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Math Help - Assistance sought on how to go about solving graphing problem.

  1. #1
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    Assistance sought on how to go about solving graphing problem.

    The text asks: 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.

    I know that the equation for finding the area of a triangle is \frac{1}{2}bh. How do i go about solving this algebraically? The answers in the book are \frac{-3}{2}, \frac{-3}{8}. I do not know how they arrived at these answers. I don't know how you could graph it without first determining more than one set of points. Any help would be greatly appreciated. Thanks.
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  2. #2
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    Re: Assistance sought on how to go about solving graphing problem.

    Let (X, 0) and (0, Y) be the points at which such a line intercepts the x and y axes, respectively. Then the area of the triangle formed is (1/2)XY= 27 we we must have XY= 54.

    Any line through (4,3) can be written y= m(x- 4)+ 3. That means we must have y= m(0-4)+ 3= -4m+ 3= Y and 0= m(X- 4)+ 3 so that mX- 1= 0, mX= 1, and X= 1/m. XY= (-4m+ 3)(1/m)= 54. Solve that for m.
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  3. #3
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    Re: Assistance sought on how to go about solving graphing problem.

    Hello, KhanDisciple!

    Here's another approach . . .


    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.

    Answers: \text{-}\tfrac{3}{2},\;\text{-}\tfrac{3}{8}
    Code:
          |
         b*
          |  *
          |     *  (4,3)
          |        o
          |           *
          |              *
      - - + - - - - - - - - * - -
          |                 a
    A line through (4,3) with slope m
    . . has the equation: y - 3\:=\:m(x-4) \quad\Rightarrow\quad y \:=\:mx + 3 - 4m

    It has x-intercept: \left(\frac{4m-3}{m},\:0\right)

    It has y-intercept: \big(0,\:-[4m-3]\big)

    The area of the triangle is: . A \:=\:\tfrac{1}{2}bh \:=\:\tfrac{1}{2}\left(\frac{4m-3}{m}\right)\big(-[4m-3]\big)


    The area is 27: . -\tfrac{1}{2}\frac{(4m-3)^2}{m} \;=\;27

    . . . . . . . . (4m-3)^2 \:=\:-54m \quad\Rightarrow\quad 16m^2 - 24m + 9 \:=\:-54m

    . . . . . . . 16m^2 + 30m + 9 \:=\:0 \quad\Rightarrow\quad (2m+3)(8m+3) \:=\:0


    Therefore: . \begin{Bmatrix}2m+3 \:=\:0 & \Rightarrow & m \:=\:\text{-}\frac{3}{2} \\ \\[-4mm] 8m+3 \:=\:0 & \Rightarrow & m \:=\:\text{-}\frac{3}{8} \end{Bmatrix}
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    Re: Assistance sought on how to go about solving graphing problem.

    Thanks a ton hallsofivy, and soroban, you guys are the best. Thank you for showing me how to solve this algebraically (the image of the graph helped also), although in hallsofivy's answer I don't see how solving for the x-intercept 0=m(x-4)+3 can be reduced to mx-1=0. I really appreciate it thank you guys.
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