How do you use the Section Formula to solve a problem?

To solve, use the Section Formula for external or internal division according to the problem and put in the given coordinates and ratio.

What if a point is dividing the line segment in a ratio of 1:1?

The point is the midpoint of the line segment, and its coordinates are the average of the coordinates of the endpoints.

What is the importance of internal division in the Section Formula?

Internal division happens when a point divides a line segment between its ends.

How is the external division different from the internal division in the Section Formula?

External division happens when the point of division is outside the line segment, past one of its ends.

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Section Formula

1.0Master Internal and External Divisions in Minutes

Learn how to find the exact coordinates of a point that divides a straight line segment joining two given points into a specific ratio. Master the derivation using similar triangles, understand the midpoint shortcut, and solve high-yield board exam questions—all formatted in clean, plain text for simple copy-pasting.

Class: 10 Mathematics (CBSE)
Chapter: Coordinate Geometry
Estimated Learning Time: 15–20 Minutes

2.0Learning Outcomes

After completing this lesson, you will be able to:

State the Section Formula for internal division of a line segment.
Understand the geometric concept of ratio division on a Cartesian plane.
Apply the Midpoint Formula as a special case of the section formula.
Solve board exam problems to find unknown coordinates or division ratios (m1:m2 or k:1).

3.0Introduction to Section Formula

Suppose you are on a trail and you have a map with two points marked on it: the start and end of the trail. Somewhere between these points, locate a resting spot that splits the trail in a ratio of m:n. How would you determine its location on the map? The answer is the section formula. In coordinate geometry, the section formula is an interesting tool that helps find a point that lies between any two locations, either internally or externally.

4.0Insight into Section Formula

The Section Formula is a mathematical tool to locate a point that divides a line segment connecting two given points in a certain ratio. It offers a means of calculating the position of the dividing point from the known positions of the endpoints and the specified ratio in which the segment is divided. The formula finds extensive application in coordinate geometry in solving problems that involve dividing line segments internally as well as externally.

5.0Types of Section Formula

Based on how a point divides a line segment, the section formula can be divided into two types. These two types include:

Internal Division

Internal division in a line segment specifically means when a point divides it between the two endpoints. In other words, a point lies on the segment; dividing it into two parts in a given ratio is known as internal division. Understand it like this: for a line segment joining the points A(x₁,y₁) and B(x₂,y₂), if point P divides the segment AB in the ratio m:n, then the coordinates of point P(x,y) can be calculated with the section formula for internal division. Which is expressed as:

$x = \frac{m x _{2} + n x _{1}}{m + n}, y = \frac{m y _{2} + n y _{1}}{m + n}$

Internal division is an important part of section formula class 10, while studying coordinate geometry, which forms a base for studying these concepts in higher standards.

External Division

External division refers to the division of a line segment by a point outside the segment. This simply means the point doesn’t actually divide the segment itself; it extends beyond the endpoints. For calculating the coordinates of this point, consider a point P(x,y) that divides the line segment joining two points A(x₁,y₁) and B(x₂,y₂) externally in the ratio m:n, then the coordinates of point P are given by the section formula of external division:

$x = \frac{m x _{2} - n x _{1}}{m - n}, y = \frac{m y _{2} - n y _{1}}{m - n}$

6.0Section Formula Proofs

The section formula for a point dividing a line segment can be derived using the similarity rules for a pair of triangles. In the figure given below, two right-angled triangles, ACP and PDB, are given. The hypotenuse of these triangles is the ratio of lines AB, which is m:n. Here, the sides AC and PD of the triangles are parallel to each other.

In $△ A CP an d △ P D B$

∠PAC = ∠BPD (Corresponding angles)

∠ACP = ∠PDB (Right angle)

$△ A CP \sim △ P D B (AA)$

$\frac{A P}{BP} = \frac{A C}{P D} = \frac{PC}{B D} (Corresponding parts of similar triangle)$

Since,

$\frac{A P}{BP} = \frac{m}{n}$

Hence,

$\frac{A P}{BP} = \frac{A C}{P D} = \frac{PC}{B D} = \frac{m}{n} \dots$

According to the figure,

AC = x – x₁ …(2)

PD = x₂ – x …(3)

From equations 1, 2, and 3:

