Results 1 to 13 of 13
Like Tree10Thanks
  • 1 Post By topsquark
  • 2 Post By HallsofIvy
  • 2 Post By greg1313
  • 2 Post By MarkFL
  • 2 Post By MarkFL
  • 1 Post By harpazo

Thread: Equation of the Line

  1. #1
    Super Member harpazo's Avatar
    Joined
    Sep 2014
    From
    NYC
    Posts
    995
    Thanks
    42

    Equation of the Line

    Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

    What does it mean that cuts off the least area from the first quadrant?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,477
    Thanks
    890
    Awards
    1

    Re: Equation of the Line

    Quote Originally Posted by harpazo View Post
    Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

    What does it mean that cuts off the least area from the first quadrant?
    Okay, a couple of things.

    First, you certainly can use the usual y = mx + b for this but it's a bit less transparent at how to use the equation. I'm going to use what is called the intercept-intercept form. This will more easily help us with the area thing.

    The intercept-intercept form talks about both the y-intercept (0, b) and the x-intercept (a, 0). (These are simply points at which the line is crossing the relevant axis.) This equation has the form $\displaystyle y = -\left ( \dfrac{b}{a} \right ) x + b$. Why use this form? Look at the graph. (It looks better blown up.)
    Equation of the Line-area.jpg

    The two points on the axes are the x and y intercepts. By doing things this way we can easily come up with the area of the triangle: $\displaystyle A = \dfrac{1}{2}bh$ (b as in "base length of the triangle" and not y-intercept b.) In the example I have set a = 1, b = 5. So what's the area of the triangle? It's $\displaystyle A = \dfrac{1}{2}(1)(5) = \dfrac{5}{2}$.

    Using different a and b you get different triangular areas. What is the smallest you can get?

    See what you can do with this.

    -Dan
    Last edited by topsquark; Dec 24th 2018 at 09:03 AM.
    Thanks from harpazo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: Equation of the Line

    Any line, other than a vertical line, that passes through (3, 5), crosses both x-axis and y-axis. Let's say that the x-intercept is $\displaystyle (x_0, 0)$ and the y-intercept is $\displaystyle (0, y_0)$. The are "cut off" by that line in the first quadrant is the triangle with vertices $\displaystyle (0, 0)$, $\displaystyle (x_0, 0)$, and $\displaystyle (0, y_0)$. That has area $\displaystyle \frac{1}{2}x_0y_0$. Any non-vertical line through (3, 5) is of the form y= a(x- 3)+ 5. That crosses the x-axis at (3- 5/a, 0) and the y-axis at (0, 5- 3a) so the area of the triangle "cut off" is $\displaystyle \frac{1}{2}(3- 5/a)(5- 3a)= frac{1}{2}(30- 9a- 25/a)$. The derivative of that with respect to a is $\displaystyle \frac{1}{2}(-9+ 25/a^2$. That will be 0 when $\displaystyle 9a^2= 25$ so (a must, or course, be positive) a= 5/3.
    Thanks from topsquark and harpazo
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,477
    Thanks
    890
    Awards
    1

    Re: Equation of the Line

    Oops! I forgot about the (3, 5) thing. (I'll go get the fifty lashings with the wet textbook now.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    2,256
    Thanks
    500

    Re: Equation of the Line

    Didn't you tell us you've decided to concentrate on basic algebra?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2016
    From
    Earth
    Posts
    274
    Thanks
    142

    Re: Equation of the Line

    Quote Originally Posted by HallsofIvy View Post
    Any line, other than a vertical line, that passes through (3, 5), crosses both x-axis and y-axis.
    No, a horizontal line that passes through (3, 5), also does not cross both the x-axis and the y-axis.
    Last edited by greg1313; Dec 24th 2018 at 02:12 PM.
    Thanks from HallsofIvy and harpazo
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member harpazo's Avatar
    Joined
    Sep 2014
    From
    NYC
    Posts
    995
    Thanks
    42

    Re: Equation of the Line

    Quote Originally Posted by DenisB View Post
    Didn't you tell us you've decided to concentrate on basic algebra?
    A friend requested math help. I am currently working on algebra 1 material.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member harpazo's Avatar
    Joined
    Sep 2014
    From
    NYC
    Posts
    995
    Thanks
    42

    Re: Equation of the Line

    Quote Originally Posted by topsquark View Post
    Oops! I forgot about the (3, 5) thing. (I'll go get the fifty lashings with the wet textbook now.)

    -Dan
    We all make mistakes and forget things.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member harpazo's Avatar
    Joined
    Sep 2014
    From
    NYC
    Posts
    995
    Thanks
    42

    Re: Equation of the Line

    Thank you everyone.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,313
    Thanks
    1033

    Re: Equation of the Line

    Quote Originally Posted by harpazo View Post
    A friend requested math help. I am currently working on algebra 1 material.
    I would use Lagrange Multipliers for this problem. Let a and b be positive real numbers representing the x and y intercepts respectively. Suppose the line is constrained to pass though the quadrant I point:

    $\displaystyle \left(x_1,y_1\right)$

    We have the objective function (the area bounded by the line and the coordinate axes in the first quadrant):

    $\displaystyle f(a,b)=\frac{1}{2}ab$

    Subject to the constraint:

    $\displaystyle g(a,b)=\frac{x_1}{a}+\frac{y_1}{b}-1=0$

    This implies the system:

    $\displaystyle \frac{1}{2}b=\lambda\left(-\frac{x_1}{a^2}\right)$

    $\displaystyle \frac{1}{2}a=\lambda\left(-\frac{y_1}{b^2}\right)$

    And this system implies:

    $\displaystyle b=\frac{y_1a}{x_1}$

    Substituting into the constraint, there results:

    $\displaystyle \frac{x_1}{a}+\frac{x_1}{a}-1=0\implies a=2x_1\implies b=2y_1$

    And so we find the line we want:

    $\displaystyle \frac{x}{2x_1}+\frac{y}{2y_1}=1$

    Or:

    $\displaystyle y=-\frac{y_1}{x_1}x+2y_1$

    In this problem, we are given:

    $\displaystyle \left(x_1,y_1\right)=(3,5)$

    Hence:

    $\displaystyle y=-\frac{5}{3}x+10$

    Equation of the Line-mhf_0011.png
    Thanks from topsquark and harpazo
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member harpazo's Avatar
    Joined
    Sep 2014
    From
    NYC
    Posts
    995
    Thanks
    42

    Re: Equation of the Line

    This is over my head. Back to Algebra 1.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,313
    Thanks
    1033

    Re: Equation of the Line

    Quote Originally Posted by harpazo View Post
    This is over my head. Back to Algebra 1.
    Hopefully, you pointed your friend to the help posted in this thread, and it is not over their head, as it is a problem likely given in a calculus course.
    Thanks from topsquark and harpazo
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member harpazo's Avatar
    Joined
    Sep 2014
    From
    NYC
    Posts
    995
    Thanks
    42

    Re: Equation of the Line

    Quote Originally Posted by MarkFL View Post
    Hopefully, you pointed your friend to the help posted in this thread, and it is not over their head, as it is a problem likely given in a calculus course.
    I told my friend about the MHF.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Mar 17th 2014, 07:18 PM
  2. Replies: 1
    Last Post: Aug 30th 2012, 12:56 PM
  3. Replies: 1
    Last Post: Mar 18th 2011, 10:36 AM
  4. Replies: 11
    Last Post: Jun 2nd 2009, 06:08 PM
  5. Replies: 5
    Last Post: Oct 13th 2008, 10:16 AM

/mathhelpforum @mathhelpforum