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Thread: 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 $\displaystyle (x_1,y_1),(x_2,y_2)$: $\displaystyle \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

    Heron's formula for area of a triangle: $\displaystyle A = \sqrt{p(p-a)(p-b)(p-c)}$ where $\displaystyle p = \dfrac{a+b+c}{2}$, and the three sides of the triangle have lengths $\displaystyle a,b,c$.
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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 $\displaystyle (A_x,A_y),(B_x,B_y),(C_x,C_y)$, the area is $\displaystyle \left| \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} \right|$
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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:

    $\displaystyle y = 2.41x + 120.45$
    $\displaystyle y = 1.94x + 134.26$
    $\displaystyle y = mx + 25.45$

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

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

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

    So, those are your three points.

    Now, plug those three points into the equation I gave you and solve for $\displaystyle m$.
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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 $\displaystyle \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} = 0.1$ or $\displaystyle \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} = -0.1$.
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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: $\displaystyle m$.
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