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anonymous
 3 years 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).
anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1With coordinate geometry proofs, you need the coordinate plane.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1377488818630:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is what you are proving right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In 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
 3 years ago
Best ResponseYou've already chosen the best response.1You need to label your drawing. Where are the points you are referring to?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1377488962191:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1377489047397:dw

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

mathstudent55
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay so it would be a new point, C.

mathstudent55
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay...Well I cant draw a picture to answer it has to be in words only.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay...Im a bit lost now...

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

mathstudent55
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So how would I do that, thats the part im not getting...

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ohhhh...Okay makes sense!!!

mathstudent55
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right, So use the Pythagorean theorem with the coordinates?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(AF)^{2}=(BE)^{2}+(FE)^{2}\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im a little confused on how you got from the 2nd to the 3rd...

mathstudent55
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it because of what you said before?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was but then i read over it and got it!

mathstudent55
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay thanks!! i have one more question, but i know the answer but not how to get their...
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