LynFran
  • LynFran
Find the distance from (0,0) to the line 4x-4y=9
Mathematics
katieb
  • katieb
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LynFran
  • LynFran
ok so i work this problem ..but keep getting a negative distance...can that be??
Vocaloid
  • Vocaloid
nope, distance is always positive
LynFran
  • LynFran
i know thats what bothers mee

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anonymous
  • anonymous
probably be easier if you write the line as \[y=x-\frac{9}{4}\]
LynFran
  • LynFran
but dont we have to use the normal/ perpendicular formula \[d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }\]
anonymous
  • anonymous
oh okay not familiar with that what are x and y in that formula ?
LynFran
  • LynFran
the point given (0,0)..
anonymous
  • anonymous
OOH
anonymous
  • anonymous
then this gotta be simple right?
LynFran
  • LynFran
but i keep getting a negative distance
anonymous
  • anonymous
\[\frac{9}{\sqrt{4^2+4^2}}\]
anonymous
  • anonymous
the numerator should be the absolute value your formula has a silly \(\pm\) in it which just means make sure it is positive
LynFran
  • LynFran
so 1.5909 would be the distance??
anonymous
  • anonymous
\[d=\frac{ Ax+By+C }{ \pm \sqrt{A^2+B^2} }\] should be \[d=\frac{ |Ax_0+By_0+C| }{ \sqrt{A^2+B^2} }\]
anonymous
  • anonymous
idk are you supposed to give a decimal approximation, or the answer as a radical?
LynFran
  • LynFran
i didnt say
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=9%2F%28sqrt%2832%29%29
LynFran
  • LynFran
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
LynFran
  • LynFran
@sleepyhead314 .. wat do u think?
sleepyhead314
  • sleepyhead314
well for your new question, if you plug in what they gave into the equation... \[\frac{ 9 }{ \sqrt{k^2 + 1} } = 6\]
LynFran
  • LynFran
where did u get that equation frm..?
LynFran
  • LynFran
did u derive that ?
sleepyhead314
  • sleepyhead314
ehh? the equation you gave before...? \[\frac{ Ax + By + C}{ \sqrt{A^2 + B^2} }\]
LynFran
  • LynFran
o its the same so 9=C but what about A and B?
sleepyhead314
  • sleepyhead314
I just plugged in A = -k B = 1 C = 9 x = 0 y = 0 because y = km + 9 --> -km + y = 9
LynFran
  • LynFran
ok so we now solve for K?
sleepyhead314
  • sleepyhead314
yep :)
LynFran
  • LynFran
\[9=6(\sqrt{K^2+1})\]\[\frac{ 3 }{ 2}=\sqrt{k^2+1}\]\[\frac{ 9 }{ 4 }=k^2+1\]\[\frac{ 5 }{ 4 }=K^2\]\[\frac{ \sqrt{5} }{ 2 }=k\]
sleepyhead314
  • sleepyhead314
that's what I got :)
LynFran
  • LynFran
ok is there a way to check if its correct though?
sleepyhead314
  • sleepyhead314
http://www.wolframalpha.com/input/?i=distance+between+%280%2C0%29+and+y+%3D+%28sqrt%285%29%2F2%29x+%2B+9
LynFran
  • LynFran
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...^^
sleepyhead314
  • sleepyhead314
I suppose you could plug in A = (sqrt(5)/2) B = 1 C = 9 x = 0 y = 0 and make sure you get 6 ?
LynFran
  • LynFran
ok cool.. thanks to all

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