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-   -   Vectors and magnitude (https://www.askmehelpdesk.com/showthread.php?t=200765)

  • Mar 31, 2008, 01:36 PM
    Felix2008
    Vectors and magnitude
    I have two questions for you can you help me solve them and then show me how
    you solved it.

    First Question:
    Find

    a) the x-intercept A and y-intercept B of the line.

    b) the coordinates of the point P so that lin AP is perpendicular to
    the line and the magnitude of the line AP = 1. (There are two answers.)

    of the equation (3x)-(4y)=12


    Second Question:

    Is it possible to find a point on a line that is exactly one point from the x-intercept on a line without using a calculator. Here's a better question, using the
    equation Y=(-4/3)X+(16/3)find a point on the line that is exactly one unit away
    from the x-intercept without using a calculator. (There is only two answers)
  • Mar 31, 2008, 04:12 PM
    galactus
    1 Attachment(s)
    Quote:

    Is it possible to find a point on a line that is exactly one point from the x-intercept on a line without using a calculator.Yes, it is here's a better question, using the
    equation Y=(-4/3)X+(16/3)find a point on the line that is exactly one unit away
    from the x-intercept without using a calculator. (There is only two answers)


    We can use the distance formula.

    We can easily find the x-intercept by setting y=0 and solving for x.

    We find quickly that x=4.

    Using the coordinates of the x-intercept which are (4,0).





    But, , sub that in:





    Multiply by 9 to get rid of the fractions:





    Factor:



    Now, it's easy to find the two solutions. Plug those back into the line equation to find their corresponding y values. Those are the two points which are 1 unit along the line from the x-intercept. And no calculator. :)

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