Class 10 · Chapter 7 · Coordinate Geometry unit (6 of 80 marks)

Coordinate Geometry Class 10: notes and important questions

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 40 board-style questions.

Practise this chapter Join freeFree account, no card: +5 practice questions on every chapter, with instant feedback.

40 questions
12 multiple choice, 3 assertion–reason, 8 very short answer, 9 short answer, 5 long answer, 3 case study
About 9 hours to master

Coordinate Geometry unit: 6 of 80 theory marks (Coordinate Geometry is the only chapter in this unit).

NCERT textbook: Chapter 7 on ncert.nic.in (PDF)

Revision notes

Coordinate Geometry — revision notes

1. Distance formula

$P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Distance from the origin: $\sqrt{x^{2} + y^{2}}$ . Distance from the $x$ -axis is $| y |$ ; from the $y$ -axis is $| x |$ .
Points on the $x$ -axis are $(x, 0)$ ; points on the $y$ -axis are $(0, y)$ .

2. Section formula (internal division)

The point dividing the segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m_{1} : m_{2}$ is $(\frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}})$

Finding an unknown ratio: take it as $k : 1$ and use whichever coordinate is known (e.g. $y = 0$ for the $x$ -axis).
Trisection points: ratios $1 : 2$ and $2 : 1$ . Four equal parts: mid-point, then mid-points of each half.

3. Mid-point formula

$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

4. Identifying shapes

Collinear: sum of two distances equals the third.
Right triangle: $a^{2} + b^{2} = c^{2}$ for the side lengths (converse of Pythagoras, known from earlier classes).
Parallelogram: diagonals bisect each other (same mid-point). Rhombus: all sides equal. Rectangle: opposite sides equal and diagonals equal. Square: all sides equal and diagonals equal.

Worked example 1

Distance between $(2, - 1)$ and $(- 4, 7)$ : $\sqrt{36 + 64} = 10$ .

Worked example 2

Point dividing $A (1, 7)$ , $B (9, - 5)$ in $3 : 1$ : $(\frac{27 + 1}{4}, \frac{- 15 + 7}{4}) = (7, - 2)$ .

Worked example 3

Ratio in which the $y$ -axis divides $P (- 2, 3)$ , $Q (6, - 1)$ : $\frac{6 k - 2}{k + 1} = 0 \Rightarrow k = \frac{1}{3}$ , ratio $1 : 3$ ; point $(0, 2)$ .

Common errors

Swapping $m_{1}$ and $m_{2}$ in the section formula — $m_{1}$ multiplies the coordinates of the second point.
Forgetting the square root, or squaring a negative difference incorrectly.
Concluding “square” from equal sides alone (a rhombus also has equal sides); check the diagonals.
Leaving the final distance unsimplified ( $\sqrt{72} = 6 \sqrt{2}$ ).

Board-exam tips

Write the formula first, then substitute — method marks are given for correct substitution.
For shape proofs, compute all four sides and both diagonals and then state the conclusion with a reason.
Note: area of a triangle using coordinates is not in the current syllabus.

Topics in this chapter: Distance formula · Mid-point formula · Section formula · Applications to geometric shapes.

Formulas for this chapter: Chapter 7 on the Class 10 formula sheet (free PDF and A3 poster).

Coordinate Geometry Class 10: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 6 pages · 570 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceDistance formula

How far apart are the points $A (- 2, 5)$ and $B (6, - 1)$ ?

Coordinate grid with A(−2, 5) and B(6, −1) joined by a dashed line. — A(−2, 5) and B(6, −1)

Show worked answer

$\begin{aligned} A B & = \sqrt{(6 + 2)^{2} + (- 1 - 5)^{2}} \\ = \sqrt{64 + 36} \\ = 10 \end{aligned}$ A1
Correct option: (b) $10$ units

Q2·1 mark·Multiple choiceMid-point formula

The mid-point of the segment joining $(2, - 6)$ and $(- 4, 8)$ is

Q3·1 mark·Assertion–reasonDistance formula

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): The point $(0, - 7)$ lies on the $y$ -axis.

Reason (R): Every point on the $x$ -axis has $y$ -coordinate $0$ .

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerSection formula

In what ratio does the $y$ -axis divide the segment joining $A (- 4, 7)$ and $B (6, - 3)$ ? Find the point of division.

Show worked answer

Answer:

2 : 3

;

(0, 3)

Ratio $k : 1$ : $\begin{aligned} x = \frac{6 k - 4}{k + 1} = 0 \\ \Rightarrow & k = \frac{2}{3} \end{aligned}$ , i.e. $2 : 3$ M1
$y = \frac{2 (- 3) + 3 (7)}{5} = 3$ : the point is $(0, 3)$ A1

Q5·2 marks·Short answerDistance formula

The point $P (x, - 1)$ is at a distance of $5$ units from $Q (2, 3)$ . Find $x$ .

