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Math Help - coordinate geometry problem

  1. #1
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    Hi all,

    could someone please help with the following.

    1. Find h if the point (h,0) is equidistant from the points (-1,7) and (-3,-2)

    thanks

    i tried using the perpendicular but this seems to be flawed.

    y=mx+c

    and

    y=-\frac{1}{m}x+c

    m=\frac{y{2}-y{1}}{x{2}-x{1}}

    the functions were then

    y=\frac{9}{2}x+\frac{23}{2}

    and

    y=-\frac{2}{9}x+\frac{23}{2}

    therefore when y=0

    \frac{2x}{9}=\frac{23}{2}\equiv 4x=207

    h=\frac{207}{4}

    should i be using the midpoint or something?
    Last edited by mr fantastic; May 12th 2009 at 04:10 AM. Reason: Merged posts
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  2. #2
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    1. Find h if the point (h,0) is equidistant from the points (-1,7) and (-3,-2)

    sqrt((h+1)^2+(0-7)^2)=sqrt((h+3)^2+(0+2)^2)
    h=37/4
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  3. #3
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    Thanks for the answer learnmath

    i understand what you did, but have trouble visualising it!
    Attached Thumbnails Attached Thumbnails coordinate geometry problem-problem.jpg  
    Last edited by sammy28; May 12th 2009 at 03:24 AM. Reason: added drawing
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by sammy28 View Post
    i tried using the perpendicular but this seems to be flawed.

    y=mx+c

    and

    y=-\frac{1}{m}x+c
    .....
    ...


    should i be using the midpoint or something?


    Why do you think that the value of "c" will remain same.


    ________________________________________
    Visualizing it




    • Two points are given
    • Make a line joining these points.
    • Find mid-point of the given points in the line
    • Draw a line perpendicular to previous line and passing through midpoint
    • The second line will intersect with x-axis at some point. The x coordinate of this point is value of "h"
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  5. #5
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    Hope this will help.
    Attached Files Attached Files
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  6. #6
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    Thanks ADARSH and learnmath, i got it in the end

    im sure i was supposed to answer this your way learnmath, ie using pythagoras. the drawing does help a lot.

    i made mistake as ADARSH says of assuming c were equal.

    midpoint is \left(-\frac{4}{2},\frac{5}{2}\right)

    perpendicular line is

    y=-\frac{2}{9}x+c

    solving for c by using midpoint gives our perpendicular line equation

    y=-\frac{2}{9}x+\frac{37}{18}

    now find the x intercept

    \frac{2x}{9}=\frac{37}{18} \equiv 36x=333

    therefore

    x=\frac{37}{4}
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