Lesson 12Using Equations for Lines

Learning Goal

Let’s write equations for lines.

Learning Targets

  • I can find an equation for a line and use that to decide which points are on that line.

Lesson Terms

  • similar
  • slope

Warm Up: Missing Center

Problem 1

A dilation with scale factor 2 sends to . Where is the center of the dilation?

Two points labeled A and B with point A below and to the right of point B.

Activity 1: Writing Relationships from Two Points

Problem 1

Here is a line.

A line with three points: (5,3), (7,7), (x,y)
  1. Using what you know about similar triangles, find an equation for the line in the diagram.

  2. What is the slope of this line? Does it appear in your equation?

  3. Is also on the line? How do you know?

  4. Is also on the line?

Are you ready for more?

Problem 1

There are many different ways to write down an equation for a line like the one in the problem. Does represent the line? What about ? What about ? Explain your reasoning.

Activity 2: Dilations and Slope Triangles

Problem 1

  1. Here is triangle .

    • Draw the dilation of triangle with center and scale factor 2.

    • Draw the dilation of triangle with center and scale factor 2.5.

    Triangle ABC with coordinates: A (0,1), B (2,1), and C (2,2).
  2. Where is mapped by the dilation with center and scale factor ?

  3. For which scale factor does the dilation with center send to ? Explain how you know.

Lesson Summary

We can use what we know about slope to decide if a point lies on a line. Here is a line with a few points labeled.

A line graphed in the x y plane with the origin labeled O. The number 1 through 6 are indicated on each axis. The line begins in quadrant 3, slants upward and to the right passing through the points zero comma one, x comma y, and 2 comma 5.

The slope triangle with vertices and gives a slope of . The slope triangle with vertices and gives a slope of . Since these slopes are the same, is an equation for the line. So, if we want to check whether or not the point lies on this line, we can check that . Since is a solution to the equation, it is on the line!