Incenter of a Triangle: Formula, Properties and Examples

What is the Incenter of a Triangle?

The incenter of a triangle is one of the most important triangle centers in geometry. It is the point where the three internal angle bisectors of a triangle intersect. The incenter is always located inside the triangle and serves as the center of the inscribed circle (incircle), which touches all three sides of the triangle.

Incenter Meaning in Simple Words

In simple words, the incenter is the point inside a triangle that is equally distant from all three sides.

It can be thought of as the "center" of the triangle's largest possible circle that fits perfectly inside the triangle.

Definition of Incenter

The incenter of a triangle is the point of intersection of the three internal angle bisectors of the triangle.

If the angle bisectors of $\angle A$, $\angle B$, and $\angle C$ intersect at point $I$, then $I$ is called the incenter of the triangle.

Why the Incenter is Important

The incenter is an important geometric point because it possesses several unique properties.

Importance of the Incenter

• It is the center of the incircle.
• It is equidistant from all three sides of a triangle.
• It helps in solving geometry and mensuration problems.
• It is useful in coordinate geometry calculations.
• It is frequently used in engineering and architectural designs.
• It appears in board exams and competitive examinations.

Real-Life Applications of the Incenter

The concept of the incenter is used whenever an object must be placed at an equal distance from all sides of a triangular region.

Incenter of a Triangle Formula

The coordinates of the incenter can be calculated using the vertices and side lengths of a triangle.

Coordinate Formula for Incenter

For a triangle with vertices:

$A(x_1,y_1)$

$B(x_2,y_2)$

$C(x_3,y_3)$ and side lengths: $a$, $b$, and $c$ the coordinates of the incenter are: $\left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right)$

This is the standard incenter formula used in coordinate geometry.

Incenter Using Side Lengths

The incenter is determined by taking a weighted average of the vertex coordinates, where the weights are the lengths of the sides opposite those vertices.

For triangle $ABC$:

• $a = BC$
• $b = AC$
• $c = AB$

The side lengths act as weights in the coordinate formula.

Meaning of Variables in the Formula

NOTE:

If ΔABC is an equilateral triangle, then a = b = c

Coordinates of Incentre (I) = Coordinates of Centroid (G) = $\left(\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{3}, \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}\right)$

Derivation of the Incenter Formula

The incenter lies at the intersection of the three angle bisectors.

According to the Angle Bisector Theorem, an angle bisector divides the opposite side in the ratio of the adjacent sides.

Using the equations of two angle bisectors and solving them simultaneously leads to the coordinate formula:

$\left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right)$

This formula gives the exact location of the incenter in the coordinate plane.

Angle Bisector Method

The simplest method for locating the incenter is to draw the internal angle bisectors of any two angles of the triangle.

The point where they intersect is the incenter.

Geometrical Construction

The geometric construction uses only:

• Ruler
• Compass
• Pencil

No coordinate calculations are required.

The procedure is based entirely on angle bisector construction.

Diagram of Incenter Construction

For triangle $ABC$:

• Draw angle bisector of $\angle A$.
• Draw angle bisector of $\angle B$.
• Let them intersect at $I$.

Point $I$ is the incenter.

A circle drawn with center $I$ and radius equal to the perpendicular distance from $I$ to any side will touch all three sides of the triangle.

Properties of the Incenter of a Triangle

The incenter possesses several important geometric properties.

Intersection of Angle Bisectors

The incenter is formed by the intersection of the three internal angle bisectors.

Every triangle has exactly one incenter.

Equidistant from All Sides

The incenter is equally distant from all three sides of the triangle.

If the perpendicular distances from the incenter to the sides are:

$d_1$, $d_2$, and $d_3$

then: $d_1=d_2=d_3$

This common distance is called the inradius.

Center of the Incircle

The incenter serves as the center of the incircle.

The incircle touches all three sides of the triangle exactly once.

Position of the Incenter in Different Triangles

Unlike some other triangle centers, the incenter is always located inside the triangle.

Incenter and Incircle

The incenter and incircle are closely related geometric concepts.

Relationship Between Incenter and Incircle

The incenter is the center of the unique circle that can be drawn inside a triangle touching all three sides.

This circle is called the incircle.

Radius of the Incircle

The radius of the incircle is called the inradius.

It is the perpendicular distance from the incenter to any side of the triangle.

