Section Formula Class 10 ICSE — Bansal Chapter 13
The section formula class 10 ICSE chapter from the Bansal book builds the coordinate geometry tools you need for straight lines. Chapter 13, Section and Mid Point Formula, shows how to find a point that divides a segment in a given ratio, the midpoint of two points, and the centroid of a triangle. All three results rest on similar triangles.
Download Section and Mid Point Formula Class 10 ICSE Mathematics Chapter 13 PDF
Download the Section and Mid Point Formula Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 13 pages, 4 sections, 64 questions.
What is inside Chapter 13
13 pages · 4 sections · 14 worked examples · 64 exercise questions
| Section | Title |
|---|---|
| 13.1 | Introduction |
| 13.2 | The Section Formula |
| 13.4 | Mid-Point Formula |
| 13.5 | Centroid of a triangle |
- EXERCISE 13(A): 25 questions
- EXERCISE 13(B): 18 questions
- EXERCISE 13(C): 21 questions
13.1 Introduction
Section 13.1 opens the chapter by listing what coordinate geometry can do for any two given points in a Cartesian plane. The book names five uses. This list sets the frame for everything that follows.
- the distance between the given points
- the coordinates of a point that divides the join in a given ratio
- the coordinates of the midpoint of the segment
- the equation of the straight line through the points
- the equation of the perpendicular bisector of the segment
This ICSE chapter develops items (ii) and (iii) in full. The section formula answers (ii), and the midpoint formula answers (iii). The remaining three uses belong to later chapters of the Class 10 Maths book. Knowing which problem each formula solves saves time in the exam.
13.2 The Section Formula
The section formula finds the coordinates of a point P that lies on segment AB and divides it in a fixed ratio. If P divides AB in the ratio m_1 : m_2, then \frac{AP}{PB} = \frac{m_1}{m_2}. The ratio is always read from the first named point.
To derive the formula, draw AL, PM and BN perpendicular to the x-axis. Triangles APR and PBS are similar, so their corresponding sides are proportional. Solving the two proportions gives the coordinates of P:
x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \quad y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}
Here A = (x_1, y_1) and B = (x_2, y_2). The formula is a weighted average. The point sits closer to whichever endpoint carries the larger multiplier.
Worked pattern: divide the join of A(1, 2) and B(10, 8) in the ratio 2 : 3. Put m_1 = 2, m_2 = 3, with A as (x_1, y_1) and B as (x_2, y_2).
x = \frac{2 \times 10 + 3 \times 1}{2 + 3} = \frac{23}{5}, \quad y = \frac{2 \times 8 + 3 \times 2}{2 + 3} = \frac{22}{5}
So P = \left(\frac{23}{5}, \frac{22}{5}\right). The point sits closer to A, because A carries the smaller part of the ratio.
Finding the ratio (converse). When the dividing point is known but the ratio is not, set m_1 : m_2 = k : 1. This reduces two unknowns to one. The formula becomes x = \frac{k x_2 + x_1}{k + 1} and y = \frac{k y_2 + y_1}{k + 1}. Solve for k, then write the ratio as k : 1.
Axis and line intersections. A point on the x-axis has y = 0; on the y-axis, x = 0. A point on the line y = 3 has coordinates (x, 3). Use the known coordinate in the y-equation to find k, then use the x-equation for the point.
Points of trisection. If AP = PQ = QB, the first point P uses the ratio 1 : 2 and the second point Q uses 2 : 1. Apply the section formula twice.
Common mistakes:
- Swapping which endpoint is (x_1, y_1) — fix by always writing the first named point first.
- Exchanging m_1 and m_2 — the ratio starts from the first named point, so m_1 belongs to it.
- Forgetting the trisection order — the first point is 1 : 2, not 2 : 1.
- Sign errors with negative coordinates — substitute with brackets, for example (x_1, y_1) = (-2, 5).
13.4 Mid-Point Formula
The midpoint is the section formula with the ratio 1 : 1. Substituting m_1 = m_2 = 1 into the section formula gives:
x = \frac{x_1 + x_2}{2}, \quad y = \frac{y_1 + y_2}{2}
The midpoint is the average of the two x-coordinates and the average of the two y-coordinates. There is no weighting, because neither endpoint is favoured.
