Centroid of a Triangle
The centroid of a triangle is the point where all three medians intersect. It is one of the four classical centres of a triangle (centroid, orthocentre, circumcentre, incentre).
In Chapter 7 (Coordinate Geometry) of the NCERT Class 10 textbook, the centroid formula is derived using the section formula. The centroid always divides each median in the ratio 2:1 from the vertex.
The centroid is also called the centre of gravity of the triangle. If you cut a triangle from a uniform sheet of cardboard, it will balance perfectly on a pin placed at the centroid.
The centroid has several remarkable properties. It always lies inside the triangle, regardless of whether the triangle is acute, obtuse, or right-angled. This distinguishes it from the circumcentre (which can be outside for obtuse triangles) and the orthocentre (which is also external for obtuse triangles).
The centroid formula is elegantly simple: the coordinates of the centroid are the arithmetic mean of the x-coordinates and the arithmetic mean of the y-coordinates of the three vertices. This simplicity makes it one of the most computationally efficient formulas in coordinate geometry.
The concept of centroid extends beyond triangles. In physics, the centre of mass of a system of particles is calculated using the same averaging principle. In data science, the centroid of a cluster of points is the average of all point coordinates, forming the basis of the k-means clustering algorithm.
What is Centroid of a Triangle - Formula, Properties, Derivation & Examples?
Definition: The centroid of a triangle is the point of concurrence (intersection) of its three medians.
Median: A line segment joining a vertex to the midpoint of the opposite side.
Properties of the centroid:
- The centroid divides each median in the ratio 2:1 from the vertex.
- The centroid always lies inside the triangle.
- The centroid is the centre of gravity (balance point) of the triangle.
- A triangle has exactly one centroid.
- The centroid divides the triangle into six smaller triangles of equal area.
Comparison of triangle centres:
| Centre | Intersection Of | Always Inside? |
|---|---|---|
| Centroid | Medians | Yes (always) |
| Incentre | Angle bisectors | Yes (always) |
| Circumcentre | Perpendicular bisectors | No (outside for obtuse) |
| Orthocentre | Altitudes | No (outside for obtuse) |
Special cases:
- In an equilateral triangle, all four centres (centroid, incentre, circumcentre, orthocentre) coincide at the same point.
- In an isosceles triangle, the centroid lies on the axis of symmetry.
- In a right triangle with legs along the axes, the centroid is at (a/3, b/3) where a and b are the leg lengths.
Centroid of a Triangle Formula
Centroid Formula:
If the vertices of a triangle are A(x1, y1), B(x2, y2), C(x3, y3), then:
Centroid G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)
In words:
- The x-coordinate of the centroid = average of all three x-coordinates
- The y-coordinate of the centroid = average of all three y-coordinates
Key relationship:
The centroid G divides the median from vertex A to the midpoint M of BC in the ratio AG:GM = 2:1.
Derivation and Proof
Derivation of the Centroid Formula:
Let the vertices be A(x1, y1), B(x2, y2), C(x3, y3).
- Find the midpoint of BC:
Let M be the midpoint of BC.
M = ((x2 + x3)/2, (y2 + y3)/2) - The median AM:
The median from A goes to M. - The centroid G divides AM in the ratio 2:1 (from A).
Using the section formula:
G = ((2 x (x2+x3)/2 + 1 x x1) / (2+1), (2 x (y2+y3)/2 + 1 x y1) / (2+1)) - Simplify:
G = ((x2 + x3 + x1)/3, (y2 + y3 + y1)/3) - Therefore:
G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)
The same result is obtained regardless of which vertex and which median we use, confirming that all three medians meet at this single point.
Alternative derivation using vector approach:
The centroid can also be understood as the average position vector of the three vertices. If the position vectors of A, B, C are represented as vectors a, b, c, then the centroid G has position vector (a + b + c)/3. This is exactly the arithmetic mean, which explains why the formula is so simple.
Why the centroid divides each median in ratio 2:1:
Consider the median from A to the midpoint M of BC. The midpoint M = ((x2+x3)/2, (y2+y3)/2). The point G that divides AM in ratio 2:1 from A is:
- G = ((1 x x1 + 2 x (x2+x3)/2)/(1+2), (1 x y1 + 2 x (y2+y3)/2)/(1+2))
- G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)
The same result holds for medians from B and C, proving that all three medians are concurrent at this point.
