Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

OpenSessame

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).

  • 7 months ago
  • 7 months ago

  • This Question is Closed
  1. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Ill draw it out?

    • 7 months ago
  2. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    With coordinate geometry proofs, you need the coordinate plane.

    • 7 months ago
  3. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1377488818630:dw|

    • 7 months ago
  4. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    That is what you are proving right?

    • 7 months ago
  5. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    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...

    • 7 months ago
  6. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    You need to label your drawing. Where are the points you are referring to?

    • 7 months ago
  7. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1377488962191:dw|

    • 7 months ago
  8. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1377489047397:dw|

    • 7 months ago
  9. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1377489011250:dw|

    • 7 months ago
  10. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    The way you labeled it, D is not the midpoint of AB because D is not even on segment AB.

    • 7 months ago
  11. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay so it would be a new point, C.

    • 7 months ago
  12. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.

    • 7 months ago
  13. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay...Well I cant draw a picture to answer it has to be in words only.

    • 7 months ago
  14. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1377489315487:dw|

    • 7 months ago
  15. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay...Im a bit lost now...

    • 7 months ago
  16. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1377489369615:dw|

    • 7 months ago
  17. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Why the new F point?

    • 7 months ago
  18. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 7 months ago
  19. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?

    • 7 months ago
  20. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes, but this needs to use the coordinates. That is what a coordinate proof is.

    • 7 months ago
  21. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    So how would I do that, thats the part im not getting...

    • 7 months ago
  22. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    AE = |-a - 0| = |-a| = a BE = |a - 0| = |a| = a AE = BE = a

    • 7 months ago
  23. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    FE = |d - 0| = |d| = d

    • 7 months ago
  24. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Ohhhh...Okay makes sense!!!

    • 7 months ago
  25. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    We want to show that AF = BF. AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.

    • 7 months ago
  26. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Right, So use the Pythagorean theorem with the coordinates?

    • 7 months ago
  27. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    Exactly. Pythagoras with coordinates.

    • 7 months ago
  28. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    \((AF)^2 = (AE)^2 + (FE)^2\) \( AF = \sqrt{ (AE)^2 + (FE)^2 } \) \( AF = \sqrt{ a^2 + d^2 } \)

    • 7 months ago
  29. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    Now we do the same for FB.

    • 7 months ago
  30. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay...

    • 7 months ago
  31. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(AF)^{2}=(BE)^{2}+(FE)^{2}\]

    • 7 months ago
  32. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 7 months ago
  33. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Im a little confused on how you got from the 2nd to the 3rd...

    • 7 months ago
  34. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    We have shown that both AF and BF are equal to \( \sqrt{a^2 + d^2}\), so AF and BF are equal.

    • 7 months ago
  35. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Is it because of what you said before?

    • 7 months ago
  36. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    You mean confused on going from (BE)^2 to a^2?

    • 7 months ago
  37. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    I was but then i read over it and got it!

    • 7 months ago
  38. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 7 months ago
  39. OpenSessame
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay thanks!! i have one more question, but i know the answer but not how to get their...

    • 7 months ago
  40. mathstudent55
    Best Response
    You've already chosen the best response.
    Medals 1

    new post pls

    • 7 months ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.