Centroid, Incentre, Orthocentre and Circumcentre | NIMCET

Quick answer: For a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$ and opposite sides $a, b, c$: the centroid is $G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$ and the incentre is $I\left(\frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c}\right)$. The circumcentre O is equidistant from all vertices, and the orthocentre H is where the altitudes meet. O, G and H lie on the Euler line, and G divides OH so that $OG : GH = 1 : 2$.

These formulas rely on the distance and section formulas.

The four centres

Coordinate formulas

Special triangles

Euler line

O, G and H are collinear, and G divides OH internally in the ratio $1 : 2$:

$G = \frac{2O + H}{3}$

So if two of them are known, the third follows: $H = 3G - 2O$.

Locus

A locus is the path traced by a point that satisfies a given condition.

Method:

Step 1: Let the moving point be $(h, k)$.

Step 2: Write the condition as an equation in $h$ and $k$.

Step 3: Eliminate any parameter, then replace $h$ by $x$ and $k$ by $y$.

Worked examples

Example 1: centroid

Find the centroid of $(1, 2)$, $(3, -4)$ and $(5, 8)$.

$G = \left(\frac{9}{3}, \frac{6}{3}\right) = (3, 2)$

Example 2: incentre

Find the incentre of the triangle with vertices $A(0, 0)$, $B(4, 0)$ and $C(0, 3)$.

Step 1: Sides: $a = BC = 5$, $b = CA = 3$, $c = AB = 4$.

Step 2:

$I = \left(\frac{5 \cdot 0 + 3 \cdot 4 + 4 \cdot 0}{12}, \frac{5 \cdot 0 + 3 \cdot 0 + 4 \cdot 3}{12}\right) = (1, 1)$

Check: the inradius of a 3-4-5 triangle is 1, and $(1, 1)$ is 1 unit from both axes ✓.

Example 3: right triangle centres

For the same triangle, the right angle is at A, so $H = (0, 0)$ and O is the midpoint of BC, $\left(2, \frac{3}{2}\right)$. Then

$G = \frac{2O + H}{3} = \left(\frac{4}{3}, 1\right)$

which matches $\left(\frac{0 + 4 + 0}{3}, \frac{0 + 0 + 3}{3}\right)$ ✓.

Example 4: locus

Find the locus of a point equidistant from $(1, 0)$ and $(-1, 0)$.

$(h - 1)^2 + k^2 = (h + 1)^2 + k^2$ gives $-4h = 0$, so the locus is $x = 0$, the y-axis.

Summary table

Exam tips

• In the incentre formula, each vertex is weighted by the side opposite it.
• For a right triangle, write down H and O immediately.
• Use $H = 3G - 2O$ to avoid finding two altitudes.
• In locus problems, finish by replacing $(h, k)$ with $(x, y)$.

Next, read straight lines: slope and forms of equation, or browse the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

• Weighting the incentre by adjacent sides instead of opposite sides.
• Assuming the orthocentre is always inside the triangle; it lies outside for obtuse triangles.
• Reversing the Euler ratio; G is closer to O than to H.
• Leaving the locus in terms of h and k.

Frequently Asked Questions

What is the formula for the centroid of a triangle?

G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).

What is the formula for the incentre?

I = ((ax₁ + bx₂ + cx₃)/(a + b + c), (ay₁ + by₂ + cy₃)/(a + b + c)), where a, b, c are the sides opposite A, B, C.

Where is the orthocentre of a right-angled triangle?

At the vertex where the right angle is.

What is the Euler line?

It is the line through the circumcentre, centroid and orthocentre. The centroid divides the segment from circumcentre to orthocentre in the ratio 1 : 2.

What is a locus in coordinate geometry?

A locus is the set of all points that satisfy a given condition, expressed as an equation in x and y.