Types and Properties
Types of problems involving the centroid:
| Problem Type | Approach |
|---|---|
| Find centroid given vertices | Apply G = ((x1+x2+x3)/3, (y1+y2+y3)/3) |
| Find a vertex given centroid and two vertices | Set up equations from the centroid formula and solve |
| Median-related problems | Find the midpoint, then use the 2:1 ratio property |
| Area-related problems | The centroid divides the triangle into 6 equal-area sub-triangles |
| Verify centroid property | Show that all three medians pass through the same point |
Important properties for problem-solving:
- The centroid of a triangle with vertices on the coordinate axes has predictable locations — for example, if A = (a, 0), B = (0, b), C = (0, 0), the centroid is (a/3, b/3).
- If all three vertices are moved by the same vector (translation), the centroid also moves by the same vector.
- If all coordinates are multiplied by the same factor k (scaling from origin), the centroid coordinates are also multiplied by k.
- The centroid of a degenerate triangle (where all three vertices are collinear) lies on the line containing the vertices and is the average of their positions.
Methods
Method 1: Direct Formula
- Identify coordinates of the three vertices.
- Add all x-coordinates and divide by 3.
- Add all y-coordinates and divide by 3.
- The result is the centroid.
Method 2: Using Section Formula on a Median
- Find the midpoint M of one side (e.g., BC).
- The centroid divides the median from the opposite vertex to M in ratio 2:1.
- Apply the section formula with ratio 2:1.
Method 3: Finding a Missing Vertex
- Let the unknown vertex be (a, b).
- Use (x1 + x2 + a)/3 = Gx and (y1 + y2 + b)/3 = Gy.
- Solve for a and b.
Method 4: Centroid and Area Relationship
The centroid divides the triangle into:
- Three smaller triangles (formed by connecting the centroid to the three vertices), each having area = (1/3) x area of original triangle.
- Six smallest triangles (formed by the three medians), each having area = (1/6) x area of original triangle.
This property can be used to solve problems where the area of a sub-triangle formed with the centroid is given.
Method 5: Centroid of Triangle with Midpoints
The centroid of a triangle is the same as the centroid of the triangle formed by the midpoints of its sides. This is because the medial triangle has the same centroid as the original triangle.
Solved Examples
Example 1: Finding the Centroid
Problem: Find the centroid of the triangle with vertices A(2, 4), B(6, 2), C(4, 8).
Solution:
Given: (x1, y1) = (2, 4), (x2, y2) = (6, 2), (x3, y3) = (4, 8)
Centroid G:
- Gx = (2 + 6 + 4)/3 = 12/3 = 4
- Gy = (4 + 2 + 8)/3 = 14/3
Answer: Centroid = (4, 14/3).
Example 2: Centroid with Negative Coordinates
Problem: Find the centroid of the triangle with vertices P(-1, 3), Q(5, -2), R(2, 7).
Solution:
- Gx = (-1 + 5 + 2)/3 = 6/3 = 2
- Gy = (3 + (-2) + 7)/3 = 8/3
Answer: Centroid = (2, 8/3).
Example 3: Finding a Missing Vertex
Problem: The centroid of a triangle is G(3, 5). Two vertices are A(1, 7) and B(5, 3). Find the third vertex C.
Solution:
Let C = (a, b).
- (1 + 5 + a)/3 = 3 → 6 + a = 9 → a = 3
- (7 + 3 + b)/3 = 5 → 10 + b = 15 → b = 5
Answer: C = (3, 5).
Example 4: Centroid at the Origin
Problem: If the centroid of a triangle with vertices A(a, -2), B(-3, b), C(4, 5) is at the origin, find a and b.
Solution:
- (a + (-3) + 4)/3 = 0 → a + 1 = 0 → a = -1
- (-2 + b + 5)/3 = 0 → b + 3 = 0 → b = -3
Answer: a = -1, b = -3.
Example 5: Verifying the 2:1 Ratio
Problem: Verify that the centroid of triangle A(0, 0), B(6, 0), C(0, 6) divides the median from A in the ratio 2:1.
Solution:
Step 1: Centroid G = (0+6+0)/3, (0+0+6)/3 = (2, 2)
Step 2: Midpoint M of BC = ((6+0)/2, (0+6)/2) = (3, 3)
Step 3: Check if G divides AM in ratio 2:1.
- Using section formula: ((2x3+1x0)/(2+1), (2x3+1x0)/(2+1)) = (6/3, 6/3) = (2, 2)
This matches G = (2, 2). Verified.
Step 4: AG = sqrt((2-0)^2 + (2-0)^2) = sqrt(8) = 2sqrt(2)
GM = sqrt((3-2)^2 + (3-2)^2) = sqrt(2)
AG/GM = 2sqrt(2)/sqrt(2) = 2/1. Verified.
Answer: The centroid divides the median from A in ratio 2:1. Confirmed.
Example 6: Centroid of a Right Triangle
Problem: Find the centroid of the right triangle with vertices at (0, 0), (12, 0), and (0, 9).
