Unit 5, Lesson 8
Distances and Triangles
Integrated Math 1
8.1
Warm-up
Instructional Routines
NoneMaterials
To Gather
Scientific calculatorsActivity Narrative
In this activity, students practice subtracting coordinates as part of a method for calculating the distance between two points. This will be helpful in an upcoming activity in which students generalize the process for finding distance on a coordinate grid.
Monitor for students who subtract pairs of coordinates directly without first finding the coordinates of point .
Launch
Arrange students in groups of 2. Provide access to scientific calculators. After quiet work time, ask students to compare their responses to their partner’s and to decide if they are both correct, even if they are different. Follow with a whole-class discussion.
Activity
NoneSegment P Q on a coordinate plane, origin O. Horizontal axis scale 0 to 50 by 10’s. Vertical axis scale 0 to 20 by 10’s. Points P(8 comma 12) and Q(53 comma 22) form segment P Q. Point N(53 comma 12) forms segments N Q and P N with dashed segments.
Andre says, “I know that I can find the distance between two points in the plane by drawing in a right triangle and using the Pythagorean Theorem. But I’m not sure how to find the lengths of the legs of the triangle when I can’t just count the squares on the graph.”
Explain to Andre how he can find the lengths of the legs in the triangle in the image. Then, calculate the distance between points and .
Activity Synthesis
Invite a previously identified student, who subtracted coordinates directly, to share that process for finding the lengths of the legs. Emphasize the use of subtraction to find the lengths of the legs. Consider displaying the image from the task and labeling the legs “” and “.”
Student Lesson Summary
The diagram shows the point , along with several points that are 5 units away from . For some points, it is easy to see their distance from . For example, we can count or subtract to see that , , and are all 5 units away from .
For other points, we need to calculate the distance. Let's look at the point . To calculate the distance from to , we can let stand for the distance, and set up the Pythagorean Theorem: Evaluate the left-hand side to find that . Now is the positive number that squares to make 25, which means that really is 5 units away from .
Triangle and 6 points on coordinate plane, origin o. Horizontal axes from negative 2 to 9 by 1’s. Vertical axis from negative 5 to 6 by 1’s. Right triangle vertices at 3 comma 1, 7 comma 1, and 7 comma 4. Bottom leg labeled 7 minus 3 equals 4. Right leg labeled 4 minus 1 equals 3. Hypotenuse labeled D. 6 points at 3 comma negative 4, 6 comma negative 3, 3 comma 6, negative 1 comma 4, negative 2 comma 1, and 1 comma negative 2.
The point also looks like it could be 5 units away from . To find its distance from , we can do a similar calculation: . Evaluating the left side, we get . This means that must be a little more than 5. So is not exactly 5 units away from .
To check if any point is 5 units away from the point , we can use the Pythagorean Theorem to see if is equal to 52 or 25.
By the same reasoning, we can check to see if the distance between any points, and , is equal to a distance, , using the equation . When we solve this equation for , we sometimes call this the distance formula: