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aroub

  • 2 years ago

Given a triangle ABC such that B and C are fixed. Point A varies such that segment AB is congruent to segment AC. Find the locus of A. Umm, an angle bisector maybe?

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  1. AbhimanyuPudi
    • 2 years ago
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    perpendicular bisector of BC

  2. aroub
    • 2 years ago
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    Aoh -.- I get it now! Because they said that the sides are congruent so they are equidistant so it's perpendicular bisector!! Let me draw it and tell me if it's right or wrong, okay? :)

  3. AbhimanyuPudi
    • 2 years ago
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    sure

  4. klimenkov
    • 2 years ago
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    Sure. You are welcome.

  5. aroub
    • 2 years ago
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    |dw:1352049851819:dw|

  6. hartnn
    • 2 years ago
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    do u know how to solve this algebraically ?

  7. hartnn
    • 2 years ago
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    how to get to 'perpendicular bisector of BC'

  8. aroub
    • 2 years ago
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    Emmm, i don't think so.. like what do you mean? show me? :)

  9. AbhimanyuPudi
    • 2 years ago
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    |dw:1352049933632:dw|

  10. aroub
    • 2 years ago
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    Does it matter?

  11. AbhimanyuPudi
    • 2 years ago
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    Yea..the point A can be above BC or below BC according to that figure

  12. aroub
    • 2 years ago
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    Okay, thank youu!

  13. hartnn
    • 2 years ago
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    ok, i choose point B as origin of my co-ordinate axes and point C on x-axis. as they are fixed i can put them wherever i want. now let co-ordinates of A be (x,y) we need a locus in terms of x and y let me draw it. |dw:1352050085145:dw| now AB=AC will give you, using distance formula, \(\huge x^2+y^2=(x-a)^2+y^2\) can u simplify this and tell me what u get ?

  14. AbhimanyuPudi
    • 2 years ago
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    welcome :-)

  15. klimenkov
    • 2 years ago
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    |dw:1352050205624:dw|

  16. aroub
    • 2 years ago
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    Oh, oh! No.. not yet at least. Let me know how to find the locus first :P Thank you though!!

  17. hartnn
    • 2 years ago
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    if u simplify that u get x=a/2 which is a line that bisects BC , hence 'bisector' and is perpendicular to BC (as it is vertical line, its perpendicular to x-axis) hence, 'perpendicular' did u get my explanation ?

  18. AbhimanyuPudi
    • 2 years ago
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    A small correction @hartnn It is \[x ^{2}+y ^{2} = x ^{2} + (y-a)^{2}\]

  19. AbhimanyuPudi
    • 2 years ago
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    Ok got it..that isn't the correction..the correction is B=(a,0)

  20. hartnn
    • 2 years ago
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    correction is in diagram! not there |dw:1352050580220:dw|

  21. AbhimanyuPudi
    • 2 years ago
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    Ya ya..that's what I was talking about

  22. aroub
    • 2 years ago
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    Yes I did! So you can get it with solving.. Not by memorizing the conditions or cases. Nice!

  23. hartnn
    • 2 years ago
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    always!

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