Worked pattern: the midpoint of (4, 9) and (-2, 3) is \left(\frac{4 + (-2)}{2}, \frac{9 + 3}{2}\right) = (1, 6).
Reverse pattern. When the midpoint and one endpoint are known, double the midpoint and subtract the known end. If the midpoint is (2, -1) and one end is (5, 7), the other end is (2 \times 2 - 5, 2 \times (-1) - 7) = (-1, -9).
The midpoint formula in this ICSE chapter is applied to three problem types. In a parallelogram, the diagonals bisect each other, so the midpoint of AC equals the midpoint of BD. Equating the two midpoints gives the fourth vertex. For a triangle, when the three side midpoints are known, write the six linear equations and add them in pairs to find the vertices.
The third type mixes in unknown values such as m and n when the midpoint contains an expression like (2m - 1).
13.5 Centroid of a triangle
The centroid is the point where the three medians of a triangle meet. Each median runs from a vertex to the midpoint of the opposite side. The centroid divides each median in the ratio 2 : 1, counted from the vertex.
The book gives two routes. First, find the midpoint D of side BC, then apply the section formula to median AD with ratio 2 : 1. The direct method averages the three vertices:
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
Worked pattern: for A(2, 4), B(4, 8) and C(6, 0), the centroid is \left(\frac{2 + 4 + 6}{3}, \frac{4 + 8 + 0}{3}\right) = (4, 4).
Reverse problem. Suppose the centroid is (3, 2), with A(1, 4), B(5, b) and C(a, 1). Averaging gives (1 + 5 + a)/3 = 3, so a = 3. The y-coordinates give (4 + b + 1)/3 = 2, so b = 1. Side BC runs from (5, 1) to (3, 1), so BC = 2 units.
Exercise 13(C) question 21 records a useful result. The centroid of the triangle formed by the side midpoints is the same point as the centroid of the original triangle.
| Formula | Ratio used | Result |
|---|---|---|
| Section formula | m_1 : m_2 | Point dividing a segment |
| Midpoint formula | 1 : 1 | Middle of a segment |
| Centroid formula | 2 : 1 on each median | Balance point of a triangle |
What to practise from Section Formula Class 10 ICSE
The chapter carries three exercise blocks. Work through them in order, because each builds on the previous one.
| Exercise | Questions | What it practises |
|---|---|---|
| EXERCISE 13(A) | 25 | Section formula: finding the dividing point, finding the ratio, axis and line intersections, trisection |
| EXERCISE 13(B) | 18 | Midpoint and its applications, medians, first centroid questions |
| EXERCISE 13(C) | 21 | Mixed application of both formulas, centroid, ratio combined with distance |
Several questions in the book carry board year tags — 2006, 2008, 2011, 2012, 2013 and 2014 all appear. The tagged questions cluster around axis division, midpoint and centroid problems. Treat them as repeated board questions worth extra attention.
Browse the Class 10 ICSE books list for the chapters that follow this one. The official CISCE website carries the syllabus and examination pattern for verification.
Section Formula Class 10 ICSE frequently asked questions
What is the section formula class 10 ICSE and when do I use it?
The section formula gives the coordinates of a point P that divides the segment joining A and B in a given ratio m_1 : m_2. Use it whenever a problem states a ratio such as 2 : 3 and asks for the dividing point.
How do I find the ratio in which a point divides a line segment?
Set the ratio as k : 1 and substitute the known point into the section formula. Solve for k, then write the answer as k : 1, for example 3 : 2.
Is the midpoint formula a special case of the section formula?
Yes. Put m_1 : m_2 = 1 : 1 into the section formula and the ratio terms cancel, leaving the average x = \frac{x_1 + x_2}{2}, y = \frac{y_1 + y_2}{2}.
How do I find the centroid of a triangle in ICSE maths?
Average the three vertices: G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right). The centroid also divides each median in the ratio 2 : 1 from the vertex.
What are the points of trisection of a line segment?
Trisection splits a segment into three equal parts. The first point divides the segment in the ratio 1 : 2 and the second in the ratio 2 : 1. Apply the section formula twice.
Reference: ICSE Class 10 Mathematics textbook material, chapter 13: Section and Mid Point Formula — CISCE-aligned.
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