Results 1 to 7 of 7

Math Help - Cartesian plane

  1. #1
    Junior Member
    Joined
    Jun 2010
    Posts
    69

    Cartesian plane

    Hi

    To graph in the cartesian plane
    a ---> x^2+y^2 = 9
    would that be a graph where the x,y co-ordinates are (0,0) and the radius of the circle on the graph is 3 ???
    (btw, while I know 0^2+0^2 != 3^2 the only example I have is an equation where (x-10)^2 + y^2  = 16, therefore (x,y) co-ords are (10,0) and radius is 4. From this I deduce the results from the equation above (a))

    b ---> x<=y has co-ords (x,y) = (0,0), which is a diagonal line through the intersection of x and y, from top right to bottom left, shaded area on upper side marking x<=y???

    am i on the right track?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Seems you are wrong.

    from x^2+y^2 = 9 we mean set of points (x,y) on the 2-d carteisian plane which satisfiy the given equation. (0,0) definately doesn't belong to this graph.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2010
    Posts
    69
    I'm just guessing,
    but would this work?
    x^2 = 3^2 - y^2
    x = 3 - y

    then x = 0 and y = 3

    ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    No, not even close

    (3^2-y^2)\neq (3-y)^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2010
    Posts
    69
    Well, i have no idea then
    I would have thought
    if x = 3, or -3
    then y = 0

    or if
    y = 3 or -3
    then
    x = 0

    because
    3^2 + 0^2 = 9 or any combination with 3, -3 and 0

    so the points on the graph could be

    (-3,0), (3,0), (0,-3), (0,3)

    i wish I could draw it here, but basically it looks like a circle with center (0,0) and radius 3. (i am sure you can picture what I mean)

    I don't know how else to approach it?

    edit (sorry in my original post I said the x,y co-ords are (0,0) but now see this is the center and not the (x,y) co-ordinates)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Apr 2009
    Posts
    677
    yes - [tex]x^2+y^2 = 9[\math] is a circle with raidus = 3 and center = (0,0)

    so all the points you said
    (0,3),(0,-3),(3,0),(-3,0)
    are on the circle and there are many more

    But note (0,0) is not on that circle.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2010
    Posts
    69
    aman_cc and pickslides thank you very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cartesian plane....
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 28th 2009, 08:27 AM
  2. Cartesian eqn of a plane
    Posted in the Geometry Forum
    Replies: 2
    Last Post: June 21st 2009, 02:57 PM
  3. cartesian plane
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 17th 2008, 05:29 AM
  4. cartesian plane
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 14th 2008, 05:48 AM
  5. Cartesian Plane
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 12th 2008, 01:54 PM

Search Tags


/mathhelpforum @mathhelpforum