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ValentinaT Group Title

What is the length of the longest side of a triangle that has vertices at (-5, 2) (1,-6) and (1,2)?

  • one year ago
  • one year ago

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  1. v4xN0s Group Title
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    Use the distance formula

    • one year ago
  2. ValentinaT Group Title
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    Distance formula?

    • one year ago
  3. v4xN0s Group Title
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    LOL

    • one year ago
  4. v4xN0s Group Title
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    ok its \[d=\sqrt{(x _{1}-x _{2})^2+(y _{1}-y _{2})^2} \]

    • one year ago
  5. MathLegend Group Title
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    I would also suggest labeling your coordinates

    • one year ago
  6. v4xN0s Group Title
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    MathLegend ur not a math legend at all

    • one year ago
  7. MathLegend Group Title
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    A= (-5, 2) B = (1,-6) C = (1,2)

    • one year ago
  8. MathLegend Group Title
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    So when using the distance formula... you can solve for side AB first.

    • one year ago
  9. ValentinaT Group Title
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    Thank you.

    • one year ago
  10. MathLegend Group Title
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    So using that formula, can you label the coordinates as (x1, y1) & (x2,y2)

    • one year ago
  11. hba Group Title
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    |dw:1355599903259:dw|

    • one year ago
  12. MathLegend Group Title
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    A= (-5, 2) B = (1,-6) So lets solve for side AB first... so, "A" comes first right? So let that be your x1 & y1 Then, "B" can be your x2 & y2

    • one year ago
  13. MathLegend Group Title
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    Do you understand so far @ValentinaT ?

    • one year ago
  14. MathLegend Group Title
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    All I did was label them so that we can plug it into that distance formula.

    • one year ago
  15. MathLegend Group Title
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    @ValentinaT let me know when you get back so we can work this out together. :)

    • one year ago
  16. ValentinaT Group Title
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    Yeah, I'm getting it, thank you.

    • one year ago
  17. ValentinaT Group Title
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    \[\frac{ -6 - 2 }{ -5 - 1 } = \frac{ -8 }{ -6}\]

    • one year ago
  18. MathLegend Group Title
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    So for AB (1+5)^2+(-6-2)^2

    • one year ago
  19. MathLegend Group Title
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    I'm sorry the formula is actually x2-x1 and y2-y1 the above poster just mixed up the two.

    • one year ago
  20. ValentinaT Group Title
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    Okay.

    • one year ago
  21. MathLegend Group Title
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    (1+5)^2+(-6-2)^2 (6)^2+(-8)^2 36+64

    • one year ago
  22. MathLegend Group Title
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    36+64 = 100

    • one year ago
  23. MathLegend Group Title
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    \[\sqrt{100}\]

    • one year ago
  24. ValentinaT Group Title
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    10

    • one year ago
  25. MathLegend Group Title
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    Now, we need the square root... because if you notice that entire formula had the square root symbol over it

    • one year ago
  26. MathLegend Group Title
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    Good.

    • one year ago
  27. MathLegend Group Title
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    So side AB = 10

    • one year ago
  28. MathLegend Group Title
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    So that is one side. So lets go for side BC

    • one year ago
  29. MathLegend Group Title
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    B = (1,-6) x1 y1 C = (1,2) x2 y2

    • one year ago
  30. MathLegend Group Title
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    @ValentinaT do you feel comfortable trying it out on your own? Tell me what you get and I'll check to see if your answer for side BC is correct.

    • one year ago
  31. ValentinaT Group Title
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    \[\frac{ 2 - -6 }{ 1 - 1 } = \frac{ 8 }{ 0 }\]

    • one year ago
  32. MathLegend Group Title
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    Remember we are not trying to find a slope.

    • one year ago
  33. MathLegend Group Title
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    We are looking for the distance between the vertices.

    • one year ago
  34. MathLegend Group Title
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    (1-1)^2 + (2+6)^2

    • one year ago
  35. MathLegend Group Title
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    To get that all I did was plug it into the formula. (x2-x1)^2+(y2-y1)^2

    • one year ago
  36. MathLegend Group Title
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    (1-1)^2 + (2+6)^2 Try solving that.

    • one year ago
  37. ValentinaT Group Title
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    Okay, 0 + 64?

    • one year ago
  38. MathLegend Group Title
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    Good

    • one year ago
  39. MathLegend Group Title
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    0 + 64 = 64 \[\sqrt{64}\]

    • one year ago
  40. ValentinaT Group Title
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    So, 8 as the square root.

    • one year ago
  41. MathLegend Group Title
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    Yes, so side BC = 8

    • one year ago
  42. MathLegend Group Title
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    Now, try side AC

    • one year ago
  43. MathLegend Group Title
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    (x2-x1)^2+(y2-y1)^2

    • one year ago
  44. ValentinaT Group Title
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    A = -5, 2 = x1, y1 C = 1, 2 = x2, y2 \[\frac{ 1 - -5^2}{ 2 - 2^2 } = \frac{ 6 }{ 1 } \] so 36?

    • one year ago
  45. MathLegend Group Title
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    (x2-x1)^2+(y2-y1)^2 (1+5)^2 + (2-2)^2 Do you understand this step?

    • one year ago
  46. ValentinaT Group Title
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    36 + 0 Yeah I get it, I just couldn't figure out how to write it.

    • one year ago
  47. MathLegend Group Title
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    Good so if you took the square root of 36 \[\sqrt{36}\]

    • one year ago
  48. MathLegend Group Title
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    So now, we know the length of the longest side is AB

    • one year ago
  49. ValentinaT Group Title
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    Okay, thank you!

    • one year ago
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