$\frac{A C}{P D} = \frac{m}{n} \frac{x - x _{1}}{x _{2} - x} = \frac{m}{n} n (x - x_{1}) = m (x_{2} - x) x = \frac{m x _{2} + n x _{1}}{m + n} \dots .. (a)$

Now, for y coordinates

CP = y – y₁ ….(4)

BD = y₂ – y …..(5)

From equations 1, 4, and 5

$\frac{PC}{B D} = \frac{m}{n} \frac{y - y _{1}}{y _{2} - y} = \frac{m}{n} n (y - y_{1}) = m (y_{2} - y) y = \frac{m y _{2} + n y _{1}}{m + n} \dots (b)$

According to equations a and b, the coordinates P(x,y), we have:

$P (x, y) = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$

Similar to the derivation of internal division, the formula for external division can also be derived, which gives the result:

$P (x, y) = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n})$

7.0Section Formula: Special Cases

Midpoint of a Line Segment: The midpoint divides the segment in a ratio of 1:1. Thus, if point P is the midpoint of AB, then:

$P (x, y) = (\frac{x _{2} + x _{1}}{2}, \frac{y _{2} + y _{1}}{2})$

Centroid of a Triangle: The centroid separates the medians of the triangle in a ratio of 2:1. The centroid G of a triangle whose vertices are A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃) can be determined by using the section formula, taking the ratio 2:1 along each median.

8.0Uses of the Section Formula

The section formula, either internal or external, is used in a large number of topics of mathematics; these are:

First, it is used to find the coordinates of a point dividing a line segment in a specific ratio.
Section formulas are also used to find the centroids, midpoints, and other special points of geometry problems.
In analytical and vector geometry, section formulae help to find locations and midpoints in a plane.
The formulas are also involved in deriving various straight lines, triangles, and conic sections formulas of geometry.

9.0Solved Problems for Section Formula

Problem 1: Find the coordinates of the point P that divides the segment joining A(2,3) and B(4,6) in the ratio 4:3, internally.

Solution: Given that, A(2,3) and B(4,6) and m:n = 4:3

The section formula for internal division is

$P (x, y) = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n}) P (x, y) = (\frac{4 ( 4 ) + 3 ( 2 )}{4 + 3}, \frac{4 ( 6 ) + 3 ( 3 )}{4 + 3}) P (x, y) = (\frac{22}{7}, \frac{33}{7})$

Problem 2: Find the coordinates of the point P that divides the line segment joining A(1,2) and B(4,5) externally in the ratio 2:3, externally.

Solution: Given that, A(1,2) and B(4,5) and m:n = 2:3

Using the section formula for external division:

$P (x, y) = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n}) P (x, y) = (\frac{2 ( 4 ) - 3 ( 1 )}{2 - 3}, \frac{2 ( 5 ) - 3 ( 2 )}{2 - 3}) P (x, y) = (\frac{5}{- 1}, \frac{4}{- 1}) P (x, y) = (- 5, - 4)$

10.0EUREKA by ALLEN - Your Smart Companion for Class 10 Success

Designed for all Class 10 students preparing for their CBSE & State Board Exam, ALLEN EUREKA is a unique digital program utilizing very specific learning techniques to enable students to master their content. It brings together real experience with professional content to ensure that students can understand and recall difficult subject areas. In addition, EUREKA provides opportunities for students to develop strong fundamental skills and boost their confidence when doing exam-focused work through the use of interactive lessons, AI doubt-clearing tools, and elaborate practice opportunities.

Key Features of EUREKA Class 10 Courses:
Interactive concept learning modules Engaging story-based video lectures CBSE & State Board exam preparation Subjective answer writing practice Instant quizzes with performance feedback Detailed progress tracking 24×7 AI-powered doubt assistance NCERT-aligned curriculum Learn anytime, anywhere

Explore Now

11.0Supporting Study Materials

This study material CBSE Notes and NCERT Solutions for the Chapter "Coordinate Geometry" on Section Formula Topics is designed according to the latest CBSE Class 10 Mathematics syllabus and NCERT guidelines. It provides clear explanations of key concepts, definitions, formulas, and important questions to help students understand internal division of a line segment, finding coordinates of a point using ratio, the midpoint formula, and prepare effectively for examinations.

