RSS

Tag Archives: CCSS-M G-GPE.6

Partitioning a Segment

CCSS-M G-GPE.B.6 Find the point on a directed line segment between two given points that partitions the segment in a given ratio.

When I started writing this lesson last year, I did a Google search for ideas. What I found felt out of control to me, to say the least. Should our focus for this standard be on the formula that I found at this site?

Screen Shot 2014-04-05 at 10.25.36 AM

Or the explanation from this link?

Screen Shot 2014-04-05 at 10.59.36 AM

My search also led me to the curriculum site of the government of India, where I found some good and interesting tasks.

We started our lesson with partitioning vertical and horizontal segments.

Screen Shot 2014-03-27 at 9.31.27 AM

And we talked about the need for directed distance.

Screen Shot 2014-04-05 at 10.36.36 AM

When we were ready to move to an oblique segment, it occurred to me that maybe we should estimate where the point would be before we tried to calculate the coordinates of the point. I set up a Drop Points Quick Poll.

03-27-2014 Image002

Screen Shot 2014-03-27 at 9.23.59 AM Screen Shot 2014-03-27 at 9.24.05 AM

75% of my students had the correct answer, although they didn’t all just estimate. (I had not marked a correct answer yet on purpose so that we could look at the graph and decide whether (3,-1) was the desired point.

Screen Shot 2014-03-27 at 9.23.39 AM

How did you get your answer? What did you notice? Some had calculated using the Pythagorean Theorem. They realized that the given segment was 3√13. So we needed to find Q such that QR was 2√13 and QS was √13. Another student was trying to come up with what he called a “modified midpoint” formula. Instead of finding the mean of the x coordinates of the endpoint, he wanted some kind of weighted mean to get the 2:1 ratio.

Screen Shot 2014-03-27 at 3.55.42 PM

I asked students what they saw in the picture.

Endpoints. A segment.

What else?

A triangle.

What kind of a triangle?
A right triangle. The segment is the hypotenuse of a right triangle.

Screen Shot 2014-03-27 at 3.55.49 PM

We found the lengths of the vertical and horizontal sides of the right triangle. Can that help us find the point that we need?

Students began to recognize several right triangles in the diagram that would be helpful. Each with the hypotenuse on the segment. Each with a vertical length of 2 (which is one-third of the whole length 6) and a horizontal length of 3 (which is one-third of the whole length 9).

We tried another problem, for which I had them enter the coordinates instead of dropping a point on the graph.

03-27-2014 Image001

Only 7 students have the correct response. What happened? Fractions, of course.

Screen Shot 2014-04-05 at 10.48.02 AM

Those who got it correct talked about what they did.

Screen Shot 2014-04-05 at 10.49.44 AM

Screen Shot 2014-04-05 at 11.17.53 AM

And so we tried again.

This time we have 10 with the correct response until the students were dismissed to the faculty-student basketball game :-/

Screen Shot 2014-04-05 at 10.50.31 AM

At some point during class, I found what looked like a modified midpoint formula on another site.

Screen Shot 2014-03-27 at 3.56.13 PM

I included it in the students’ notes, but as I reviewed their summative assessments, I notice that no one used it. Instead, they made sense of partitioning a segment using similar triangles and proportional relationships. My hope, as the journey continues, is that they have a better chance of remembering those connections than they would a formula …

 
5 Comments

Posted by on April 5, 2014 in Coordinate Geometry, Geometry

 

Tags: , , , , ,