MathsCBSE Class 9 · Chapter 1

Orienting Yourself: The Use of Coordinates, Class 9 Notes and Solved Questions

Q: Is Coordinate Geometry in the CBSE Class 9 syllabus for 2026-27?

Yes. It is Unit III of the CBSE Class 9 Mathematics syllabus for 2026-27 and is taught through this chapter of Ganita Manjari Part I. The unit covers the Cartesian system, the distance between two points and the midpoint of a segment.

Q: The starred problems in the NCERT chapter are on midpoints and collinearity. Do I need them?

NCERT says starred (*) problems are for enrichment and are not to be assessed. But the CBSE syllabus for this unit itself lists the midpoint, checking whether three points lie on a line and checking for a right angle. So learn those skills at the level shown on this page; leave the harder starred puzzles (like trisection) for extra practice.

Q: Is the section formula or the area of a triangle in Class 9?

No. Dividing a segment in a given ratio (section formula) is a Class 10 topic. Class 9 needs only the midpoint. Area of a triangle from coordinates is not in the syllabus.

Q: Do I have to rationalise or simplify a surd answer like $\sqrt{80}$?

Writing \sqrt{80} is correct. If you can, simplify it to 4\sqrt{5}, because it is neater and easier to compare with other lengths. Do not round to a decimal unless the question asks.

Q: What is the difference between 'distance along the streets' and 'distance' between two points?

Distance between two points means the straight-line distance, from the distance formula. Distance along the streets (or grid lines) is the sum of the horizontal and vertical shifts. A question will tell you which one it wants.

Q: Why does the textbook call it the Baudhāyana–Pythagoras theorem?

The result was stated in Baudhāyana's Śulba-sūtra, which is older than Pythagoras. NCERT uses both names to give credit to both traditions. It is the same theorem you learnt in Grade 8.

Updated 15 October 2026 · Fact-checked

This Class 9 chapter teaches you to locate any point on a flat surface with two numbers, its coordinates (x, y), measured from two perpendicular axes that meet at the origin. You learn the four quadrants and their signs, the distance formula from the Baudhāyana–Pythagoras theorem, and the midpoint of a segment.

What Orienting Yourself: The Use of Coordinates is about

This is the first chapter of *Ganita Manjari* Part I, and CBSE calls it Coordinate Geometry (Unit III). The idea is simple and powerful: draw two number lines at right angles, and every point on the page gets an address of two numbers, written $(x, y)$ . The chapter opens with the long history of grid thinking, from the evenly spaced streets of the Sindhu-Sarasvatī cities and Baudhāyana's East–West and North–South lines, to Brahmagupta's zero and negative numbers and the work of Fermat and Descartes.

The story of Reiaan, who cannot see, and his sister Shalini runs through the chapter. Shalini makes a pin-and-thread map of his room on a grid so he can feel where the door, bed and table are. You use the same map to read coordinates, measure door widths and place furniture.

The chapter has three skills you must have for the exam:

reading and plotting points, including points on the axes and points in each quadrant
finding the distance between two points, first along lines parallel to an axis, then in any direction using the Baudhāyana–Pythagoras theorem
finding the midpoint of a segment (listed in the CBSE syllabus), and using distances to test whether three points lie on one line or a triangle is right-angled

Questions range from 1-mark MCQs on quadrants to 5-mark questions that ask you to prove a set of points forms a square, rhombus or right-angled triangle.

Weightage: This chapter is part of Unit III, Coordinate Geometry, which carries 4 of the 80 marks in the school annual exam.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit III Coordinate Geometry

Key concepts

Why we need coordinates

NCERT section 1.1 Introduction

A system of coordinates is a grid that lets you describe exactly where something is using numbers. On a single number line one number is enough. On a flat surface (a 2-D plane) you need two numbers: how far across and how far up or down.

The idea is very old. Planned cities of the Sindhu-Sarasvatī Civilisation had streets laid out North–South and East–West, about 10 m apart, so a place could be found by counting blocks. Brahmagupta treated zero and negative numbers as numbers in their own right, which is what makes the left and lower halves of today's plane possible. Descartes later wrote any point in a plane as a pair of numbers, which joined geometry and algebra.

For the exam, you are not asked dates. You should be able to explain in a line or two why two numbers are needed and why negative numbers are essential.

