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OpenSessame

  • 2 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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  1. OpenSessame
    • 2 years ago
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    Ill draw it out?

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

  3. OpenSessame
    • 2 years ago
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    |dw:1377488818630:dw|

  4. OpenSessame
    • 2 years ago
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    That is what you are proving right?

  5. OpenSessame
    • 2 years 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
    • 2 years ago
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    You need to label your drawing. Where are the points you are referring to?

  7. OpenSessame
    • 2 years ago
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    |dw:1377488962191:dw|

  8. OpenSessame
    • 2 years ago
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    |dw:1377489047397:dw|

  9. mathstudent55
    • 2 years ago
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    |dw:1377489011250:dw|

  10. mathstudent55
    • 2 years 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
    • 2 years ago
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    Okay so it would be a new point, C.

  12. mathstudent55
    • 2 years 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
    • 2 years ago
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    Okay...Well I cant draw a picture to answer it has to be in words only.

  14. mathstudent55
    • 2 years ago
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    |dw:1377489315487:dw|

  15. OpenSessame
    • 2 years ago
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    Okay...Im a bit lost now...

  16. mathstudent55
    • 2 years ago
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    |dw:1377489369615:dw|

  17. OpenSessame
    • 2 years ago
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    Why the new F point?

  18. mathstudent55
    • 2 years 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
    • 2 years ago
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    I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?

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

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

  22. mathstudent55
    • 2 years ago
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    AE = |-a - 0| = |-a| = a BE = |a - 0| = |a| = a AE = BE = a

  23. mathstudent55
    • 2 years ago
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    FE = |d - 0| = |d| = d

  24. OpenSessame
    • 2 years ago
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    Ohhhh...Okay makes sense!!!

  25. mathstudent55
    • 2 years 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
    • 2 years ago
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    Right, So use the Pythagorean theorem with the coordinates?

  27. mathstudent55
    • 2 years ago
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    Exactly. Pythagoras with coordinates.

  28. mathstudent55
    • 2 years 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
    • 2 years ago
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    Now we do the same for FB.

  30. OpenSessame
    • 2 years ago
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    Okay...

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

  32. mathstudent55
    • 2 years 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
    • 2 years ago
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    Im a little confused on how you got from the 2nd to the 3rd...

  34. mathstudent55
    • 2 years 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
    • 2 years ago
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    Is it because of what you said before?

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

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

  38. mathstudent55
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    new post pls

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