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How do I find the equidistant point?

  1. Aug 29, 2007 #1
    1. Find the coordinates of the point on the line y=3x+1 that is equidistant from (0,0) and (-3,4)



    2. distance formula



    3. I have no idea how to do this. X_X
     
  2. jcsd
  3. Aug 29, 2007 #2
    does anyone know?
     
  4. Aug 29, 2007 #3
    nobody?
     
  5. Aug 29, 2007 #4
    Any point on the line has coordinates (x,3x+1). Set the distance from that to (0,0) equal to the distance from that to (-3,4)
     
  6. Aug 29, 2007 #5
    huh? can u show me? step by step
     
  7. Aug 29, 2007 #6
    The distance between a point (x,y) and the point (-3,4) is [tex]d1=\sqrt{(x+3)^2 + (y-4)^2}[/tex]. The distance between a point (x,y) and the point (0,0) is [tex]d2 = \sqrt{x^2 + y^2}[/tex]. You know what y is from the equation of the line, so substitute that, set d1=d2, and solve for x.
     
  8. Aug 29, 2007 #7

    cristo

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    Staff Emeritus
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    In future, please note that you must show some work before help can be provided-- hence the presence of #3 in the homework posting template. Furthermore, this is not calculus and should be in the precalculus forum.
     
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