Area Formula Using Inradius

The area of a triangle can be calculated using the inradius.

If: $r$ = inradius

$s$ = semiperimeter

then:

A=rsA=rsA=rs

where:

$s=\frac{a+b+c}{2}$

Applications of the Incircle

The incircle is useful in:

• Geometry constructions
• Engineering design
• Surveying
• Architecture
• Computer graphics
• Optimization problems

How to Find the Incenter of a Triangle?

There are multiple methods to locate the incenter.

Using Coordinate Geometry

When the coordinates of the vertices are known, use:

$\left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right)$

Using Angle Bisectors

Construct any two internal angle bisectors.

Their point of intersection is the incenter.

Step-by-Step Method

1. Find the side lengths.
2. Apply the coordinate formula.
3. Calculate the weighted average of coordinates.
4. Obtain the incenter coordinates.

Solved Examples

Example

Find the incenter of an equilateral triangle.

Since all sides are equal:

$a=b=c$

The formula simplifies to:

$x=\frac{x_1+x_2+x_3}{3}$

$y=\frac{y_1+y_2+y_3}{3}$

Thus, the incenter coincides with the centroid, circumcenter, and orthocenter.

Difference Between Incenter, Centroid, Circumcenter, and Orthocenter

A triangle has four important centers, each obtained differently.

Incenter vs Centroid

Incenter vs Circumcenter

Incenter vs Orthocenter

Comparison Table of Triangle Centers

Understanding these triangle centers is essential for mastering geometry, coordinate geometry, mensuration, and competitive examination mathematics.

How to Construct the Incenter of a Triangle?

The construction of the incenter of a triangle is possible with the help of a compass.

Steps to construct the incenter of a triangle:

Step 1: Place one of the compass's ends at one of the triangle vertices. The other side of the compass is on one side of the triangle.

Step 2: Draw two arcs on two sides of the triangle using the compass.

Step 3: By using the same width as before, draw two arcs inside the triangle so that they cross each other from the point where each arc crosses the side.

Step 4: Draw a line from the vertex of the triangle to where the two arcs inside the triangle cross.

Step 5: Repeat the same process from the other vertex of the triangle.

Step 6: The point at which the two lines meet or intersect is the incenter of a triangle

Learn more in detail about Straight Lines - Definition, Equations, Properties

Best Books for Incenter of a Triangle

The incenter is an important concept in geometry, coordinate geometry, and mensuration. These books help develop a strong understanding of triangle centers and geometric constructions.

Shortcut Tips and Tricks for Incenter Problems

Many geometry questions involving the incenter can be solved faster by remembering its fundamental properties and formulas.

Important Formula Table

The following formulas are essential for solving incenter and incircle-related questions in geometry.

Solved Examples Based on Incentre of a Triangle

Example 1. Let $A(0,1)$, $B(1,1)$, and $C(1,0)$ be the midpoints of the sides of a triangle whose incenter is $D$. If the focus of the parabola $y^2=4ax$ passing through $D$ is $(\alpha+\beta\sqrt{3},0)$, where $\alpha$ and $\beta$ are rational numbers, find the value of $\frac{\alpha}{\beta^2}$.

Solution:

The given midpoints correspond to a triangle with vertices:

$P(0,0),\ Q(0,2),\ R(2,0)$

Side lengths are:

$a=QR=2\sqrt2$

$b=PR=2$

$c=PQ=2$

Using the incenter formula,

$I\left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right)$

Substituting the coordinates,

$D=\left(\frac{4}{2+2+2\sqrt2},\frac{4}{2+2+2\sqrt2}\right)$

$=\left(\frac{2}{2+\sqrt2},\frac{2}{2+\sqrt2}\right)$

Since $D$ lies on the parabola,

$y^2=4ax$

Substituting the coordinates of $D$,

$\left(\frac{2}{2+\sqrt2}\right)^2=4a\left(\frac{2}{2+\sqrt2}\right)$

$\Rightarrow 4a=\frac{2}{2+\sqrt2}$

$\Rightarrow a=\frac{1}{4}(2-\sqrt2)$

Therefore,

$\alpha=\frac12,\quad \beta=-\frac14$

Hence,

$\frac{\alpha}{\beta^2}$

$=\frac{\frac12}{\left(-\frac14\right)^2}$

$=\frac{\frac12}{\frac1{16}}$

$=8$

Correct Answer: $8$

Example 2. The coordinates of the midpoints of the sides of a triangle are $(0,1)$, $(1,1)$, and $(1,0)$. Find the $x$-coordinate of the incenter of the triangle.

