standard_description: Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.
problem_activity: Activity
Students generalize their observations and explanation from the previous activity to prove the medians of a triangle always coincide. It’s okay if groups don’t include all necessary details in the proof. Anything likely to be missed will be discussed in the activity synthesis.
Launch
Arrange students in groups of 3–4. Give students 2 minutes of work time to read and interpret the beginning of the problem, monitoring for groups that draw a diagram. Then, pull the class back together briefly, inviting a group to share its diagram.
Representing, Conversing: MLR7 Compare and Connect.
Use this routine to prepare students for the whole-class discussion about the proof that the medians of a triangle always coincide. After groups prove that the medians of triangle intersect at a single point, invite them to create a visual display of their work. Then ask students to quietly circulate and observe at least two other visual displays in the room. Give students quiet think time to consider what is the same and what is different about their proofs. Next, ask students to find a partner to discuss what they noticed. Listen for and amplify the language students use to justify why the proof applies to any triangle on the coordinate plane.
Design Principle(s): Cultivate conversation
Engagement: Develop Effort and Persistence.
Provide reminders or checklists that focus on increasing the length of on-task orientation in the face of distractions. For example, provide a checklist of the responsibilities of each group member with pre-assigned sides and medians.
Supports accessibility for: Attention; Social-emotional skills
Student Facing
The goal is to prove that the medians of any triangle intersect at a point. Suppose the vertices of a triangle are
\((0,0), (w,0),\)
and
\((a,b)\)
.
Each student in the group should choose 1 side of the triangle. If your group has 4 people, 2 can work together. Write an expression for the midpoint of the side you chose.
Each student in the group should choose a median. Write an expression for the point that partitions each median in a
\(2:1\)
ratio from the vertex to the midpoint of the opposite side.
Compare the coordinates of the point you found to those of your groupmates. What do you notice?
Explain how these steps prove that the 3 medians of any triangle intersect at a single point.
Student Response
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Anticipated Misconceptions
A diagram is a helpful tool.
Suggest that students share their results with their groups after each step.
Make sure students understand that the
\(2:1\)
ratio applies to the segment that
begins
with the vertex and
ends
with the midpoint of the opposite side of the triangle.
If students struggle to find the point that partitions the median in a
\(2:1\)
ratio, remind them of the notation they used in earlier activities:
\(\frac13A + \frac23B\)
. How does that apply to this problem?
Activity Synthesis
The goal of the discussion is to ensure students have incorporated all the necessary elements in their proof. Here are some questions for discussion:
“What can we conclude if we know that a single point is on 3 different lines?” (It means either the lines coincide, or the lines intersect at that point.)
“How do we know the medians don’t coincide?” (The midpoints are all different because the sides of the triangles don’t coincide.)
“Why did we test a
\(2:1\)
ratio?” (That’s the ratio we measured for the triangle in the warm-up, so we thought it might apply to all triangles.)
“Does this proof apply to a triangle that isn’t positioned like the one in the diagram?” (Yes. We can use a sequence of rigid motions to take any triangle to an image in which a vertex is at
\((0,0)\)
and one side coincides with the
\(x\)
-axis.)


aligns: no