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LynFran

  • one year ago

Find the distance from (0,0) to the line 4x-4y=9

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  1. LynFran
    • one year ago
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    ok so i work this problem ..but keep getting a negative distance...can that be??

  2. Vocaloid
    • one year ago
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    nope, distance is always positive

  3. LynFran
    • one year ago
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    i know thats what bothers mee

  4. anonymous
    • one year ago
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    probably be easier if you write the line as \[y=x-\frac{9}{4}\]

  5. LynFran
    • one year ago
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    but dont we have to use the normal/ perpendicular formula \[d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }\]

  6. anonymous
    • one year ago
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    oh okay not familiar with that what are x and y in that formula ?

  7. LynFran
    • one year ago
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    the point given (0,0)..

  8. anonymous
    • one year ago
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    OOH

  9. anonymous
    • one year ago
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    then this gotta be simple right?

  10. LynFran
    • one year ago
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    but i keep getting a negative distance

  11. anonymous
    • one year ago
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    \[\frac{9}{\sqrt{4^2+4^2}}\]

  12. anonymous
    • one year ago
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    the numerator should be the absolute value your formula has a silly \(\pm\) in it which just means make sure it is positive

  13. LynFran
    • one year ago
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    so 1.5909 would be the distance??

  14. anonymous
    • one year ago
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    \[d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }\] should be \[d=\frac{ |Ax_0+By_0+C| }{ \sqrt{A^2+B^2} }\]

  15. anonymous
    • one year ago
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    idk are you supposed to give a decimal approximation, or the answer as a radical?

  16. LynFran
    • one year ago
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    i didnt say

  17. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=9%2F%28sqrt%2832%29%29

  18. LynFran
    • one year ago
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    ok i have another question Determine K so that the distance from the origin to the line y=km+9; p=6 will be stated

  19. LynFran
    • one year ago
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    @sleepyhead314 .. wat do u think?

  20. sleepyhead314
    • one year ago
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    well for your new question, if you plug in what they gave into the equation... \[\frac{ 9 }{ \sqrt{k^2 + 1} } = 6\]

  21. LynFran
    • one year ago
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    where did u get that equation frm..?

  22. LynFran
    • one year ago
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    did u derive that ?

  23. sleepyhead314
    • one year ago
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    ehh? the equation you gave before...? \[\frac{ Ax + By + C}{ \sqrt{A^2 + B^2} }\]

  24. LynFran
    • one year ago
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    o its the same so 9=C but what about A and B?

  25. sleepyhead314
    • one year ago
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    I just plugged in A = -k B = 1 C = 9 x = 0 y = 0 because y = km + 9 --> -km + y = 9

  26. LynFran
    • one year ago
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    ok so we now solve for K?

  27. sleepyhead314
    • one year ago
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    yep :)

  28. LynFran
    • one year ago
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    \[9=6(\sqrt{K^2+1})\]\[\frac{ 3 }{ 2}=\sqrt{k^2+1}\]\[\frac{ 9 }{ 4 }=k^2+1\]\[\frac{ 5 }{ 4 }=K^2\]\[\frac{ \sqrt{5} }{ 2 }=k\]

  29. sleepyhead314
    • one year ago
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    that's what I got :)

  30. LynFran
    • one year ago
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    ok is there a way to check if its correct though?

  31. LynFran
    • one year ago
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    like when we solve equation we can sub the value we find back into the equation to see that its correct..eg 4x+2y=0 so y=0 and x=0 4(0)+2(0)=0 like this...^^

  32. sleepyhead314
    • one year ago
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    I suppose you could plug in A = (sqrt(5)/2) B = 1 C = 9 x = 0 y = 0 and make sure you get 6 ?

  33. LynFran
    • one year ago
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    ok cool.. thanks to all

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