Solution:
- Gx = (0 + 12 + 0)/3 = 4
- Gy = (0 + 0 + 9)/3 = 3
Answer: Centroid = (4, 3).
Example 7: Distance from Centroid to a Vertex
Problem: Find the distance from the centroid to vertex A for the triangle A(1, 1), B(5, 1), C(3, 7).
Solution:
Step 1: Centroid G = ((1+5+3)/3, (1+1+7)/3) = (3, 3)
Step 2: Distance AG = sqrt((3-1)^2 + (3-1)^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2) units
Answer: Distance from centroid to A = 2sqrt(2) units (approximately 2.83 units).
Example 8: Centroid and Area Relationship
Problem: The triangle has vertices A(0, 0), B(6, 0), C(3, 6). Find the area of the triangle formed by the centroid and two vertices A and B.
Solution:
Step 1: Centroid G = ((0+6+3)/3, (0+0+6)/3) = (3, 2)
Step 2: Area of triangle AGB with vertices A(0,0), G(3,2), B(6,0):
- Area = (1/2)|0(2-0) + 3(0-0) + 6(0-2)|
- = (1/2)|0 + 0 - 12| = 6 sq units
Step 3: Area of original triangle ABC:
- = (1/2)|0(0-6) + 6(6-0) + 3(0-0)|
- = (1/2)|0 + 36 + 0| = 18 sq units
Step 4: Ratio = 6/18 = 1/3. The centroid divides the triangle into 3 pairs of equal triangles, each with area = (1/3) of the original.
Answer: Area of triangle AGB = 6 sq units = (1/3) of original triangle area.
Example 9: Finding Centroid with Fractional Coordinates
Problem: Find the centroid of the triangle with vertices (1/2, 3/2), (5/2, 1/2), (3/2, 7/2).
Solution:
- Gx = (1/2 + 5/2 + 3/2)/3 = (9/2)/3 = 9/6 = 3/2
- Gy = (3/2 + 1/2 + 7/2)/3 = (11/2)/3 = 11/6
Answer: Centroid = (3/2, 11/6).
Example 10: Centroid Given Midpoints of Sides
Problem: The midpoints of the sides of a triangle are D(3, 1), E(5, 3), F(1, 3). Find the centroid of the original triangle.
Solution:
Key Property: The centroid of a triangle = centroid of the triangle formed by the midpoints of its sides.
- Gx = (3 + 5 + 1)/3 = 9/3 = 3
- Gy = (1 + 3 + 3)/3 = 7/3
Answer: Centroid = (3, 7/3).
Example 11: Proving the Centroid Divides the Triangle into Equal-Area Sub-Triangles
Problem: Triangle ABC has vertices A(0, 0), B(6, 0), C(0, 9). Verify that the centroid divides the triangle into three sub-triangles of equal area.
Solution:
Step 1: Centroid G = ((0+6+0)/3, (0+0+9)/3) = (2, 3)
Step 2: Area of original triangle ABC = (1/2)|0(0-9) + 6(9-0) + 0(0-0)| = (1/2)(54) = 27 sq units
Step 3: Area of triangle AGB:
- = (1/2)|0(0-0) + 2(0-0) + 6(0-3)| = (1/2)|(-18)| = 9 sq units
Step 4: Area of triangle BGC:
- = (1/2)|6(3-0) + 2(9-0) + 0(0-3)| = (1/2)|18+18| = ... recalculate
- = (1/2)|6(3-9) + 2(9-0) + 0(0-3)| = (1/2)|(-36)+18+0| = 9 sq units
Step 5: Area of triangle AGC:
- = (1/2)|0(3-9) + 2(9-0) + 0(0-3)| = (1/2)|0+18+0| = 9 sq units
Each sub-triangle has area 9 sq units = 27/3. Verified.
Example 12: Finding the Centroid of an Equilateral Triangle
Problem: An equilateral triangle has one vertex at A(0, 0) and another at B(6, 0). The third vertex C is above the x-axis. Find the centroid.
Solution:
Step 1: For an equilateral triangle with side 6, the third vertex C has coordinates:
- x = (0+6)/2 = 3 (midpoint of AB horizontally)
- y = 6 x sqrt(3)/2 = 3sqrt(3) (height of equilateral triangle)
- So C = (3, 3sqrt(3))
Step 2: Centroid G = ((0+6+3)/3, (0+0+3sqrt(3))/3) = (3, sqrt(3))
Answer: Centroid = (3, sqrt(3)), which is approximately (3, 1.732).
Example 13: Centroid and the Coordinate Axes
Problem: A triangle has vertices on the x-axis at (2, 0) and (8, 0), and the third vertex at (5, 9). Find the centroid and verify it lies inside the triangle.