Solution:

The triangle corresponding to these midpoints has vertices:

$(0,0),\ (0,2),\ (2,0)$

$I\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}\right)$

- wherein

a,b,c is the length of the sides of ∆ ABC, and $A(x_1,y_1) B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices.

On solving h=0, k=2

Triangle becomes

The side lengths are:

$a=2\sqrt2$

$b=2$

$c=2$

Using the incenter formula,

$x=\frac{ax_1+bx_2+cx_3}{a+b+c}$

Substituting values,

$x=\frac{(2\sqrt2)(0)+(2)(0)+(2)(2)}{2\sqrt2+2+2}$

$=\frac{4}{4+2\sqrt2}$

Rationalizing,

$x=\frac{4(4-2\sqrt2)}{16-8}$

$=2-\sqrt2$

Correct Answer: $2-\sqrt2$

Example 3. Let the tangent to the circle $x^2+y^2=25$ at the point $R(3,4)$ meet the coordinate axes at points $P$ and $Q$. If $r$ is the radius of the circle passing through the origin and centered at the incenter of triangle $OPQ$, find $r^2$.

Solution:

The tangent to

$x^2+y^2=25$

at $(3,4)$ is

$3x+4y=25$

The intercepts are:

$P\left(\frac{25}{3},0\right)$

$Q\left(0,\frac{25}{4}\right)$

and

$O(0,0)$

Using the incenter formula, the incenter is

$I=\left(\frac{25}{12},\frac{25}{12}\right)$

Since the circle passes through the origin,

$r=OI$

Therefore,

$r^2=\left(\frac{25}{12}\right)^2+\left(\frac{25}{12}\right)^2$

$=\frac{625}{144}+\frac{625}{144}$

$=\frac{1250}{144}$

$=\frac{625}{72}$

Correct Answer: $\frac{625}{72}$

Example 4. Find the incenter of the triangle formed by the lines

$y=15,\quad 12y=5x,\quad 3x+4y=0$

Solution:

The vertices of the triangle are:

$(36,15),\ (0,0),\ (-20,15)$

The lengths of the opposite sides are:

$25,\ 56,\ 39$

Using the incenter formula,

$I=\left(\frac{25(36)+56(0)+39(-20)}{25+56+39},\frac{25(15)+56(0)+39(15)}{25+56+39}\right)$

$=\left(\frac{900-780}{120},\frac{375+585}{120}\right)$

$=\left(\frac{120}{120},\frac{960}{120}\right)$

$=(1,8)$

Correct Answer: $(1,8)$

Example 5. Find the incenter of the triangle with vertices

$(1,\sqrt3),\ (0,0),\ (2,0)$

Solution:

Let the vertices be:

$A(1,\sqrt3)$

$B(0,0)$

$C(2,0)$

Using the distance formula,

$AB=\sqrt{(1-0)^2+(\sqrt3-0)^2}$

$=\sqrt{1+3}$

$=2$

Similarly,

$AC=2$

and

$BC=2$

Therefore,

$AB=BC=CA$

Hence, the triangle is equilateral.

For an equilateral triangle:

Incenter = Centroid

Therefore,

$\left(\frac{1+0+2}{3},\frac{\sqrt3+0+0}{3}\right)$

$=\left(1,\frac{\sqrt3}{3}\right)$

$=\left(1,\frac1{\sqrt3}\right)$

Correct Answer: $\left(1,\frac1{\sqrt3}\right)$

NCERT Resources

Students preparing for Class 11 Mathematics can strengthen their understanding of straight lines through NCERT solutions, revision notes, and exemplar problems. These resources help build conceptual clarity, improve problem-solving skills, and provide effective exam preparation support.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

NCERT Class 11 Maths Chapter 10 Notes: Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

List of Topics related to Straight Lines

To develop a strong understanding of coordinate geometry and straight lines, it is important to study related mathematical concepts. These topics help build a solid foundation for solving problems involving slopes, equations of lines, distances, angles, and geometric relationships in two-dimensional coordinate systems.