# Lesson 11

Perpendicular Lines in the Plane

## 11.1: Revisiting Transformations (5 minutes)

### Warm-up

Students use a coordinate transformation rule to perform a 90-degree rotation. In upcoming activities, students will use transformations to prove the slope criterion for perpendicular lines.

### Student Facing

The image shows quadrilateral \(ABCD\).

Apply the transformation rule \((x,y) \rightarrow (\text-y,x)\) to quadrilateral \(ABCD\). What is the effect of the transformation rule?

### Student Response

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### Activity Synthesis

Invite students to share the effects they came up with. Listen for terms and descriptions such as *rotation*, *90 degrees*, *counterclockwise*, *the origin is the center*, and *perpendicular lines*. If perpendicularity is not mentioned by students, draw the lines \(AD\) and \(AD’\) on the solution or display the image shown here. Invite students to discuss any other observations. (A 90-degree rotation means the original and image lines intersect at a 90-degree angle, which means the lines are perpendicular.)

## 11.2: Make a Conjecture (15 minutes)

### Activity

Students use the results of several slope calculations to make a conjecture that the slopes of perpendicular lines are opposite reciprocals. In an upcoming activity, they will prove their conjecture.

Making a spreadsheet available gives students an opportunity to choose appropriate tools strategically (MP5).

Monitor for students whose conjecture is that the product of the slopes is -1, and for those who mention something about the structure of the two slopes, such as the fact that they have opposite signs or that they are “flipped.”

### Student Facing

- Complete the table with the slope of each segment from the warm-up.
original figure slope image slope product \(AB\) \(BC\) \(CD\) \(DA\) - The image in the warm-up is a 90-degree rotation of the original figure, so each line in the original figure is perpendicular to the corresponding line in the image. Use your slope calculations to make a conjecture about slopes of perpendicular lines.

### Student Response

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### Anticipated Misconceptions

Students may need to be reminded that a *product* is the result of multiplication.

### Activity Synthesis

Select previously identified students to share in this order:

- First, invite a student whose conjecture involved the structures of the slopes to share their idea. Challenge students to use precise language. For example, if they say the slopes are “flipped” or “upside down,” ask them if there is a mathematical term for this concept (reciprocal). Students may want to describe the slopes of perpendicular lines as
*negative*reciprocals.**Opposite reciprocals**is clearer language—if the original slope is negative (such as \(\text-\frac52\)), we can avoid the awkward \(\text-\left(\text-\frac{5}{2}\right)\) and instead jump directly to \(\frac52\). - Next, invite a student whose conjecture involved the product of the slopes being -1 to share. Ask students if this conjecture is different from the previous one. Explore this idea by presenting a slope of \(\frac{a}{b}\) to students, and asking them to provide the opposite reciprocal (\(\text-\frac{b}{a}\)). Instruct students to find the product of these fractions (-1). The conjectures are saying the same thing, because the value that multiplies with \(\frac{a}{b}\) to create -1 is \(\text-\frac{b}{a}\).

If it hasn’t come up, tell students there is at least one exception that doesn’t fit the conjecture: a pair of horizontal and vertical lines. Because a vertical line has no slope, the idea of an opposite reciprocal doesn’t make sense.

Finally, challenge students to calculate the opposite reciprocal of several values, such as \(\frac47\), -5, and \(\frac13\).

## 11.3: Prove It (15 minutes)

### Activity

Students use transformation arguments to prove that the slopes of perpendicular lines that pass through the origin are opposite reciprocals. The proof is extended to all pairs of non-vertical and non-horizontal perpendicular lines in the whole-class synthesis.

### Student Facing

Let’s prove our conjecture about slopes of perpendicular lines for the case where the lines pass through the origin.

- Find the slope of a line passing through the point \((a,b)\) and the origin. Assume the line is not horizontal or vertical.
- Suppose the line is rotated using the transformation rule \((x,y) \rightarrow (\text-y,x)\). Find the coordinates of the images of the points \((a,b)\) and the origin.
- How does the original line relate to the image?
- Find the slope of the image.
- Compare your slopes. What did you just prove?

### Student Response

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### Student Facing

#### Are you ready for more?

We can take any rhombus \(EFGH\) and translate it so that the image of \(E\) is \((0,0)\). We can then rotate using \((0,0)\) as a center so that the image of \(F\) under this sequence is on the positive \(x\)-axis at some point \((c,0)\).

- Suppose these transformations take the point \(H\) to some point \((a,b)\). What must the coordinates of \(G\) be?
- For the values of \(a,b,\) and \(c\) in this problem, why is it true that \(a^2+b^2=c^2\)?
- Prove that the diagonals of the image are perpendicular.
- Prove that the diagonals of the image bisect each other.
- Why does this imply the diagonals of
**all**rhombi are perpendicular bisectors of each other?