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerApplications to geometric shapes

Prove that the points $A (- 1, 0)$ , $B (3, 3)$ , $C (6, - 1)$ and $D (2, - 4)$ , taken in order, form a square.

Coordinate grid with A(−1, 0), B(3, 3), C(6, −1) and D(2, −4) plotted and joined in order. — A(−1, 0), B(3, 3), C(6, −1), D(2, −4)

Show worked answer

Answer:

A B C D

is a square

$A B = \sqrt{16 + 9} = 5$ , $B C = \sqrt{9 + 16} = 5$ , $C D = \sqrt{16 + 9} = 5$ , $D A = \sqrt{9 + 16} = 5$ M1
All four sides are equal A1
Diagonals: $A C = \sqrt{49 + 1} = \sqrt{50}$ and $B D = \sqrt{1 + 49} = \sqrt{50}$ are equal, so $A B C D$ is a square A1

Q7·3 marks·Short answerSection formula

Find the coordinates of the points of trisection of the segment joining $A (- 3, 5)$ and $B (6, - 4)$ .

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerMid-point formula

The vertices of $△ A B C$ are $A (2, 3)$ , $B (- 4, 1)$ and $C (6, - 5)$ . (a) Find the mid-point $D$ of $B C$ and the length of the median $A D$ . (b) Find the point $G$ on $A D$ with $A G : G D = 2 : 1$ , and check that $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

Show worked answer

Answer:

D (1, - 2)

A D = \sqrt{26}

G (\frac{4}{3}, - \frac{1}{3})

(a) $\begin{aligned} D & = (\frac{- 4 + 6}{2}, \frac{1 - 5}{2}) \\ = (1, - 2) \end{aligned}$ A1
$\begin{aligned} A D & = \sqrt{(1 - 2)^{2} + (- 2 - 3)^{2}} \\ = \sqrt{26} \end{aligned}$ A1
(b) $G = (\frac{2 (1) + 1 (2)}{3}, \frac{2 (- 2) + 1 (3)}{3})$ M1
$G = (\frac{4}{3}, - \frac{1}{3})$ A1
$(\frac{2 - 4 + 6}{3}, \frac{3 + 1 - 5}{3}) = (\frac{4}{3}, - \frac{1}{3})$ : the same point A1

Q9·5 marks·Long answerDistance formula

A point $P (x, y)$ is equidistant from $A (5, 1)$ and $B (- 1, 5)$ . (a) Find a relation between $x$ and $y$ . (b) If $P$ also lies on the line $x + y = 10$ , find $P$ and its distance from $A$ .

Section E style · Case study (4 marks)

Q10·4 marks·Case studySection formula

Sports day drill. On a sports day, positions on the field are marked on a grid with $1$ unit $= 1$ m. Riya stands at $R (2, 3)$ and Sam at $S (8, 7)$ .

Find the distance between Riya and Sam. [1 mark]
The teacher stands exactly halfway between them. Find her position. [1 mark]
A flag $F$ is placed on $R S$ so that $R F : F S = 1 : 2$ . Find the coordinates of $F$ . [2 marks]

Sports field grid in metres with Riya at R(2, 3) and Sam at S(8, 7), joined by a dashed line. — Riya at R(2, 3) and Sam at S(8, 7)

Show worked answer

$\begin{aligned} R S & = \sqrt{6^{2} + 4^{2}} \\ = \sqrt{52} \\ = 2 \sqrt{13} \end{aligned}$ m A1
Mid-point $= (\frac{2 + 8}{2}, \frac{3 + 7}{2}) = (5, 5)$ A1
$F = (\frac{1 (8) + 2 (2)}{3}, \frac{1 (7) + 2 (3)}{3})$ M1
$F = (4, \frac{13}{3})$ A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 10 assertion–reason questions · the Class 10 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can find the distance between two points and the mid-point of a segment without slips.

Read first: 1. Distance formula; 3. Mid-point formula 11 practice questions · checkpoint: 3 questions, 5 marks, pass 80%

Practise step 1

Step 2

Board standard

You can use the section formula both ways: to find the dividing point and to find the ratio (or an unknown coordinate).

Read first: 2. Section formula (internal division); 3. Mid-point formula; Worked examples 1-3 18 practice questions · checkpoint: 4 questions, 11 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can prove the type of a quadrilateral or triangle from coordinates in a full, reasoned long answer.

Read first: 4. Identifying shapes; Board-exam tips 7 practice questions · checkpoint: 3 questions, 13 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can handle trisection points, points equidistant from three given points and right-angle conditions.

