## 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). 11 months ago 11 months ago

1. OpenSessame Group Title

Ill draw it out?

2. mathstudent55 Group Title

With coordinate geometry proofs, you need the coordinate plane.

3. OpenSessame Group Title

|dw:1377488818630:dw|

4. OpenSessame Group Title

That is what you are proving right?

5. OpenSessame Group Title

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 Group Title

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

7. OpenSessame Group Title

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8. OpenSessame Group Title

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9. mathstudent55 Group Title

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10. mathstudent55 Group Title

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

11. OpenSessame Group Title

Okay so it would be a new point, C.

12. mathstudent55 Group Title

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

13. OpenSessame Group Title

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

14. mathstudent55 Group Title

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15. OpenSessame Group Title

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

16. mathstudent55 Group Title

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17. OpenSessame Group Title

Why the new F point?

18. mathstudent55 Group Title

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 Group Title

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

20. mathstudent55 Group Title

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

21. OpenSessame Group Title

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

22. mathstudent55 Group Title

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

23. mathstudent55 Group Title

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

24. OpenSessame Group Title

Ohhhh...Okay makes sense!!!

25. mathstudent55 Group Title

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

26. OpenSessame Group Title

Right, So use the Pythagorean theorem with the coordinates?

27. mathstudent55 Group Title

Exactly. Pythagoras with coordinates.

28. mathstudent55 Group Title

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

29. mathstudent55 Group Title

Now we do the same for FB.

30. OpenSessame Group Title

Okay...

31. OpenSessame Group Title

$(AF)^{2}=(BE)^{2}+(FE)^{2}$

32. mathstudent55 Group Title

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 Group Title

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

34. mathstudent55 Group Title

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

35. OpenSessame Group Title

Is it because of what you said before?

36. mathstudent55 Group Title

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

37. OpenSessame Group Title

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

38. mathstudent55 Group Title

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 Group Title

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

40. mathstudent55 Group Title

new post pls