Results 1 to 6 of 6

Thread: Linear Relations.

  1. #1
    Junior Member
    Joined
    Jan 2017
    From
    Australia
    Posts
    46

    Linear Relations.

    A parallelogram has two side lengths of 5 units. Three of its sides have equations y = 0, y = 2,y = 2x. Find the equation of the fourth side.
    Im having trouble drawing out the shape on the graph. Can someone help me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2017
    From
    Finland
    Posts
    34
    Thanks
    10

    Re: Linear Relations.

    Linear Relations.-untitled.png
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    334
    Thanks
    45

    Re: Linear Relations.

    Good morning stateside!

    As someone who has written a lot of math curricula, I am not thrilled with the lack of precision to which the wording of this problem attends. I would prefer "Three of its sides are line segments on the lines with equations y=0, y=2, and y=2x..." But since I believe we all know what the question wants us to do, I digress.

    To be rigorous, let's determine which set of parallel line segments are 5 units long. Suppose the parallel line segments on the lines y=2x+k and y=2x are 5 units in length. Specifically, let's concern ourselves with the line segment on the line y=2x. Since y=2x must intersect both y=0 and y=2, we draw a right triangle that connects y=0 to y=2 with a hypotenuse of 5, and height 2. In this case, using the Pythagorean Theorem, we would get a base of length \sqrt{21}. We have a contradiction because the rise-over-run, i.e. slope, must be 2 by virtue of y=2x; but \frac{2}{\sqrt{21}}\neq 2. If this is confusing, this part is probably optional for you.

    Now, we have two line segments of length 5 apiece that are parallel: one on the line y=0 and the other on the line y=2. The third side is on the line y=2x. How do we get the fourth line segment to complete the parallelogram? Well, we know the fourth line segment must be on a line that is PARALLEL to y=2x. Two lines are parallel when they have the same slope! So the line on which the 4th segment lies is y=2x+k.

    Finally, go back to y=2x. Since y=2x intersects y=0 and y=2, we take note of where these intersections occur on the x-axis. y=2x intersects y=0 where x=0. y=2x intersects y=2 where x=1. So, to have line segments that are five units each, we need y=2x+k to intersect y=0 at x=0+5=5 and y=2x+k to intersect y=2 at x=1+5=6. So set the equations up!

    y=2x+k \implies 0 = 2(5)+k \implies k=-10
    y=2x+k \implies 2 = 2(6)+k \implies k=-10

    The final line segment of your parallelogram lies on the line y=2x-10.
    Check your work by graphing all 4 lines and connecting the segments.

    And as the previous poster stated, there is another solution, i.e. y=2x+10.
    Last edited by abender; Sep 4th 2017 at 02:45 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2017
    From
    Australia
    Posts
    46

    Re: Linear Relations.

    Thanks. I have two questions. This one might be obvious, but why does all the line segments need to equal to 5? Also why did you do two equations at the end? Also I might sound stupid, but y = 0 and y = 2 are those the width of the rhombus or the length?
    Last edited by Ito4; Sep 4th 2017 at 02:51 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    334
    Thanks
    45

    Re: Linear Relations.

    Not a stupid question. The problem stated (more like suggested b/c of poor wording) that two of the segments of the parallelogram were five units long. I showed why the y=2x and y=2x+k segments could not be 5 units long. If I was tutoring you, I could have shown you on paper. I am sorry if my solution was a bit... unclear since it did not accompany pictures.

    y=0 and y=2 are the EQUATIONS of lines that two of the parallelogram's segments lie directly on.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    334
    Thanks
    45

    Re: Linear Relations.

    I did two equations because both got us the same answer. Only one was necessary.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear relations and graphs
    Posted in the New Users Forum
    Replies: 1
    Last Post: Jul 25th 2012, 10:22 PM
  2. Linear Relations
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Jun 4th 2011, 09:17 AM
  3. Application of Linear Law to Non-linear Relations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jan 29th 2010, 08:07 AM
  4. Replies: 1
    Last Post: Jan 29th 2010, 06:01 AM
  5. linear relations
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Jul 11th 2008, 07:13 PM

/mathhelpforum @mathhelpforum