JEE Mains Maths · Straight Lines

Centres of a Triangle

The orthocentre, circumcentre, centroid and incentre of a triangle given by its vertices or by its sides, and the Euler line that joins three of them.

All Straight Lines notes JEE Mains Maths strategy

Why this matters

Twenty-six PYQs, twenty-two of them multiple choice, and four from 2026. Fifteen use the orthocentre, found from two altitudes or given and worked back to a vertex or a side. Six find the circumcentre from perpendicular bisectors. Five use the centroid, the incentre or a point dividing a side. Three ideas cover the page.

Concept 1 of 3: The orthocentre

The orthocentre

H

is where the altitudes meet. Each altitude passes through a vertex and is perpendicular to the opposite side, so two dot products fix

H

. When

H

is given and a vertex is unknown, the same two dot products become equations for the vertex. In a right triangle,

H

is the vertex at the right angle. The centroid lies on the segment from

H

to the circumcentre

O

, two thirds of the way from

H

Definition

$(H-A)\cdot(C-B)=0$ and $(H-B)\cdot(A-C)=0$ .
Right angle at $A$ : $H=A$ .
Euler line: $G=\frac{H+2O}3$ , with $O$ the circumcentre.
Equilateral triangle: $H$ , $G$ and $O$ coincide.

Altitude conditions

\overrightarrow{AH}\cdot\overrightarrow{BC}=0,\qquad \overrightarrow{BH}\cdot\overrightarrow{CA}=0

Worked example

Find the orthocentre of

A(0,0)

B(6,0)

C(2,4)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q65Moderate

Example 1 · Straight Lines · Centres of a Triangle

Let the three sides of a triangle are on the lines

4x- 7y+ 10 = 0,x+y=5

and

7x+ 4y= 15

. Then the distance of its orthocenter from the orthocenter of the triangle formed by the lines

x= 0,y= 0

and

x+y= 1

is.

Check for a right angle first

If two sides have slopes whose product is

-1

, the orthocentre is their common vertex, and no altitude needs writing. Missing this turns a one-line question into a page of algebra.

Drill 14 more on the orthocentre

Concept 2 of 3: The circumcentre

The circumcentre

O

is equally far from the three vertices, so it lies on the perpendicular bisector of each side, and two of them fix it. Setting

|OA|^2=|OB|^2

removes the squares and leaves a straight line. In a right triangle,

O

is the midpoint of the hypotenuse. When the triangle is given by its sides, find the vertices first by solving the lines in pairs.

Definition

$|OA|=|OB|=|OC|=R$ .
Perpendicular bisector of $AB$ : through the midpoint, perpendicular to $AB$ .
Right angle at $C$ : $O$ is the midpoint of $AB$ , and $R=\frac{AB}2$ .
Area $=\frac{abc}{4R}$ .

Equal distances

|OA|^2=|OB|^2=|OC|^2

Worked example

Find the circumcentre of

A(0,0)

B(8,0)

C(2,6)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q62Moderate

Example 2 · Straight Lines · Centres of a Triangle

Let the vertex A of a triangle ABC be (1, 2), and the mid-point of the side AB be

(5, - 1)

. If the centroid of this triangle is

(3,4)

and its circumcenter is

(\alpha,\beta)

, then

21(\alpha+\beta)

is equal to:

Midpoint alone is not enough

The perpendicular bisector needs the midpoint and the perpendicular direction. A median also passes through the midpoint, but it goes through

O

only when the triangle is isosceles there.

Drill 5 more on the circumcentre

Concept 3 of 3: Centroid, incentre and dividing a side

The centroid is the average of the vertices. The midpoints of the sides give it too, since the triangle of midpoints has the same centroid, and they give the vertices:

A=E+F-D

, where

D,E,F

are the midpoints of

BC,CA,AB

. The incentre is the average of the vertices weighted by the opposite side lengths. A point dividing

BC

in the ratio

m:n

\frac{nB+mC}{m+n}

Definition

Centroid: $G=\frac{A+B+C}3$ .
From the midpoints $D,E,F$ of $BC,CA,AB$ : $A=E+F-D$ , $B=F+D-E$ , $C=D+E-F$ .
Incentre: $I=\frac{aA+bB+cC}{a+b+c}$ , with $a=BC$ , $b=CA$ , $c=AB$ .
$P$ divides $BC$ as $m:n$ : $P=\frac{nB+mC}{m+n}$ .

Incentre

I=\frac{aA+bB+cC}{a+b+c}

Worked example

Find the incentre of

A(0,0)

B(4,0)

C(0,3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q59Moderate

Example 3 · Straight Lines · Centres of a Triangle

Let the mid points of the sides of a triangle

ABC

\left( \frac{5}{2},7 \right),\left( \frac{5}{2},3 \right)

and

(4,5)

. If its incentre is

(h,k)

, then

3\text{ }h + k

is equal to :

Weight by the opposite side

In the incentre formula,

A

is weighted by

a=BC

, the side opposite it, not by a side through it. Weighting by the adjacent sides gives a different point.

Drill 4 more on centroid, incentre and dividing a side

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The orthocentre
Altitude conditions
$\overrightarrow{AH}\cdot\overrightarrow{BC}=0,\qquad \overrightarrow{BH}\cdot\overrightarrow{CA}=0$
The circumcentre
Equal distances
$|OA|^2=|OB|^2=|OC|^2$
Centroid, incentre and dividing a side
Incentre
$I=\frac{aA+bB+cC}{a+b+c}$

Watch out for (3)

Check for a right angle first
→ The orthocentre
Midpoint alone is not enough
→ The circumcentre
Weight by the opposite side
→ Centroid, incentre and dividing a side

Test yourself on Straight Lines

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