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Math Help - Determining slope of third line

  1. #1
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    Determining slope of third line

    Hi, Here's an interesting problem:


    I have 2 lines with slopes and y-intercepts.

    I have a third line with just a y-intercept and I need to determine it's slope in the following condition.

    The condition:

    What slope of the third line creates a triangle with an area of 0.1?

    There should be two answers. One on each side of the intersection point of the first two lines.

    How would I set this up to take in multiple line data? Can I reduce the calculation of the slope of the third line to one formula?

    sample data:

    Line 1: 2.41, 120.45
    Line 2: 1.94, 134.26
    Line 3: ?, 25.45

    any help is appreciated...
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  2. #2
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    Re: Determining slope of third line

    You have three lines: y = 2.41x + 120.45, y = 1.94x + 134.26, and y = mx + 25.45

    See where the three lines intersect, and that will give you three points. Find the distance between each point. Then use Heron's formula to calculate area.

    Distance between two points (x_1,y_1),(x_2,y_2): \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

    Heron's formula for area of a triangle: A = \sqrt{p(p-a)(p-b)(p-c)} where p = \dfrac{a+b+c}{2}, and the three sides of the triangle have lengths a,b,c.
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  3. #3
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    Re: Determining slope of third line

    Another way (without the distance formula and Heron's formula) is once you have the three points of intersection (A_x,A_y),(B_x,B_y),(C_x,C_y), the area is \left| \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} \right|
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  4. #4
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    Re: Determining slope of third line

    perfect. thanks
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  5. #5
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    Re: Determining slope of third line

    still can't figure this out though, what's the answer?
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  6. #6
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    Re: Determining slope of third line

    Here are the three equations:

    y = 2.41x + 120.45
    y = 1.94x + 134.26
    y = mx + 25.45

    Let's figure out where the first two lines intersect. Then 2.41x + 120.45 = 1.94x + 134.26. We find x = \dfrac{1381}{47}. If we plug that value in for either line, we will get the y-coordinate. So, our first intersection point is \left(\dfrac{1381}{47}, 2.41\dfrac{1381}{47} + 120.45\right) = \left(A_x,A_y\right).

    The next point of intersection occurs when 2.41x + 120.45 = mx + 25.45. That occurs at \left(\dfrac{9500}{100m-241},2.41\dfrac{9500}{100m-241}+120.45\right) = \left(B_x,B_y\right).

    The last point of intersection occurs when 1.94x + 134.26 = mx + 25.45. This occurs at \left(\dfrac{10881}{100m-194}, 1.94\dfrac{10881}{100m-194}+134.26\right) = \left(C_x,C_y\right).

    So, those are your three points.

    Now, plug those three points into the equation I gave you and solve for m.
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  7. #7
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    Re: Determining slope of third line

    I need more than the slope for a triple intersection. I need the slope that produces a triangle of area 0.1. There should be 2 answers, One triangle on either side of the first intersection point of the two known lines.
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  8. #8
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    Re: Determining slope of third line

    The formula I gave you for area uses an absolute value. That means the two solutions are \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} = 0.1 or \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} = -0.1.
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  9. #9
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    Re: Determining slope of third line

    I only have one intersection point though, too many unknowns
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  10. #10
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    Re: Determining slope of third line

    I gave you three intersection points. Read above. So, you have only one unknown: m.
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