CBSE Class 10 Maths Notes Chapter 7 Coordinate Geometry	NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry

12.0Previous Year Question on Section Formula

Problem: The point P(x,y) divides the line segment joining A(2,3) and B(8,7) in a certain ratio. If the coordinates of point P are (5,5), find the ratio.

Solution: Given that A(2,3), B(8,7), and P(5,5)

Let the ratio of the line divided by a point P = k:1

Now, using the section formula:

$x = \frac{m x _{2} + n x _{1}}{m + n} 5 = \frac{k ( 8 ) + 1 ( 2 )}{k + 1} 5 (k + 1) = 8 k + 2 5 k + 5 = 8 k + 2 3 k = 3 k = 1$

Hence, the required ratio is 1:1, which also means that P is a midpoint to AB.

13.030-Second Revision: Section Formula

Section Formula: Finds point P(x, y) splitting line AB into ratio m1 : m2.
X-Coordinate: x = (m1*x2 + m2*x1) / (m1 + m2)
Y-Coordinate: y = (m1*y2 + m2*y1) / (m1 + m2)
Midpoint Shortcut: Average the endpoints: ( (x1 + x2)/2, (y1 + y2)/2 ).
Finding Unknown Ratio: Always assume the ratio is k : 1 to solve the algebra faster.
Axis Intercepts: On X-axis, set y = 0. On Y-axis, set x = 0.

14.0 Recommended Next Topics

Prime Factorisation

Quadratic Formula

Sum of n Terms of Arithmetic Progression

Distance Formula

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Section Formula

1.0Master Internal and External Divisions in Minutes

Learn how to find the exact coordinates of a point that divides a straight line segment joining two given points into a specific ratio. Master the derivation using similar triangles, understand the midpoint shortcut, and solve high-yield board exam questions—all formatted in clean, plain text for simple copy-pasting.

Class: 10 Mathematics (CBSE)
Chapter: Coordinate Geometry
Estimated Learning Time: 15–20 Minutes

2.0Learning Outcomes

After completing this lesson, you will be able to:

State the Section Formula for internal division of a line segment.
Understand the geometric concept of ratio division on a Cartesian plane.
Apply the Midpoint Formula as a special case of the section formula.
Solve board exam problems to find unknown coordinates or division ratios (m1:m2 or k:1).

3.0Introduction to Section Formula

Suppose you are on a trail and you have a map with two points marked on it: the start and end of the trail. Somewhere between these points, locate a resting spot that splits the trail in a ratio of m:n. How would you determine its location on the map? The answer is the section formula. In coordinate geometry, the section formula is an interesting tool that helps find a point that lies between any two locations, either internally or externally.

4.0Insight into Section Formula

The Section Formula is a mathematical tool to locate a point that divides a line segment connecting two given points in a certain ratio. It offers a means of calculating the position of the dividing point from the known positions of the endpoints and the specified ratio in which the segment is divided. The formula finds extensive application in coordinate geometry in solving problems that involve dividing line segments internally as well as externally.

5.0Types of Section Formula

Based on how a point divides a line segment, the section formula can be divided into two types. These two types include:

Internal Division

Internal division in a line segment specifically means when a point divides it between the two endpoints. In other words, a point lies on the segment; dividing it into two parts in a given ratio is known as internal division. Understand it like this: for a line segment joining the points A(x₁,y₁) and B(x₂,y₂), if point P divides the segment AB in the ratio m:n, then the coordinates of point P(x,y) can be calculated with the section formula for internal division. Which is expressed as:

$x = \frac{m x _{2} + n x _{1}}{m + n}, y = \frac{m y _{2} + n y _{1}}{m + n}$

Internal division is an important part of section formula class 10, while studying coordinate geometry, which forms a base for studying these concepts in higher standards.

External Division

External division refers to the division of a line segment by a point outside the segment. This simply means the point doesn’t actually divide the segment itself; it extends beyond the endpoints. For calculating the coordinates of this point, consider a point P(x,y) that divides the line segment joining two points A(x₁,y₁) and B(x₂,y₂) externally in the ratio m:n, then the coordinates of point P are given by the section formula of external division:

$x = \frac{m x _{2} - n x _{1}}{m - n}, y = \frac{m y _{2} - n y _{1}}{m - n}$

6.0Section Formula Proofs

The section formula for a point dividing a line segment can be derived using the similarity rules for a pair of triangles. In the figure given below, two right-angled triangles, ACP and PDB, are given. The hypotenuse of these triangles is the ratio of lines AB, which is m:n. Here, the sides AC and PD of the triangles are parallel to each other.

