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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
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  1. OpenSessame Group Title
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    Ill draw it out?

    • one year ago
  2. mathstudent55 Group Title
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    With coordinate geometry proofs, you need the coordinate plane.

    • one year ago
  3. OpenSessame Group Title
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    |dw:1377488818630:dw|

    • one year ago
  4. OpenSessame Group Title
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    That is what you are proving right?

    • one year ago
  5. OpenSessame Group Title
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    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...

    • one year ago
  6. mathstudent55 Group Title
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    You need to label your drawing. Where are the points you are referring to?

    • one year ago
  7. OpenSessame Group Title
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    |dw:1377488962191:dw|

    • one year ago
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    |dw:1377489047397:dw|

    • one year ago
  9. mathstudent55 Group Title
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    |dw:1377489011250:dw|

    • one year ago
  10. mathstudent55 Group Title
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    The way you labeled it, D is not the midpoint of AB because D is not even on segment AB.

    • one year ago
  11. OpenSessame Group Title
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    Okay so it would be a new point, C.

    • one year ago
  12. mathstudent55 Group Title
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    My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.

    • one year ago
  13. OpenSessame Group Title
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    Okay...Well I cant draw a picture to answer it has to be in words only.

    • one year ago
  14. mathstudent55 Group Title
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    |dw:1377489315487:dw|

    • one year ago
  15. OpenSessame Group Title
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    Okay...Im a bit lost now...

    • one year ago
  16. mathstudent55 Group Title
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    |dw:1377489369615:dw|

    • one year ago
  17. OpenSessame Group Title
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    Why the new F point?

    • one year ago
  18. mathstudent55 Group Title
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    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
  19. OpenSessame Group Title
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    I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?

    • one year ago
  20. mathstudent55 Group Title
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    Yes, but this needs to use the coordinates. That is what a coordinate proof is.

    • one year ago
  21. OpenSessame Group Title
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    So how would I do that, thats the part im not getting...

    • one year ago
  22. mathstudent55 Group Title
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    AE = |-a - 0| = |-a| = a BE = |a - 0| = |a| = a AE = BE = a

    • one year ago
  23. mathstudent55 Group Title
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    FE = |d - 0| = |d| = d

    • one year ago
  24. OpenSessame Group Title
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    Ohhhh...Okay makes sense!!!

    • one year ago
  25. mathstudent55 Group Title
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    We want to show that AF = BF. AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.

    • one year ago
  26. OpenSessame Group Title
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    Right, So use the Pythagorean theorem with the coordinates?

    • one year ago
  27. mathstudent55 Group Title
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    Exactly. Pythagoras with coordinates.

    • one year ago
  28. mathstudent55 Group Title
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    \((AF)^2 = (AE)^2 + (FE)^2\) \( AF = \sqrt{ (AE)^2 + (FE)^2 } \) \( AF = \sqrt{ a^2 + d^2 } \)

    • one year ago
  29. mathstudent55 Group Title
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    Now we do the same for FB.

    • one year ago
  30. OpenSessame Group Title
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    Okay...

    • one year ago
  31. OpenSessame Group Title
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    \[(AF)^{2}=(BE)^{2}+(FE)^{2}\]

    • one year ago
  32. mathstudent55 Group Title
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    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
  33. OpenSessame Group Title
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    Im a little confused on how you got from the 2nd to the 3rd...

    • one year ago
  34. mathstudent55 Group Title
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    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
  35. OpenSessame Group Title
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    Is it because of what you said before?

    • one year ago
  36. mathstudent55 Group Title
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    You mean confused on going from (BE)^2 to a^2?

    • one year ago
  37. OpenSessame Group Title
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    I was but then i read over it and got it!

    • one year ago
  38. mathstudent55 Group Title
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    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
  39. OpenSessame Group Title
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    Okay thanks!! i have one more question, but i know the answer but not how to get their...

    • one year ago
  40. mathstudent55 Group Title
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    new post pls

    • one year ago
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