Worked example

In a planned town, the main roads cross at the clock tower. Take East as the positive x-direction and North as the positive y-direction, with 1 unit = 1 block. A bakery is 3 blocks East and 2 blocks South of the clock tower. Write its coordinates. If each block is 100 m, how far does a person walk along the streets from the clock tower to the bakery?

Try it first, then show the solutionSolution

East is positive, so the x-coordinate is $+ 3$ .
South is the negative y-direction, so the y-coordinate is $- 2$ .
The bakery is at $(3, - 2)$ .
Walking along the streets means 3 blocks East and then 2 blocks South: $3 + 2 = 5$ blocks.
$5 \times 100 = 500$ m.

Answer: The bakery is at $(3, - 2)$ , and the walk along the streets is 500 m.

Axes, origin and points on the axes

NCERT section 1.3 The 2-d Cartesian Coordinate System

Two number lines are drawn at right angles. The horizontal one is the x-axis and the vertical one is the y-axis. They cross at the origin O, whose coordinates are $(0, 0)$ . Distances are marked in equal units on both axes.

Right of O and above O are positive; left of O and below O are negative.
Every point on the x-axis has the form $(x, 0)$ : its y-coordinate is 0.
Every point on the y-axis has the form $(0, y)$ : its x-coordinate is 0.

So $(4.5, 0)$ is on the x-axis, 4.5 units right of O, and $(0, - 4.5)$ is on the y-axis, 4.5 units below O. When plotting, we often write P $(x, y)$ instead of $P = (x, y)$ .

Worked example

State where each point lies and on which side of the origin: $P (0, - 3.5)$ , $Q (- 6, 0)$ , $R (2.5, 0)$ , $S (0, 7)$ .

Try it first, then show the solutionSolution

P has x-coordinate 0, so it is on the y-axis. Its y-coordinate is negative, so it is 3.5 units below O.
Q has y-coordinate 0, so it is on the x-axis, 6 units to the left of O.
R has y-coordinate 0, so it is on the x-axis, 2.5 units to the right of O.
S has x-coordinate 0, so it is on the y-axis, 7 units above O.

Answer: P: y-axis, below O. Q: x-axis, left of O. R: x-axis, right of O. S: y-axis, above O.

The four quadrants and their sign pattern

NCERT section 1.3 The 2-d Cartesian Coordinate System

The axes split the Cartesian plane (also called the coordinate plane or xy-plane) into four quadrants, numbered anticlockwise starting from the top right.

Quadrant	Sign of x	Sign of y
I	+	+
II	−	+
III	−	−
IV	+	−

In $(x, y)$ , x is the perpendicular distance of the point from the y-axis (with a sign), and y is its perpendicular distance from the x-axis. The order matters: $(x, y)$ and $(y, x)$ are the same point only when $x = y$ . A point on an axis is not in any quadrant.

Worked example

(i) Name the quadrant or axis of each point: $A (- 2, 7)$ , $B (5, - 1)$ , $C (- 3, - 3)$ , $D (0, 4)$ . (ii) For what value of a is the point $(a, 6)$ the same as the point $(6, a)$ ?

Try it first, then show the solutionSolution

A: x negative, y positive, so Quadrant II.
B: x positive, y negative, so Quadrant IV.
C: both negative, so Quadrant III.
D: x-coordinate 0, so D is on the y-axis and is in no quadrant.
$(a, 6) = (6, a)$ only when the first coordinates match and the second coordinates match, so $a = 6$ .

Answer: (i) A: II, B: IV, C: III, D: on the y-axis. (ii) $a = 6$ .

Distance along a line parallel to an axis

NCERT section 1.3 The 2-d Cartesian Coordinate System

When two points have the same y-coordinate, the segment joining them is horizontal (parallel to the x-axis). Its length is just the difference of the x-coordinates, taken as positive:

$distance = ∣ x_{2} - x_{1} ∣$

When two points have the same x-coordinate, the segment is vertical and its length is $∣ y_{2} - y_{1} ∣$ .

This is how you find the width of a door on a floor plan. The absolute value bars mean the answer is never negative, whichever point you call the first one.

Worked example

On a floor plan (1 unit = 1 ft), a room door runs from $D_{1} (7, 0)$ to $D_{2} (10.5, 0)$ and a cupboard door runs from $C_{1} (0, 3)$ to $C_{2} (0, 5.5)$ . Find both widths and say which is wider.

