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Slope of Linear function

  1. Nov 26, 2014 #1
    • Member warned about posting problems without the template and with no effort
    Hello,
    I'm having a hard time solving a linear function slope problem.So I would be thankful if someone could provide me with an answer and explanation of the problem.

    The problem is the following: Find the possible slopes of a line that passes through(4,3) so that the portion of the line in the first quadrant forms a triangle of area 27 with the positive coordinate axes.
     
  2. jcsd
  3. Nov 26, 2014 #2

    BvU

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    To determine the triangle area you will need to calculate where the line crosses the axes.
    Start with a drawing. Give the general form of a line through (4,3) with slope m and solve x for y = 0 and y for x = 0.
     
  4. Nov 26, 2014 #3
    Are you able to see how a line can form the hypotenuse of a right triangle with the positive coordinate axes?

    Chet
     
  5. Nov 26, 2014 #4
    I've tried to find where the line crosses the axes, but I ended up with many variables with no answer.

    I would really appreciate it if you could provide the answer with an explanation.
     
  6. Nov 26, 2014 #5
    Show us the details of what you did. Then we can help you. That's what this is all about.

    Chet
     
  7. Nov 26, 2014 #6
    Sorry for not posting my solution attempt. I'm Kind of new to this website.

    I assumed that one of the possible lines crosses the x-axis at (a,0) and the y-axis at (0,b) and that a*b=54 since (Area of triangle*2 )=27*2=ab. I used the the two-intercept form of the line equation.

    The solution attempt: (-b*x/a)+b=y → bx+ay=ab →4b+3a=54
    this is where I ended up
    I tried to assigning numbers to satisfy the equation 4b+3a=54 , and the results I got was( a=12 and b=4.5) and (b=9 and a=6)
    so the slopes are (-3/2) and (-3/8)

    I don't know if there is any other possible slopes or how to make sure there isn't. Also, is there a way to find the slopes without ending up assigning numbers to satisfy 4b+3a=ab; that is, a way to find the answer without dealing with two variables.
     
  8. Nov 26, 2014 #7

    You were very close to having it solved.

    You can still use ab= 54 again by substituting for either b or a in the above equation. This will give you an equation exclusively in terms of either a or b.

    Chet
     
  9. Nov 26, 2014 #8

    Ray Vickson

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    Those are the only two solutions. The easiest way to see this is to use a single variable (slope = ##-s##), and write the equation of the line as
    [tex] y = 3 - s(x-4)[/tex]
    Note that when ##x = 4## we have ##y = 3##, as we need; and the slope is ##-s##, as stated. You can find the x- and y-intercepts in terms of ##s##: to find the x-intercept ##B## (= "base"), set ##y = 0## and solve for ##x## in terms of ##s##. To find the y-intercept ##H## (= "height"), just put ##x = 0##. Now the area ##A = \frac{1}{2} BH## becomes a function of ##s##. Setting ##A = 27## yields a quadratic equation for ##s##, so has at most two roots. In this case it does have exactly two positive, real roots that you can find using the quadratic-root formula.
     
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