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Math Help - Graphing Linear Equations Problem

  1. #1
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    Graphing Linear Equations Problem

    A triangle is formed by the intersection of the lines 2x +3y = 14, 4x -5y = -16 and the x-axis. Find the area of the triangle.

    I started by attempting to convert both given equations into slope intercept form, so I could begin to graph the lines

    2x -2x +3y = 14 -2x

    3y = -2x +14
    3 3

    y = -2x +4.666666667
    3

    4x -4x -5y = -16 -4x

    -5y = -4x -16
    -5 -5

    y = -4x + 3.2
    5


    The y intercept, 4.666666667 is impossible to graph by hand and this is stopping me from going any further.

    What am i doing wrong?
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  2. #2
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    Hi

    You can start by computing the coordinates of the vertices of the triangle ABC
    4x -5y = -16 and y = 0 gives A(-4,0)
    2x +3y = 14 and y = 0 gives B(7,0)
    2x +3y = 14 and 4x -5y = -16 gives C(1,4)
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    You can start by computing the coordinates of the vertices of the triangle ABC
    4x -5y = -16 and y = 0 gives A(-4,0)
    2x +3y = 14 and y = 0 gives B(7,0)
    2x +3y = 14 and 4x -5y = -16 gives C(1,4)
    I understand how you got Vertices A and B, but how did you compute C?
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  4. #4
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    For C you have to solve the system

    2x +3y = 14
    4x -5y = -16

    2x = 14 - 3y => 4x = 28 - 6y
    4x = -16 + 5y

    28 - 6y = -16 + 5y => 11y = 44 => y = 4
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  5. #5
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    Thanks for your help so far.

    I actually managed to go back and work it out.

    I now have a triangle plotted with the area; 7cm x 2.5cm /2 = 8.75cm2


    The next part of the question asks me to find the vertices of a triangle with half the area of the triangle just found.

    Well the area will be 4.375cm2.

    How do I get the vertices from that triangle?
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  6. #6
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    Quote Originally Posted by lee1001 View Post
    Thanks for your help so far.

    I actually managed to go back and work it out.

    I now have a triangle plotted with the area; 7cm x 2.5cm /2 = 8.75cm2
    This is not what I get as area
    Are you sure ?
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  7. #7
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    Quote Originally Posted by running-gag View Post
    This is not what I get as area
    Are you sure ?
    Well i measured out the base and height of the triangle on 1cm squared paper and got 7cm for the base and 2.5cm on the height. So 7(2.5) /2 = 8.75cm2
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  8. #8
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    Just have a look to the sketch below

    The area of the triangle is AB x CH / 2

    A(-4,0) and B(7,0) therefore AB = 11
    C(1,4) and H(1,0) therefore CH = 4

    Area = 11 x 4 / 2 = 22

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  9. #9
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    Ok, it took me a little while to see how you did that calculation but the formula is exactly the same as:

    Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By) /2

    By inputting the vertices co-ordinates I get the same answer.

    Thanks!

    The next part of the problem asks me to find the co-ordinates of the vertices of a triangle with half the area of the one just found.

    Obviously the area of this new triangle will be 11, multiplied by 2 giving 22. How do i begin to deconstruct the formula to find the vertices?
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  10. #10
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    If you have no other constraint, you can use for instance C'(1,2)
    In this case the triangle ABC' has an area equal to 11 because
    AB = 11 and C'H = 2 therefore area(ABC') = 11 x 2 / 2 = 11
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  11. #11
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    Quote Originally Posted by running-gag View Post
    If you have no other constraint, you can use for instance C'(1,2)
    In this case the triangle ABC' has an area equal to 11 because
    AB = 11 and C'H = 2 therefore area(ABC') = 11 x 2 / 2 = 11
    I don't understand.

    How does that allow me to calculate the vertices?
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