Results 1 to 10 of 10

Math Help - Need help with symmetric equation

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    67
    Awards
    1

    Need help with symmetric equation

    Write a symmetric equation and parametric equation for
    a) Line passes through the point A (4,1) with direction vector m = [-3, 1]

    b) passes through the point R(-6, 2) and S(4, -2)

    2) The equation of a line is 2x-3y+12=0. Write equation for line in parametric form

    Thanks in advance

    *plz also explain them to me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    269
    Thanks
    37
    b) passes through the point R(-6, 2) and S(4, -2)


    First, we need a vector, v that will be parallel to the line. We thus let v be the vector that starts at point S and ends at point R.

    v = <-6-4, 2-(-2)> = <-10, 4>.

    Now that we have our direction vector v, all we need to do is choose of our two given points, R or S. Let's use R.

    The equation of our line r = <-6, 2> + t<-10, 4> = <-6-10t, 2+4t>.

    In parametric form:
     x = -6 - 10t
     y = 2 + 4t

    In symmetric form:
     \frac{x+6}{-10} = \frac{y-2}{4} .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2007
    Posts
    67
    Awards
    1
    Thanks for the explanation

    now need part a and 2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    269
    Thanks
    37
    a) Line passes through the point A (4,1) with direction vector m = [-3, 1]


    We have a direction vector, m and a point to work with. This should not be difficult.

    r = <4,1> + t<-3,1> = <4-3t, 1+t>

    parametric form:
     x = 4 - 3t
     y = 1 + t

    symmetric form:
     \frac{x-4}{-3} = \frac{y-1}{1}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2007
    Posts
    67
    Awards
    1
    Thanks again

    now need help with last question
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    269
    Thanks
    37
    2) The equation of a line is 2x-3y+12=0. Write equation for line in parametric form.

    What we are given here is an implicit equation of a line. To convert from an implicit representation to a parametric representation, follow my steps below:

    We let  \vec r be the position vector,  <x, y> and let  \vec n be the vector of coefficients,  <2, -3> .

    That is,
     \vec r = <x, y>
     \vec n = <2, -3>

    Now, we can rewrite 2x-3y+12=0 in terms of  \vec r and  \vec n .

     \vec n \cdot \vec r = -12

    To write this in vector form, we must now do two things. A) we must find a point,  \vec{r_{0}} on the line and B) find a vector  \vec v parallel to the line.

    To find any point on 2x-3y+12=0, we can simply just use the x-intercept. That is, we simultaneously solve for 2x-3y+12=0 and x=0.

     2x - 3y + 12 = 0

     x = 0

     -3y + 12 = 0

     -3y = -12

     3y = 12

     y = \frac{12}{3} = 4

    Therefore, we let point  \vec{r_{0}} = <0, 4> .

    Now, for part B that I mentioned above, find a vector  \vec v parallel to the line.

     \vec n = <2, -3>

     \vec v = \vec{n_{\perp}} = <-3, -2>

    The general parametric equation in vector form is  \vec{r}(t) = \vec{r_{0}} + t \vec v .

    Thus,  \vec{r}(t) = <0, 4> + t<-3, -2> .

    Finally, in parametric non-vector form, we have:

    x = -3t
    y = 4 - 2t

    Good luck!
    -Andy
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    269
    Thanks
    37
    And to double-check my solution for question 2, let's substitute x=-3t and y=4-2t in 2x-3y+12 = 0:

     2x-3y+12 = 0

     2(-3t) - 3(4-2t) + 12 = 0

     -6t -12 + 6t + 12 = 0

    And we know we are correct...

    -Andy
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jul 2007
    Posts
    67
    Awards
    1
    Thanks Andy
    Follow Math Help Forum on Facebook and Google+

  9. #9
    K-D
    K-D is offline
    Newbie
    Joined
    May 2009
    Posts
    1

    Vectors symmetric equations

    This question is givin me trouble big tim

    Que/

    Find symmetric equations for the line

    x=2+3t, y=4-4t, z=-1+6t


    If anyone can solve this, itll be very greatful..
    thanx
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by K-D View Post
    This question is givin me trouble big tim

    Que/

    Find symmetric equations for the line

    x=2+3t, y=4-4t, z=-1+6t


    If anyone can solve this, itll be very greatful..
    thanx
    As abender has shown in the first reply you only have to solve each equation for t and set them equal:

    x=2+3t~\implies~\boxed{t=\dfrac{x-2}3}
    y=4-4t~\implies~\boxed{t=\dfrac{y-4}4}
    z=-1+6t~\implies~\boxed{t=\dfrac{z+1}6}

    Thus the line in symmetric form is:

    \dfrac{x-2}3 = \dfrac{y-4}4 = \dfrac{z+1}6
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 22nd 2010, 02:46 AM
  2. Replies: 4
    Last Post: May 26th 2010, 10:48 AM
  3. Find symmetric equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 4th 2009, 12:10 AM
  4. symmetric equation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 12th 2008, 12:10 PM
  5. Graph of the equation is symmetric
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: September 30th 2007, 02:40 PM

Search Tags


/mathhelpforum @mathhelpforum