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Math Help - Locus of Points

  1. #1
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    Locus of Points

    Find an equation that represents the locus of points equidistant from the lines whose equations are y = 3x + 8 and y = 3x - 6?
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    For equations like
    y= mx+ c

    c is the y- intercept (value of y at x=0)

    y = 3x + 8 and y = 3x - 6

    thus y interceept of our two lines are 8 and (-6)

    So if we want a point on y axis equidistant from these lines
    we need to have a value of y-intercept in between them

    -- this means the value of y at x=0 of our answer should be in between 8 and (-6)

    And the line should be parralel so that its equidistant throughout (see the attatchment)

     <br />
y=3x+\frac{8+(-6)}{2} <br /> <br />

    thus answer is
    y=3x+1
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  3. #3
    MHF Contributor
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    Very easy......

    Quote Originally Posted by ADARSH View Post
    For equations like
    y= mx+ c

    c is the y- intercept (value of y at x=0)

    y = 3x + 8 and y = 3x - 6

    thus y interceept of our two lines are 8 and (-6)

    So if we want a point on y axis equidistant from these lines
    we need to have a value of y-intercept in between them

    -- this means the value of y at x=0 of our answer should be in between 8 and (-6)

    And the line should be parralel so that its equidistant throughout (see the attatchment)

     <br />
y=3x+\frac{8+(-6)}{2} <br /> <br />

    thus answer is
    y=3x+1
    Thank you for the easy explanation and diagram.
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