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DLS Group Title

If two vertices of an equilateral triangle have integral coordinates,then the third vertex willl have

  • one year ago
  • one year ago

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  1. DLS Group Title
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    a)integral co-ordinates b)co-ordinates which are rational c)at least one co-ordinate irrational d)co-ordinates which are irrational

    • one year ago
  2. DLS Group Title
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    @LogicalApple

    • one year ago
  3. mathstudent55 Group Title
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    If a is an odd integer, b will not be an integer. |dw:1356846803078:dw|

    • one year ago
  4. mathstudent55 Group Title
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    |dw:1356846856691:dw|

    • one year ago
  5. mathstudent55 Group Title
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    Height of triangle c, is also the y-coordinate of the upper vertex. It's irrational.

    • one year ago
  6. DLS Group Title
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    how did u get a/2?

    • one year ago
  7. mathstudent55 Group Title
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    The triangle is equilateral, and all sides measure "a" long. The vertical height is the perpendicular bisector of the side that lies on the x-axis. Each half, therefore, measure a/2 long.

    • one year ago
  8. DLS Group Title
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    can u possibly explain this?i got confused.. http://gyazo.com/077e2faa197346bf8450280343875ee2

    • one year ago
  9. mathstudent55 Group Title
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    Yes, I'll explain the whole thing. The problem states "two vertices of an equilateral triangle have integral coordinates" That means those coordinates are integers.

    • one year ago
  10. mathstudent55 Group Title
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    To simplify the calculations, I set one vertex at (0, 0), and the orther one at (a, 0), with the condition that a is an integer.

    • one year ago
  11. DLS Group Title
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    okay

    • one year ago
  12. mathstudent55 Group Title
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    Because the triangle is equilateral, the third vertex lies in the perpendicular bisector of the first side which has coordinates (0, 0) and (a, 0).

    • one year ago
  13. DLS Group Title
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    yes

    • one year ago
  14. mathstudent55 Group Title
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    The x-coordinate of the third vertex is the same as the x-coordinate of the midpoint of the first side. As such it's (x1 + x2)/2, but in our case, x1 is 0. So if x2, which is a, is odd, then the x-coordinate of the third vertex will not be an integer.

    • one year ago
  15. DLS Group Title
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    yes..

    • one year ago
  16. mathstudent55 Group Title
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    So at least one coordinate may be not an integer. That rules out the first choice.

    • one year ago
  17. DLS Group Title
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    is it C?

    • one year ago
  18. mathstudent55 Group Title
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    That is b, the x-coordinate of the third vertex. b = a/2, and if a is odd, a/2 is not an integer.

    • one year ago
  19. DLS Group Title
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    answer=b?

    • one year ago
  20. mathstudent55 Group Title
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    No, not yet. I was referring to small letter b in my drawing. The answer is C, you're right.

    • one year ago
  21. DLS Group Title
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    yes,thanks!

    • one year ago
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