Read first: 2. Section formula; 4. Identifying shapes; Common errors 4 practice questions · checkpoint: 3 questions, 11 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 4 of the 40 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceDistance formula

The distance between the points $(- 3, 4)$ and $(5, - 2)$ is

(a) $8$
(b) $14$
(c) $\sqrt{28}$
(d) $10$

Q2·1 mark·Multiple choiceDistance formula

The distance of the point $(5, - 12)$ from the origin is

(a) $13$
(b) $17$
(c) $7$
(d) $\sqrt{119}$

Q3·1 mark·Multiple choiceDistance formula

If the distance between $(4, p)$ and $(1, 0)$ is $5$ , then $p$ equals

(a) $4$ only
(b) $\pm 3$
(c) $- 4$ only
(d) $\pm 4$

Worked solution·2 marks·Very short answerSection formula

In what ratio does the point $P (2, 3)$ divide the segment joining $A (- 2, 5)$ and $B (8, 0)$ ?

Full worked solution Free, no sign-in

Let the ratio be $k : 1$ : $\frac{8 k - 2}{k + 1} = 2 \Rightarrow 6 k = 4$ M1
$k = \frac{2}{3}$ : ratio $2 : 3$ (check: $y = \frac{0 + 15}{5} = 3$ ✓) A1

Every question here has a full worked solution. See them all →

Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I18 · Problem I86 · Problem I93 · Problem I100. All 28 →

Themed lesson: Halloween maths lesson · Women in maths maths lesson · Maths art maths lesson (starter, activity and worked answers). All themed maths

Knowledge organiser: Coordinate Geometry on one page (key facts, formulas and mistakes to avoid; printable PDF)

Formulas in this topic: Distance formula · Midpoint formula · Section formula. All formulas A–Z

Where marks are lost in Coordinate Geometry

Swapping m₁ and m₂ in the section formula. Fix: m₁ multiplies the coordinates of the second point; write ((m₁x₂ + m₂x₁)/(m₁ + m₂), …) before substituting.
Calling a quadrilateral a square from four equal sides. Fix: compute all four sides and both diagonals; equal sides and equal diagonals give a square, equal sides alone a rhombus.
Squaring negative differences wrongly or forgetting the square root. Fix: write (x₂ − x₁)² with brackets and take the root on the last line.
Leaving surds unsimplified (√72 instead of 6√2). Fix: simplify at the end; some schemes award the A1 only for the simplified form.
For 'ratio in which the y-axis divides', not using x = 0 (or y = 0 for the x-axis). Fix: take the ratio as k : 1 and set the relevant coordinate to zero.
Wasting time on area of a triangle by coordinates. Fix: it is deleted from the current syllabus; use distances and Pythagoras instead.

Exam strategy: how answers are step-marked · competency-based questions

Examiner Insights: common mistakes in CBSE Class 10 Maths, with fixes (our analysis of public sources) →

Coordinate Geometry in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 6, 7, 37
Sample paper 2 (Board standard): questions 7, 8, 20, 28
Sample paper 3 (Above board): questions 8, 9, 37
Sample paper 4 (Above board): questions 8, 23, 28
Sample paper 5 (95+ challenge): questions 7, 8, 37

Want it against the clock? Take a timed 30-mark chapter test on Coordinate Geometry, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Coordinate Geometry: common questions

What is in CBSE Class 10 Maths Chapter 7, Coordinate Geometry?

The topics are: Distance formula, Mid-point formula, Section formula, Applications to geometric shapes. This page has revision notes for each, and 40 board-style practice questions with step-marked solutions.

How many marks is Coordinate Geometry worth?

It is in the Coordinate Geometry unit, which carries 6 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Coordinate Geometry?

Swapping m₁ and m₂ in the section formula. Fix: m₁ multiplies the coordinates of the second point; write ((m₁x₂ + m₂x₁)/(m₁ + m₂), …) before substituting. Calling a quadrilateral a square from four equal sides. Fix: compute all four sides and both diagonals; equal sides and equal diagonals give a square, equal sides alone a rhombus.

Are the Coordinate Geometry important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

Work through the Route to 95 with your progress saved

Save your answers and track your progressA free account remembers the questions you have done and brings you back to this page.

Join free

Already have an account? Sign in

Try it first

Revision notes

Coordinate Geometry — revision notes

1. Distance formula

2. Section formula (internal division)

3. Mid-point formula

4. Identifying shapes

Worked example 1

Worked example 2

Worked example 3

Common errors

Board-exam tips

Coordinate Geometry Class 10: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Coordinate Geometry

Coordinate Geometry in our sample papers

Coordinate Geometry: common questions

Work through the Route to 95 with your progress saved