# Thread: Determining slope of third line

1. ## 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...

2. ## 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 $a,b,c$.

3. ## 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 $\left| \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} \right|$

4. ## Re: Determining slope of third line

perfect. thanks

5. ## Re: Determining slope of third line

still can't figure this out though, what's the answer?

6. ## 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 $m$.

7. ## 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.

8. ## 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 $\dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} = -0.1$.

9. ## Re: Determining slope of third line

I only have one intersection point though, too many unknowns

10. ## Re: Determining slope of third line

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