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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).

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  1. OpenSessame
    • one year ago
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    Ill draw it out?

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

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

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

  5. OpenSessame
    • one year ago
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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...

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

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

  8. OpenSessame
    • one year ago
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    |dw:1377489047397:dw|

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

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

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

  12. mathstudent55
    • one year ago
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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.

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

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

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

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

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

  18. mathstudent55
    • one year ago
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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.

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

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

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

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

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

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

  25. mathstudent55
    • one year ago
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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.

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

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

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

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

  30. OpenSessame
    • one year ago
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    Okay...

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

  32. mathstudent55
    • one year ago
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    Yes, but we are doing BF now, not AF. \( (BF)^2=(BE)^2+(FE)^2 \) \( BF=\sqrt{(BE)^2+(FE)^2 } \) \( BF=\sqrt{a^2+d^2} \)

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

  34. mathstudent55
    • one year ago
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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.

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

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

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

  38. mathstudent55
    • one year ago
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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.

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

  40. mathstudent55
    • one year ago
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    new post pls

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