# Reflections Follow-up

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## Objective

SWBAT apply what they've learned about reflections to solve problems

#### Big Idea

Uh..anybody seen the reflection line?...'Cuz it's missing. In this lesson students hone their sleuthing skills as they pursue the equation of the unknown reflection line.

## Activating Prior Knowledge

20 minutes

Where We've Been: Students have just finished a guided discovery activity on reflections in the coordinate plane. They should know, or at least be acquainted with, the fact that the line of reflection is the perpendicular bisector of every segment that joints a point on the preimage to the corresponding point on the image.

Where We're Going: I want students to become more versatile and fluent in their application of what they know about reflections. For example, given preimage and image, find the reflection line. Given preimage and reflection line, find image.

This section is basically a recap of what happened in the last lesson: Defining Reflection.

I remind students how we found that the midpoints of all of the segments joining corresponding pairs of points on the preimage and image were collinear. The line on which these midpoints fell was the line of reflection. So the reflection line goes through all of the midpoints of the segments joining corresponding pairs of points. We also found that the slope of the reflection line was the opposite reciprocal of the slope of the segments joining corresponding pairs of points on the preimage and image. Opposite reciprocal slopes implies perpendicularity (i.e.., intersection at right angles). So the reflection line goes through the midpoint of the segments, and it's perpendicular to the segments. That makes it the perpendicular bisector of all the segments joining corresponding pairs of points on the preimage and image.

Next I model how we used the graph of the reflection line (method 1) and point-slope formula (method 2) to find the equation of the reflection line. I point out that using the graphical method is great, except sometimes the y-intercept and/or the slope cannot be discerned from the graph. Other times, there is no graph at all. So that gives me an excuse to review point-slope formula. I start off by saying that the point-slope formula is a tool for writing equations of lines. Then I show how to use it to derive the equation of the perpendicular bisector from the Defining Reflection lesson. Next, I verify that this is the correct equation via a Sketchpad demo. I basically graph ABCD, and then graph the reflection line. Then I reflect ABCD over the reflection line to see if I get the coordinates for A'B'C'D' that were given in the original problem.

Hopefully, after this section is over, students understand the connection between the perpendicular bisector and reflections and they know how to apply the point-slope formula.

## Pre-Flight Check

30 minutes

Before we go all out and deal with the perpendicular bisector nested within a reflection problem, I check to see if students have the necessary understanding of the perpendicular bisector, the slope criteria for perpendicular lines, and the application of the point-slope formula.

To do this, I give students the Perpendicular Bisector Equation student resource. Students' first reflex seems to be to use A in the point-slope formula. They assume that the first point must be (x1,y1). And consistent with that logic, students tend to throw in the first slope they can find in the place of m.

At this point, I have to take a dramatized moment to remind students that we do not do random plugging and chugging in geometry. In geometry, the geometric ideas and concepts dictate everything.

To illustrate, I lead the class in an analysis of the term perpendicular bisector. To make a long story short, it has to cut a segment in half and make a 90 degree angle when it does so. So suppose we had used A for our point...we would have been finding the equation of a line passing through A, but the perpendicular bisector, geometrically speaking, can't go through A. And what if we randomly used the slope of segment AB? We'd be finding the equation of a line with the same slope as segment AB and passing through A....Hold on, that would be line AB....that's not what we want. We want the perpendicular bisector.

I put the coordinate plane up on the doc cam and start with segment AB. So we're looking for a line that cuts it in half (i.e. (hopefully a student says this before I do) it has to go through the midpoint of segment AB) and is perpendicular to it (i.e., it has a slope that is opposite and reciprocal to the slope of segment AB.

To be clear, in order to use the point-slope formula  to correctly write the equation of a line, we need a point that the line actually passes through, and we need the slope of the line we want to write an equation for.

Hopefully, this little push is enough to get most students headed in the right direction. For the stragglers, I go around and provide additional support.

Before moving on from this section, I come back to the document camera and present the problem from start to finish. Since students have done the problem and should understand the process, I involve the students in my presentation and don't just present as if it were a new problem. I am careful, though, to model the entire thought process so that students can get a coherent view of the problem. Some students have pieced together the components of the solution, but haven't really made sense of how it all fits together.