### Student Response

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### Anticipated Misconceptions

If students struggle to find the first slope, suggest that they draw a picture of a line passing through the origin and choose a point on the line to label \((a,b)\).

### Activity Synthesis

Ask students if their proof for lines through the origin can be extended to other lines that do not pass through the origin. (Yes. For a line that doesn’t pass through the origin, we could first translate it so it passes through the origin. Translation of lines results in parallel lines, so this translation would not affect the slope. The proof doesn’t apply if we start with a vertical line, though, because it has no slope.)

Then, ask students what the difference is between what they did in the previous activity and what they did in this one. (In the previous activity, we showed that the slopes of perpendicular lines were opposite reciprocals for a few specific cases. In this activity, we showed that this is true for *all* pairs of non-vertical and non-horizontal lines.)

Tell students that the converse is also true: If 2 lines have slopes that are opposite reciprocals, then they are perpendicular. Ask students to add this theorem to their reference charts as you add it to the class reference chart:

Lines are perpendicular if and only if their slopes are opposite reciprocals. (*Theorem*)

Finally, invite students to write the equation of a line that passes through the point \((\text-9, 3)\) and is perpendicular to the line given by \(y-5=\frac27 (x+2)\). The result is \(y-3=\text-\frac72 (x-(\text-9))\).

*Speaking: MLR8 Discussion Supports.*As students share how they know that the slopes of perpendicular lines are opposite reciprocals, press for details by asking students how they know that the slopes of the lines are \(\frac{b}{a}\) and \(\frac{\text-a}{b}\). Show concepts multi-modally by drawing the original line passing through the origin and a point \((a,b)\). Also, draw the image of the line passing through through the origin and the point \((\text-b,a)\). This will help students justify why the slopes of perpendicular lines are opposite reciprocals.

*Design Principle(s): Support sense-making; Optimize output (for justification)*

*Action and Expression: Develop Expression and Communication.*Invite students to talk about their ideas with a partner before sharing them with the class. Display sentence frames to support students when they explain their ideas. For example: “In the previous activity, I . . . , but in this activity, I . . .”, or “I noticed _____ so I . . . .”

*Supports accessibility for: Language; Organization*

## Lesson Synthesis

### Lesson Synthesis

Display this image for all to see.

Ask students to write an equation for any line perpendicular to line \(\ell\), and to sketch a graph of their line. If students struggle to begin, ask them what two pieces of information are required to write the equation of a line (a point and the slope, or the slope and the \(y\)-intercept). Explain that the point or the \(y\)-intercept for the new line can be of their choosing.

Choose several completed equations and graphed lines to display for all to see. Ask students what they notice about the equations and lines. (Sample responses: They are all parallel; they all have slope -4; they have different \(y\)-intercepts.)

If time permits, ask students to find an equation of the image of the line if it was rotated 90 degrees using the origin as a center. One possible answer is: \(y-4=\text-4(x-(\text-3))\).

## 11.4: Cool-down - Another Perpendicular Line (5 minutes)

### Cool-Down

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## Student Lesson Summary

### Student Facing

The diagram shows triangle \(ABC\) and its image, triangle \(AB'C'\), under a 90-degree rotation counterclockwise using the origin as the center. Since the rotation was through 90 degrees, all line segments in the image are perpendicular to the corresponding segments in the original triangle. For example, segment \(AC\) is horizontal, while segment \(AC’\) is vertical.

Look at segments \(AB\) and \(AB’\), which, like the other pairs of segments, are perpendicular. The slope of segment \(AB\) is \(\frac25\), while the slope of segment \(AB’\) is \(\text-\frac{5}{2}\). Notice the relationship between the slopes: They are reciprocals of one another, and have opposite signs. The product of the slopes, \(\frac25 \boldcdot \left(\text-\frac{5}{2}\right)\), is -1. As long as perpendicular lines are not horizontal or vertical, their slopes will be **opposite reciprocals** and have a product of -1.

We can use this fact to help write equations of lines. For example, try writing the equation of a line that passes through the point \((23, \text-30)\) and is perpendicular to a line \(\ell\) represented by \(y=3x+5\). The slope of line \(\ell\) is 3. The slope of any line perpendicular to line \(\ell\) is the opposite reciprocal of 3, or \(\text-\frac13\). Substitute the point \((23, \text-30)\) and the slope \(\text-\frac13\) into the point-slope form to get the equation \(y-(\text-30)=\text-\frac13 (x-23)\).