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Math Help - Pre-Calc

  1. #1
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    Pre-Calc

    Find a relationship between x and y such that (x,y) is equidistant from the two points.

    (3,-2) and (-7, 1)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ss103 View Post
    Find a relationship between x and y such that (x,y) is equidistant from the two points.

    (3,-2) and (-7, 1)
    Solve this equation for y

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  3. #3
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    I get part of that. but, i dont understand how you would solve for y when you dont know x..?
    i think i need a little clear up on that :]
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    I get part of that. but, i dont understand how you would solve for y when you dont know x..?
    i think i need a little clear up on that :]
    I posted another reply, but took it down when I realized it did not yield the answer in the form you need. Mathstud28's equation is the way to go. You are not solving for an exact value of (x,y) but rather for the relationship between them -- y in terms of x. You dont need to know the value of x, because you don't need to eliminate it.

    To solve this equation:

    1. Square both sides of the equation
    2. Use subtraction to get both of the y terms on one side and both of the x terms on the other
    3. Expand the squares and simplify
    4. Solve for y in terms of x


    Check your answer against (-2, -0.5) -- the exact values, obtained by the midpoint formula --by plugging them into the resulting function.
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  5. #5
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    Quote Originally Posted by ss103 View Post
    Find a relationship between x and y such that (x,y) is equidistant from the two points.

    (3,-2) and (-7, 1)
    Have you had Geometry? If so, simply construct the perpendicular bisector of the line segment joining the two points.
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  6. #6
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    Thanks for the help everyone!! I solved this using the distance formula as mathstud and sinewave suggested. I got the answer y=10/3x+37/6.
    Does that seem correct?
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  7. #7
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    You are correct. There is an alternative, simpler and easier way to do this btw:

    1. Using midpoint formula, find the point that is equidistant between these two points is:

    ((x1+x2)/2, (y1+y2)/2)

    2. Find the slope of the two given points. (-3/10)

    3. Take the negative reciprocal of the slope (10/3).

    4. Plug in point-slope form (Use the negative reciprocal slope and the point we got from midpoint formula):

    y - y1 = m(x - x1)

    There, you got yourself the equation for a line that is perpendicular to the midpoint.
    Last edited by Chop Suey; July 25th 2008 at 01:30 PM.
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  8. #8
    Member sinewave85's Avatar
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    Quote Originally Posted by ss103 View Post
    Thanks for the help everyone!! I solved this using the distance formula as mathstud and sinewave suggested. I got the answer y=10/3x+37/6.
    Does that seem correct?
    That's what I got, too, and the values check out. Glad to be of assistance!

    p.s. Don't forget to hit the "Thanks" button under usefull posts -- that way the posters get positive reputation points.
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  9. #9
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    Quote Originally Posted by ss103 View Post
    Thanks for the help everyone!! I solved this using the distance formula as mathstud and sinewave suggested. I got the answer y=10/3x+37/6.
    Does that seem correct?
    It does not matter what it "seem"s. It matters that it is correct.

    You could have used the construction of the perpendicular bisector to verify your answer. Working it two ways and getting the same answer is a great confidence builder. Too bad Chop Suey did it for you. You missed a great opportunity to explore and to learn. Don't let the next opportunity get by you.
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