Solution:
- Centroid G = ((2+8+5)/3, (0+0+9)/3) = (15/3, 9/3) = (5, 3)
Verification: The base is along the x-axis from (2,0) to (8,0). The third vertex is at (5,9) above the base. The centroid (5, 3) has y-coordinate 3, which is between 0 (base) and 9 (apex), and x-coordinate 5, which is between 2 and 8. So the centroid lies inside the triangle.
Answer: Centroid = (5, 3).
Real-World Applications
Applications of the Centroid:
- Centre of gravity: The centroid is the balance point of a triangular lamina. Engineers use this to determine where to support triangular structures.
- Civil engineering: Load distribution in triangular trusses and bridge components is calculated using the centroid.
- Computer graphics: The centroid is used for positioning labels, calculating centres of triangular meshes, and in rendering algorithms.
- Navigation: Finding the "average position" of three landmarks for triangulation purposes.
- Statistics: The concept extends to the centroid of a set of data points (the mean of coordinates), used in clustering algorithms (k-means).
- Physics: The centre of mass of a uniform triangular plate coincides with its centroid.
Key Points to Remember
- The centroid is the intersection point of all three medians of a triangle.
- Centroid G = ((x1+x2+x3)/3, (y1+y2+y3)/3).
- The centroid divides each median in the ratio 2:1 from the vertex.
- The centroid always lies inside the triangle (for all types of triangles).
- The centroid is the centre of gravity of a uniform triangular lamina.
- The centroid divides the triangle into six smaller triangles of equal area.
- Each of the three sub-triangles formed by the centroid and a side has area = (1/3) of the original.
- The centroid of a triangle equals the centroid of the triangle formed by the midpoints of its sides.
- The formula is derived using the section formula applied to a median with ratio 2:1.
- The centroid coordinates are simply the arithmetic mean of the vertex coordinates.
Practice Problems
- Find the centroid of the triangle with vertices (2, -3), (4, 5), (-6, 7).
- The centroid of a triangle is (4, -2). Two vertices are (3, -5) and (7, 1). Find the third vertex.
- If the centroid of a triangle with vertices (x, 2), (1, y), (5, -1) is (2, 3), find x and y.
- Find the centroid of a right triangle with hypotenuse endpoints at (0, 8) and (6, 0) and the right angle at the origin.
- Prove that the centroid of an equilateral triangle with one vertex at the origin and base on the x-axis lies on the y-axis.
- Show that the centroid divides the median from vertex A(1, 2) to the midpoint of BC (B(5, 0), C(3, 6)) in ratio 2:1.
- The medians of a triangle meet at (2, 4). If two vertices are at the origin and (6, 0), find the third vertex.
- Find the area of the triangle formed by the centroid and two vertices of triangle (0, 0), (8, 0), (4, 6).
Frequently Asked Questions
Q1. What is the centroid of a triangle?
The centroid is the point where all three medians of a triangle intersect. A median connects a vertex to the midpoint of the opposite side.
Q2. What is the centroid formula?
For a triangle with vertices (x1,y1), (x2,y2), (x3,y3), the centroid is G = ((x1+x2+x3)/3, (y1+y2+y3)/3).
Q3. In what ratio does the centroid divide a median?
The centroid divides each median in the ratio 2:1, measured from the vertex to the midpoint of the opposite side.
Q4. Does the centroid always lie inside the triangle?
Yes. Unlike the circumcentre or orthocentre, the centroid always lies inside the triangle, regardless of whether the triangle is acute, obtuse, or right-angled.
Q5. What is the physical significance of the centroid?
The centroid is the centre of gravity (balance point) of a uniform triangular lamina. A triangular plate will balance perfectly on a support placed at its centroid.
Q6. How is the centroid formula derived?
By finding the midpoint M of one side (say BC), then applying the section formula to divide the median AM in the ratio 2:1 from vertex A.
Q7. How does the centroid divide the triangle's area?
The three medians divide the triangle into 6 smaller triangles of equal area. Each median divides the triangle into two triangles of equal area.
Q8. Is the centroid the same as the incentre?
No. The centroid is the intersection of medians; the incentre is the intersection of angle bisectors. They coincide only in an equilateral triangle.
Related Topics
- Midpoint Formula
- Section Formula
- Area of Triangle Using Coordinates
- Medians and Altitudes of Triangle
- Introduction to Coordinate Geometry
- Cartesian Plane
- Plotting Points in Four Quadrants
- Abscissa and Ordinate
- Distance Formula
- Collinearity Using Distance Formula
- External Division (Section Formula)
- Coordinate Geometry Word Problems
- Equation of a Line (Introduction)
- Quadrilateral Using Coordinate Geometry
