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Triangle centres look like four formulas to memorise, and many students mix them up under time pressure. The incentre formula gets the weights on the wrong vertices; the circumcentre gets found by averaging. This post gives each centre with the reason behind it, works two triangles in full, and shows the Euler line relation that lets you check three centres against each other.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists "co-ordinate of the centroid, orthocentre and circumcentre of a triangle" under straight lines. The JEE (Advanced) 2026 brochure lists "Centroid, orthocentre, incentre and circumcentre of a triangle". NCERT Class 11 Chapter 9, Straight Lines covers the line equations you need to find them. Check your exam year's syllabus. For line forms and distances, see our post on straight lines in coordinate geometry.
The four centres in one place
For A(x₁, y₁), B(x₂, y₂), C(x₃, y₃), with a = BC, b = CA and c = AB:
Centroid G, where the medians meet: ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).
Incentre I, where the angle bisectors meet: ((ax₁ + bx₂ + cx₃)/(a + b + c), (ay₁ + by₂ + cy₃)/(a + b + c)). Each vertex is weighted by the side opposite it.
Circumcentre O, where the perpendicular bisectors meet: the point equally distant from A, B and C. Find it by solving OA² = OB² and OA² = OC².
Orthocentre H, where the altitudes meet: intersect two altitudes.
Worked example 1: a right triangle
Take A(0, 0), B(6, 0), C(0, 8). Then a = BC = 10, b = CA = 8, c = AB = 6.
Centroid: G = ((0 + 6 + 0)/3, (0 + 0 + 8)/3) = (2, 8/3).
Circumcentre: in a right triangle it is the midpoint of the hypotenuse BC, so O = (3, 4). Check: OA = OB = OC = 5.
Orthocentre: the two legs are altitudes, so H is the right-angle vertex, (0, 0).
Incentre: I = ((10 × 0 + 8 × 6 + 6 × 0)/24, (10 × 0 + 8 × 0 + 6 × 8)/24) = (2, 2).
Check the incentre a second way. The inradius is r = area/s = 24/12 = 2, and with the right angle at the origin and legs on the axes, the incentre must be (r, r) = (2, 2). It matches.
Worked example 2: a scalene triangle
Take A(0, 0), B(4, 0), C(1, 3).
Centroid: G = (5/3, 1).
Circumcentre: the perpendicular bisector of AB is x = 2. Then OA² = OC² gives 4 + y² = 1 + (y − 3)², so 6y = 6 and y = 1. O = (2, 1), and OA = OB = OC = √5.
Orthocentre: the altitude from C to AB (the x-axis) is x = 1. BC has slope 3/(1 − 4) = −1, so the altitude from A has slope 1: y = x. They meet at H = (1, 1).
Apply it: 3G − 2O = (5 − 4, 3 − 2) = (1, 1), which is the H we found. In example 1: 3G − 2O = (6 − 6, 8 − 8) = (0, 0), again correct.
Common mistakes and how to avoid them
Weighting the incentre by adjacent sides. The weight on A is a = BC, the side opposite A.
Averaging to find the circumcentre. Only the centroid is an average. The circumcentre needs equal distances.
Assuming the circumcentre lies inside. It is outside for an obtuse triangle and on the hypotenuse for a right one.
Assuming all centres coincide. That happens only for an equilateral triangle.
Arithmetic slips in the altitude slope. Perpendicular slopes multiply to −1; check it.
Why compute the same centre two ways?
Rittle-Johnson and Star (2007) randomly assigned 70 seventh-grade students to compare alternative solution methods side by side or to study the same methods one at a time. The comparison group made greater gains in procedural knowledge and flexibility. For triangle centres, two methods are always available: the incentre by formula and by inradius, the orthocentre by altitudes and by H = 3G − 2O.
A practice routine
Choose three triangles with integer vertices: one right, one acute, one obtuse.
Find all four centres for each.
Check every set with H = 3G − 2O, and the incentre with the inradius where you can.
Sketch each triangle and mark the centres, so you see where they fall for each type. Our post on locus problems uses the same equal-distance idea.
If one centre keeps disagreeing with the Euler check and you cannot find why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: find the centroid, circumcentre and orthocentre of A(0, 0), B(8, 0), C(2, 6), and check them with H = 3G − 2O.
