Partitioning a line segment (and finding the midpoint)

One of the Common Core standards for geometry is partitioning a directed line segment into a given ratio (G.GPE.6). This falls firmly within the category of Things I Never Learned in School.

Cursory Googling led to a nice little formula, which Shmoop calls the “section formula”: \displaystyle \left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right).

Formulas are nice, but my students don’t do so well with them. As a general rule, they aren’t willing to really commit to memorizing anything. By the end of the year, this results in a crazy mess of half-remembered formulas. I’m trying to avoid that this year, so I’m leaning more toward processes and thinking that are generalizable… even if this means students spend longer solving specific cases than they otherwise would have to.

Thus, the proportions method of finding midpoint. I really like this because last year my students had no idea what the midpoint formula meant… despite the fact that I felt like we drew it out pretty well. On the other hand, my students are really good at proportional reasoning. I also really like this method because it is generalizable: it doesn’t just work for finding the midpoint when given two endpoints, but it also works for finding one endpoint when given the midpoint and another endpoint, or for finding the point that partitions a line segment into a given ratio. It even works when given one end point, the point of division, and the ratio. In other words… it always works. One process to rule them all. Holla.

So here’s what you do: “Find point C on \overline{AB} such that AC:CB=4:3, if A=(2,3) and B=(4,5).”

First, draw a picture. The picture is very important, because when you’re doing these problems, you’re talking about directed line segments, so order matters. I always draw the lines as though they were perfectly horizontal, but you could draw them accurately if you liked. What’s important here is the order of the points.

We use (x,y) for the coordinates of C, because we love algebra and when we don’t know something, we use a variable.

Now you set up a fancy proportion to find the x-coordinate. We need to find distance from A to C considering only the x-coordinates. Hence, x-2. Again, order is important: subtract the more-right coordinate from the more-left coordinate.  Do the same for the distance from C to B and get 4-x. When all is done, you get the proportion \displaystyle \frac{AC}{CB}=\frac{4}{3}=\frac{x-2}{4-x}. Solve for x and you get the x-coordinate of C is 22/7.

Now do it again, but considering the y-coordinates. The proportion is \displaystyle \frac{AC}{CB}=\frac{4}{3}=\frac{y-3}{5-y}. Solve for y and get the y-coordinate of C is 29/7.

So \displaystyle C=(22/7,29/7) and AC:CB=4:3.

If you want to find the midpoint, use a 1:1 ratio. If you want to find an endpoint when you’re given one of the endpoints and the midpoint, just follow the exact same process, and it will all work out in the end. My students grasped this a lot quicker than they did the midpoint shenanigans last year. And they can solve problems they couldn’t have last year. Go team.

If you’d like to derive the midpoint formula, use A=(x_1,y_1), B=(x_2,y_2) and the midpoint C=(x,y). Go through the proportions and out pops \displaystyle C=\left(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2}\right)

Here’s the worksheet I used when I introduced the process. We walked through problem #1 together with me just helping to organize everyone’s thinking. Here’s a solution key that is hopefully legible.


8 thoughts on “Partitioning a line segment (and finding the midpoint)

  1. Wow. Thank you!! I’ve been searching the web for two days trying to figure out how to teach my students this standard. This is definitely the best explanation I’ve read.

  2. I understand this much better, as I am preparing for high school this year. What I don’t understand is how you got you beginning ratio of AC:CB=4:3

    • For this example, AC:CB = 4:3 because that’s what the example I was using in the post gave us: “Find point C on AB such that AC:CB = 4:3, if A = (2,3) and B = (4,5).” I don’t know if that answers your question… let me know if it doesn’t!

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