In $△ A CP an d △ P D B$

∠PAC = ∠BPD (Corresponding angles)

∠ACP = ∠PDB (Right angle)

$△ A CP \sim △ P D B (AA)$

$\frac{A P}{BP} = \frac{A C}{P D} = \frac{PC}{B D} (Corresponding parts of similar triangle)$

Since,

$\frac{A P}{BP} = \frac{m}{n}$

Hence,

$\frac{A P}{BP} = \frac{A C}{P D} = \frac{PC}{B D} = \frac{m}{n} \dots$

According to the figure,

AC = x – x₁ …(2)

PD = x₂ – x …(3)

From equations 1, 2, and 3:

$\frac{A C}{P D} = \frac{m}{n} \frac{x - x _{1}}{x _{2} - x} = \frac{m}{n} n (x - x_{1}) = m (x_{2} - x) x = \frac{m x _{2} + n x _{1}}{m + n} \dots .. (a)$

Now, for y coordinates

CP = y – y₁ ….(4)

BD = y₂ – y …..(5)

From equations 1, 4, and 5

$\frac{PC}{B D} = \frac{m}{n} \frac{y - y _{1}}{y _{2} - y} = \frac{m}{n} n (y - y_{1}) = m (y_{2} - y) y = \frac{m y _{2} + n y _{1}}{m + n} \dots (b)$

According to equations a and b, the coordinates P(x,y), we have:

$P (x, y) = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$

Similar to the derivation of internal division, the formula for external division can also be derived, which gives the result:

$P (x, y) = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n})$

7.0Section Formula: Special Cases

Midpoint of a Line Segment: The midpoint divides the segment in a ratio of 1:1. Thus, if point P is the midpoint of AB, then:

$P (x, y) = (\frac{x _{2} + x _{1}}{2}, \frac{y _{2} + y _{1}}{2})$

Centroid of a Triangle: The centroid separates the medians of the triangle in a ratio of 2:1. The centroid G of a triangle whose vertices are A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃) can be determined by using the section formula, taking the ratio 2:1 along each median.

8.0Uses of the Section Formula

The section formula, either internal or external, is used in a large number of topics of mathematics; these are:

First, it is used to find the coordinates of a point dividing a line segment in a specific ratio.
Section formulas are also used to find the centroids, midpoints, and other special points of geometry problems.
In analytical and vector geometry, section formulae help to find locations and midpoints in a plane.
The formulas are also involved in deriving various straight lines, triangles, and conic sections formulas of geometry.

9.0Solved Problems for Section Formula

Problem 1: Find the coordinates of the point P that divides the segment joining A(2,3) and B(4,6) in the ratio 4:3, internally.

Solution: Given that, A(2,3) and B(4,6) and m:n = 4:3

The section formula for internal division is

$P (x, y) = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n}) P (x, y) = (\frac{4 ( 4 ) + 3 ( 2 )}{4 + 3}, \frac{4 ( 6 ) + 3 ( 3 )}{4 + 3}) P (x, y) = (\frac{22}{7}, \frac{33}{7})$

Problem 2: Find the coordinates of the point P that divides the line segment joining A(1,2) and B(4,5) externally in the ratio 2:3, externally.

Solution: Given that, A(1,2) and B(4,5) and m:n = 2:3

Using the section formula for external division:

$P (x, y) = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n}) P (x, y) = (\frac{2 ( 4 ) - 3 ( 1 )}{2 - 3}, \frac{2 ( 5 ) - 3 ( 2 )}{2 - 3}) P (x, y) = (\frac{5}{- 1}, \frac{4}{- 1}) P (x, y) = (- 5, - 4)$

10.0EUREKA by ALLEN - Your Smart Companion for Class 10 Success

Designed for all Class 10 students preparing for their CBSE & State Board Exam, ALLEN EUREKA is a unique digital program utilizing very specific learning techniques to enable students to master their content. It brings together real experience with professional content to ensure that students can understand and recall difficult subject areas. In addition, EUREKA provides opportunities for students to develop strong fundamental skills and boost their confidence when doing exam-focused work through the use of interactive lessons, AI doubt-clearing tools, and elaborate practice opportunities.

