Section Formula Class 10 ICSE — Das Gupta Chapter 13

Written by Greya LakshmiPublished on 28 September 2026

The section formula class 10 icse chapter — Distance and Section Formulae in the A. Das Gupta Mathematics textbook — teaches two tools of coordinate geometry. First, the distance formula measures how far apart two points are. Second, the section formula locates a point that divides a line segment in a given ratio. The chapter also derives the midpoint and the centroid of a triangle as special cases. This page walks through both theorems section by section.

Download Distance and Section Formulae Class 10 ICSE Mathematics Chapter 13 PDF

Download the Distance and Section Formulae Class 10 ICSE Mathematics chapter PDF from the A. Das Gupta textbook to revise offline or print for practice. It is the full chapter as printed in the book — 30 pages, 2 sections, 68 questions.

What is inside Chapter 13

30 pages · 2 sections · 68 exercise questions

Section	Title
1	Distance Formula
2	Section Formula

EXERCISE 13 (A): 29 questions
EXERCISE 13 (B): 39 questions

1 Distance Formula

The distance formula gives the length of the straight segment joining two points. For points $P(x_1, y_1)$ and $Q(x_2, y_2)$ , that length is:

$\mathrm{PQ} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

The proof shows why the formula holds. Draw perpendiculars PM and QN to the x-axis, then a horizontal from P to NQ. That builds a right triangle whose legs are $x_2 - x_1$ and $y_2 - y_1$ . Pythagoras’ theorem turns the square of each leg into the hypotenuse PQ.

A useful special case is distance from the origin. For a point $P(x, y)$ and the origin $O(0, 0)$ , the formula collapses to:

$\mathrm{OP} = \sqrt{x^2 + y^2}.$

Worked pattern. Find the distance between $A(1, 2)$ and $B(4, 6)$ . Let B supply $x_2$ and $y_2$ in both brackets:

$\mathrm{AB} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5 \text{ units}.$

One common trap: subtract in the same order for both coordinates. The squares remove signs, so order does not change the result, but mixing which point supplies $x_2$ and which supplies $y_2$ invites arithmetic slips.

Coordinate geometry in this ICSE chapter follows the Class 10 ICSE syllabus, which the Council for the Indian School Certificate Examinations publishes on its official CISCE website. The same axes appear in the Class 10 ICSE chapter on Reflection, where the distance formula measures the gap between a point and its image.

2 Section Formula

The section formula finds the point $P$ that divides the segment from $A(x_1, y_1)$ to $B(x_2, y_2)$ internally in the ratio $AP : PB = m : n$ . The coordinates of P are:

$x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}.$

This is the formula the Class 10 ICSE syllabus ties to ratio, midpoint and centroid questions.

The proof drops perpendiculars AL, BM and PQ to the x-axis, then compares similar triangles ARP and PSB. Corresponding sides are proportional, giving $m/n = (x - x_1)/(x_2 - x)$ and $m/n = (y - y_1)/(y_2 - y)$ . Solving each proportion produces the formula. Notice the endpoint closer to P carries the larger weight.

Putting $m = n = 1$ makes P the midpoint:

$\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).$

Each coordinate is the average of the two endpoint coordinates. The midpoint is also the pivot in circle problems, where the centre is the midpoint of a diameter. The centroid of a triangle follows the same logic. If D is the midpoint of BC, the centroid G divides median AD in the ratio 2 : 1, and the section formula reduces to:

$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right).$

Worked pattern. Divide $A(0, 0)$ and $B(6, 9)$ in the ratio $1 : 2$ . Take $m = 1$ , $n = 2$ , and A as $(x_1, y_1)$ :

$x = \frac{1(6) + 2(0)}{3} = 2, \quad y = \frac{1(9) + 2(0)}{3} = 3.$

So P is $(2, 3)$ , one third of the way from A. The distances AP and PB are in the ratio $1 : 2$ .

Two errors cost the most marks in this chapter:

Swapped ratio terms. Reversing $m$ and $n$ in the example above gives $(4, 6)$ , the other trisection point. Copy $AP : PB$ exactly as the question writes it.
Wrong point order. The $m$ term multiplies $x_2$ , and the $n$ term multiplies $x_1$ . Using $x_1$ where $x_2$ belongs shifts P toward the wrong end.

The Equation of a Line chapter builds directly on this coordinate work, using the same point and ratio ideas to find line equations.

What to practise from Section Formula Class 10 ICSE

This chapter has two exercise blocks. The first trains the distance formula; the second trains the section formula class 10 icse students apply to division and centroid problems. Some questions carry the year of the board exam in which they appeared, such as 1990, 1993, 2001, 2006 and 2008.

EXERCISE 13 (A) — 29 questions on the distance formula.
EXERCISE 13 (B) — 39 questions on the section formula.

Collinearity and shape-verification questions test the distance formula. Ratio, trisection, midpoint, median and centroid questions test the section formula.

Formula	Question types it solves
Distance formula	distance between points, distance from origin, points on axes, equidistant points, collinearity, triangles, rectangles, squares, parallelograms, rhombuses, circumcentre
Section formula	internal division, midpoint, trisection, ratio problems, medians, centroid

You can work through the full list of Class 10 ICSE Maths chapters to see how this coordinate unit sits in the book.

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 finds the coordinates of a point P that divides the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $AP : PB = m : n$ . Use it whenever the question gives a ratio and asks for the point, including midpoint and trisection problems.

How do I find the point that divides a line segment in a given ratio in Class 10 ICSE Maths?

Write the ratio as $AP : PB = m : n$ , reading from the first endpoint. Then substitute into $x = \frac{m x_2 + n x_1}{m + n}$ and $y = \frac{m y_2 + n y_1}{m + n}$ . Check that your point lies between A and B — a point outside that range means the ratio was applied backwards.

What is the difference between the distance formula and the section formula in Class 10 ICSE?

The distance formula returns a length between two points, measured in units. The section formula returns the coordinates of a point on the segment. Use distance for collinearity and shape questions; use the section formula when a ratio is given.

How do I find the centroid of a triangle using the section formula in Class 10 ICSE?

Average the three vertices: $G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$ . The chapter derives this by letting the centroid divide each median in the ratio 2 : 1, then applying the section formula.

How do I find the ratio in which a point divides a line segment in this ICSE chapter?

Assume P divides AB in the ratio $k : 1$ and substitute the coordinates into the section formula, then solve for k. When a segment meets the y-axis, the x-coordinate is 0, which gives k directly.

Reference: ICSE Class 10 Mathematics textbook material, chapter 13: Distance and Section Formulae — CISCE-aligned.

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