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OpenSessame
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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).
 one year ago
 one year ago
OpenSessame Group Title
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).
 one year ago
 one year ago

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OpenSessame Group TitleBest ResponseYou've already chosen the best response.1
Ill draw it out?
 one year ago

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

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

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

OpenSessame Group TitleBest ResponseYou've already chosen the best response.1
In 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...
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

mathstudent55 Group TitleBest ResponseYou've already chosen the best response.1
Given: 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.
 one year ago

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

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

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

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

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

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

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

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

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

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

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

OpenSessame Group TitleBest ResponseYou've already chosen the best response.1
Okay...
 one year ago

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

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

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

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

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

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

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

mathstudent55 Group TitleBest ResponseYou've already chosen the best response.1
It'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.
 one year ago

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

mathstudent55 Group TitleBest ResponseYou've already chosen the best response.1
new post pls
 one year ago
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