Key Features of EUREKA Class 10 Courses:
Interactive concept learning modules Engaging story-based video lectures CBSE & State Board exam preparation Subjective answer writing practice Instant quizzes with performance feedback Detailed progress tracking 24×7 AI-powered doubt assistance NCERT-aligned curriculum Learn anytime, anywhere

Explore Now

11.0Supporting Study Materials

This study material CBSE Notes and NCERT Solutions for the Chapter "Coordinate Geometry" on Section Formula Topics is designed according to the latest CBSE Class 10 Mathematics syllabus and NCERT guidelines. It provides clear explanations of key concepts, definitions, formulas, and important questions to help students understand internal division of a line segment, finding coordinates of a point using ratio, the midpoint formula, and prepare effectively for examinations.

CBSE Class 10 Maths Notes Chapter 7 Coordinate Geometry	NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry

12.0Previous Year Question on Section Formula

Problem: The point P(x,y) divides the line segment joining A(2,3) and B(8,7) in a certain ratio. If the coordinates of point P are (5,5), find the ratio.

Solution: Given that A(2,3), B(8,7), and P(5,5)

Let the ratio of the line divided by a point P = k:1

Now, using the section formula:

$x = \frac{m x _{2} + n x _{1}}{m + n} 5 = \frac{k ( 8 ) + 1 ( 2 )}{k + 1} 5 (k + 1) = 8 k + 2 5 k + 5 = 8 k + 2 3 k = 3 k = 1$

Hence, the required ratio is 1:1, which also means that P is a midpoint to AB.

13.030-Second Revision: Section Formula

Section Formula: Finds point P(x, y) splitting line AB into ratio m1 : m2.
X-Coordinate: x = (m1*x2 + m2*x1) / (m1 + m2)
Y-Coordinate: y = (m1*y2 + m2*y1) / (m1 + m2)
Midpoint Shortcut: Average the endpoints: ( (x1 + x2)/2, (y1 + y2)/2 ).
Finding Unknown Ratio: Always assume the ratio is k : 1 to solve the algebra faster.
Axis Intercepts: On X-axis, set y = 0. On Y-axis, set x = 0.

14.0 Recommended Next Topics

Prime Factorisation

Quadratic Formula

Sum of n Terms of Arithmetic Progression

Distance Formula

Frequently Asked Questions

How do you use the Section Formula to solve a problem?

What if a point is dividing the line segment in a ratio of 1:1?

What is the importance of internal division in the Section Formula?

How is the external division different from the internal division in the Section Formula?

Frequently Asked Questions

How do you use the Section Formula to solve a problem?

What if a point is dividing the line segment in a ratio of 1:1?

What is the importance of internal division in the Section Formula?

How is the external division different from the internal division in the Section Formula?

Join ALLEN!

Section Formula

1.0Master Internal and External Divisions in Minutes

2.0Learning Outcomes

3.0Introduction to Section Formula

4.0Insight into Section Formula

5.0Types of Section Formula

Internal Division

External Division

6.0Section Formula Proofs

7.0Section Formula: Special Cases

8.0Uses of the Section Formula

9.0Solved Problems for Section Formula

10.0EUREKA by ALLEN - Your Smart Companion for Class 10 Success

11.0Supporting Study Materials

12.0Previous Year Question on Section Formula

13.030-Second Revision: Section Formula

14.0 Recommended Next Topics

Table of Contents

Join ALLEN!

Section Formula

1.0Master Internal and External Divisions in Minutes

2.0Learning Outcomes

3.0Introduction to Section Formula

4.0Insight into Section Formula

5.0Types of Section Formula

Internal Division

External Division

6.0Section Formula Proofs

7.0Section Formula: Special Cases

8.0Uses of the Section Formula

9.0Solved Problems for Section Formula

10.0EUREKA by ALLEN - Your Smart Companion for Class 10 Success

11.0Supporting Study Materials

12.0Previous Year Question on Section Formula

13.030-Second Revision: Section Formula

14.0 Recommended Next Topics

Table of Contents