1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Im not sure how to answer LOCUS question

  1. Mar 22, 2005 #1
    Decribe the locus of points that are 3 units from (-1,2)

    I plotted this point and then up down left and right I made points 3 units away I got

    (-4,2) (-1,-1) (-1,5) and (2,2)

    How do I answer this question what is the correct format to decribe the locus of points that are 3 units from (-1.2)
     
  2. jcsd
  3. Mar 22, 2005 #2
    Your in a swarm of locus! :smile:

    Start with this one since it seems to be the easiest.

    what is the formula for the distance between two points?

    what kind of geometric figure contains points that are all the same distance from a fixed point (or center)?
     
    Last edited: Mar 22, 2005
  4. Mar 22, 2005 #3
    d= square root ((x2-x1)^2+(y2-y1)^2))
    is the formula for distance between two points

    A circle !

    Anyone know any locus sites where I can see examples and problems I cant find any. :mad:
     
    Last edited: Mar 22, 2005
  5. Mar 22, 2005 #4
    plug in the distance for d, and the given point for (x1,y1) and then square it. What does that equation remind you of?

    This could have alternatively been done directly by using the equation of a circle, but doing it this way shows you where that equation comes from.
     
  6. Mar 22, 2005 #5
    What distance are u talkin about the question only gives one point is the distance 3? What am i pluging into?
     
  7. Mar 22, 2005 #6
    I apologize that my post was so terse, I thought that the hint from the first post I did would be enough to set you on your way.

    Perhaps your confused with what a locus is.

    A locus is a set of points satisfying a given condition.
    In this case, the condition is that everypoint in the locus (some set) must be at a distance of 3 units from the point (-1,2)

    in other words, if you take any point in the locus with coordinates (x,y) and compute the distance from (-1,2) using the distance equation, you should get a distance of 3.

    does that help?


    One added note:
    For a locus, you want to find all the points that satisfy the condition. In a plane, the set of all points that are at the same distance from a fixed point (the center) is a circle, so the equation you come up with better be the equation of a circle.
     
    Last edited: Mar 22, 2005
  8. Mar 23, 2005 #7
    OK yes I do understand what a locus is now but how do I come up with an equation for the locus?

    the equation of a circle is r^2=x^2+y^2

    I found an example in my text book the above equation is when the center of the circle is at (0,0) which is not the case in this question.

    I have come up with an equation I dont know if this is correct it is

    [tex] (x+1)^2 + (y-2)^2 =9 [/tex] If this is correct how do I check myself?

    The original question was describe the locus of points that are 3 units from (-1,2) What will my therefore statement be how do I answer the question?
     
    Last edited: Mar 23, 2005
  9. Mar 23, 2005 #8
    There is no set way of doing this, each problem is different, Generally you have to let your mathematical intuition guide you to the solution. I'll go through this problem in detail as an example.

    In this problem,
    You are given the condition that every point in the locus must be 3 units from (-1,2). Intuition should tell you that the locus should form a circle since a circle is defined to be the set of points in a plane that are equidistant from a fixed point. That distance is called the radius. However, knowing this ahead of time could save a little effort, but it is not mandatory to solving the problem.

    So lets forget about circles for now, our goal is to come up with an equation for every point of the locus. Lets start with whats given and see what we can come up with. I'll restate the given just to emphasize its importance.

    1. were given the point (-1,2)
    2. were told that every point in the locus must be 3 units from that point.

    Now if one doesn't know the distance formula, then this problem is virtually unsolvable. You could discover it on your own from the pythagorean theorem, but if a student at that level doesn't know the distance formula, then chances are, they wouldn't know the pythagorean theorem either. Bottom line, you have to start somehwere from the bag of tools you've accumulated thus far, and figure out what to use by what your given information is. Since were given distance between points, we would think to try the distance formula.

    Another way of saying a point is a distance of 3 units from (-1,2), is to say that it should satisfy the distance equation; That is we should get

    [tex] \sqrt{ [x-(-1)]^2 + [y-2]^2 } = 3 [/tex]

    when we plug in the coordinates of the point (x,y) for a point in the locus.

    since we want an equation for all such points, well leave x and y arbitrary (that is we wont specify a specific value for x and y).

    Now technically at this point were done. We've come up with an equation that describes all the points for the locus. It explicitly states that if a point satisfies this equation, then it is in the locus, if it doesn't satisfy the equation then it is not in the locus since it would not be a distance of 3 from (-1,2).

    To make it look nicer, and for reason's you'll see further down this post, we can square both sides to obtain the result

    [tex] (x+1)^2 + (y-2)^2 =9 [/tex]

    your result is correct!

    Your result is correct. To check yourself, recall that we think this should be the equation of a circle.

    The more general equation of a circle is defined to be.

    [tex] (x-h)^2 + (y-k)^2 = r^2 [/tex]

    with center at point (h,k) and radius r.
    notice that when we plug in the values of h=0 and k=0 in this equation we get the same equation you found for a circle with center at (0,0). The more general equation of a circle can actually be derived from the distance equation. (just use h and k for the second coordinate ).

    looking at your result, we see that it can be rewritten to resemble the equation of a circle

    [tex] [x-(-1)]^2 + [y-2]^2 = 3^2 [/tex]

    comparing this to the equation of a circle, we see that the result is an equation for a cirlce with center (-1,2) and radius 3. So our intuition was right.

    To write the answer, it is enough to say that the locus is the set of all points that satisfy the equation,

    [tex] (x+1)^2 + (y-2)^2 =9 [/tex]

    or alternatively you could say that the locus forms a cirlce with the above equation.



    I apologize that this post was so long, but I was trying to give an in depth approach to the thought process behind solving such a problem. In so doing I repeated myself many times but it seemed this was necessary due to the nature of the questions being asked. It seemed apparent that short answers were not having much of an effect.

    -MS
     
    Last edited: Mar 23, 2005
  10. Mar 23, 2005 #9
    OMG MATH STUDENT Thanks soooo much I really needed that I printed out your post for future reference. I wasnt sure if I did the question right because its an online course and they didnt show me an example or the standard form of the equation of a circle. Luckily I found a similar problem in a math text book I had and followed that to get my answer thanks so much again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Im not sure how to answer LOCUS question
Loading...