2nd to the 3rd...

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1We have shown that both AF and BF are equal to \( \sqrt{a^2 + d^2}\), so AF and BF are equal.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Is it because of what you said before?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1You mean confused on going from (BE)^2 to a^2?

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1I was but then i read over it and got it!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1It's bec of these statements above: AE = a  0 = a = a BE = a  0 = a = a AE = BE = a FE = d  0 = d = d Oh, I see that you understand already. Great.

OpenSessame
 one year ago
Best ResponseYou've already chosen the best response.1Okay thanks!! i have one more question, but i know the answer but not how to get their...
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