anonymous
  • anonymous
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).
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Ill draw it out?
mathstudent55
  • mathstudent55
With coordinate geometry proofs, you need the coordinate plane.
anonymous
  • anonymous
|dw:1377488818630:dw|

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anonymous
  • anonymous
That is what you are proving right?
anonymous
  • anonymous
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 Side-Angle-Side. 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
  • mathstudent55
You need to label your drawing. Where are the points you are referring to?
anonymous
  • anonymous
|dw:1377488962191:dw|
anonymous
  • anonymous
|dw:1377489047397:dw|
mathstudent55
  • mathstudent55
|dw:1377489011250:dw|
mathstudent55
  • mathstudent55
The way you labeled it, D is not the midpoint of AB because D is not even on segment AB.
anonymous
  • anonymous
Okay so it would be a new point, C.
mathstudent55
  • mathstudent55
My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.
anonymous
  • anonymous
Okay...Well I cant draw a picture to answer it has to be in words only.
mathstudent55
  • mathstudent55
|dw:1377489315487:dw|
anonymous
  • anonymous
Okay...Im a bit lost now...
mathstudent55
  • mathstudent55
|dw:1377489369615:dw|
anonymous
  • anonymous
Why the new F point?
mathstudent55
  • mathstudent55
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.
anonymous
  • anonymous
I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?
mathstudent55
  • mathstudent55
Yes, but this needs to use the coordinates. That is what a coordinate proof is.
anonymous
  • anonymous
So how would I do that, thats the part im not getting...
mathstudent55
  • mathstudent55
AE = |-a - 0| = |-a| = a BE = |a - 0| = |a| = a AE = BE = a
mathstudent55
  • mathstudent55
FE = |d - 0| = |d| = d
anonymous
  • anonymous
Ohhhh...Okay makes sense!!!
mathstudent55
  • mathstudent55
We want to show that AF = BF. AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.
anonymous
  • anonymous
Right, So use the Pythagorean theorem with the coordinates?
mathstudent55
  • mathstudent55
Exactly. Pythagoras with coordinates.
mathstudent55
  • mathstudent55
\((AF)^2 = (AE)^2 + (FE)^2\) \( AF = \sqrt{ (AE)^2 + (FE)^2 } \) \( AF = \sqrt{ a^2 + d^2 } \)
mathstudent55
  • mathstudent55
Now we do the same for FB.
anonymous
  • anonymous
Okay...
anonymous
  • anonymous
\[(AF)^{2}=(BE)^{2}+(FE)^{2}\]
mathstudent55
  • mathstudent55
Yes, but we are doing BF now, not AF. \( (BF)^2=(BE)^2+(FE)^2 \) \( BF=\sqrt{(BE)^2+(FE)^2 } \) \( BF=\sqrt{a^2+d^2} \)
anonymous
  • anonymous
Im a little confused on how you got from the 2nd to the 3rd...
mathstudent55
  • mathstudent55
We have shown that both AF and BF are equal to \( \sqrt{a^2 + d^2}\), so AF and BF are equal.
anonymous
  • anonymous
Is it because of what you said before?
mathstudent55
  • mathstudent55
You mean confused on going from (BE)^2 to a^2?
anonymous
  • anonymous
I was but then i read over it and got it!
mathstudent55
  • mathstudent55
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.
anonymous
  • anonymous
Okay thanks!! i have one more question, but i know the answer but not how to get their...
mathstudent55
  • mathstudent55
new post pls

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