# Determining slope of third line

• Oct 10th 2013, 03:15 PM
Stockgoblin
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...
• Oct 10th 2013, 03:57 PM
SlipEternal
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 $\displaystyle a,b,c$.
• Oct 10th 2013, 04:06 PM
SlipEternal
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 $\displaystyle \left| \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} \right|$
• Oct 13th 2013, 05:03 AM
Stockgoblin
Re: Determining slope of third line
perfect. thanks
• Oct 17th 2013, 11:49 AM
Stockgoblin
Re: Determining slope of third line
still can't figure this out though, what's the answer?
• Oct 17th 2013, 12:09 PM
SlipEternal
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 $\displaystyle m$.
• Oct 17th 2013, 12:15 PM
Stockgoblin
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.
• Oct 17th 2013, 12:19 PM
SlipEternal
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 $\displaystyle \dfrac{A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)}{2} = -0.1$.
• Oct 17th 2013, 12:28 PM
Stockgoblin
Re: Determining slope of third line
I only have one intersection point though, too many unknowns
• Oct 17th 2013, 12:41 PM
SlipEternal
Re: Determining slope of third line
I gave you three intersection points. Read above. So, you have only one unknown: $\displaystyle m$.