Try it first, then show the solutionSolution

$D_{1}$ and $D_{2}$ have the same y-coordinate, so the room door width $= ∣10.5 - 7∣ = 3.5$ ft.
$C_{1}$ and $C_{2}$ have the same x-coordinate, so the cupboard door width $= ∣5.5 - 3∣ = 2.5$ ft.
$3.5 > 2.5$ .

Answer: The room door is 3.5 ft wide and the cupboard door is 2.5 ft wide, so the room door is wider by 1 ft.

The distance formula

NCERT section 1.4 Distance Between Two Points in the 2-D Plane

For two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ that are not in line with an axis, draw a horizontal line through one and a vertical line through the other. They meet at a corner C and make a right-angled triangle.

The horizontal side is $∣ x_{2} - x_{1} ∣$ (the shift along the x-axis).
The vertical side is $∣ y_{2} - y_{1} ∣$ (the shift along the y-axis).

By the Baudhāyana–Pythagoras theorem, the hypotenuse AB is

$A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

Because each shift is squared, it does not matter which point you take first or whether a shift is negative. Distance from the origin to $(x, y)$ is $x^{2} + y^{2}$ . Leave answers like $29$ as they are unless the question asks for a decimal.

Worked example

Find the distance between $A (- 2, 3)$ and $B (4, - 5)$ .

Try it first, then show the solutionSolution

Shift along x: $4 - (- 2) = 6$ .
Shift along y: $- 5 - 3 = - 8$ .
$A B = 6^{2} + (- 8)^{2} = 36 + 64 = 100$ .
$A B = 10$ units.

Answer: $A B = 10$ units.

Reflection in an axis keeps lengths the same

NCERT section 1.4 Distance Between Two Points in the 2-D Plane

Reflecting a point in the y-axis changes the sign of its x-coordinate: $(x, y) \to (- x, y)$ . Reflecting in the x-axis changes the sign of its y-coordinate: $(x, y) \to (x, - y)$ .

If you reflect all three corners of a triangle in an axis, the shifts along each axis keep the same size (only their signs may change). So every side keeps its length. The triangle flips over but does not change shape or size.

A point and its mirror image in the x-axis are $2∣ y ∣$ apart; in the y-axis they are $2∣ x ∣$ apart.

Worked example

P is $(3, - 4)$ . Write its images $P^{'}$ in the x-axis and $P^{''}$ in the y-axis. Find $P P^{'}$ and check that $O P = O P^{''}$ .

Try it first, then show the solutionSolution

Reflect in the x-axis: change the sign of y, so $P^{'} (3, 4)$ .
Reflect in the y-axis: change the sign of x, so $P^{''} (- 3, - 4)$ .
$P$ and $P^{'}$ share the x-coordinate 3, so $P P^{'} = ∣4 - (- 4) ∣ = 8$ units.
$O P = 3^{2} + (- 4)^{2} = 5$ and $O P^{''} = (- 3)^{2} + (- 4)^{2} = 5$ .

Answer: $P^{'} (3, 4)$ , $P^{''} (- 3, - 4)$ , $P P^{'} = 8$ units, and $O P = O P^{''} = 5$ units.

Testing for a right angle with coordinates

NCERT section 1.4 Distance Between Two Points in the 2-D Plane

The converse of the Baudhāyana–Pythagoras theorem says: if the square of the longest side equals the sum of the squares of the other two sides, the triangle has a right angle opposite the longest side.

So to test a triangle given by coordinates:
1. Find the squares of all three sides with the distance formula (no need to take square roots).
2. Check whether the biggest square equals the sum of the other two.

The same squared distances tell you whether the triangle is isosceles (two equal) or equilateral (three equal).

Worked example

Show that $A (1, 2)$ , $B (4, 6)$ and $C (8, 3)$ form a right-angled isosceles triangle. Where is the right angle?

Try it first, then show the solutionSolution

$A B^{2} = (4 - 1)^{2} + (6 - 2)^{2} = 9 + 16 = 25$ .
$B C^{2} = (8 - 4)^{2} + (3 - 6)^{2} = 16 + 9 = 25$ .
$A C^{2} = (8 - 1)^{2} + (3 - 2)^{2} = 49 + 1 = 50$ .
$A B^{2} + B C^{2} = 25 + 25 = 50 = A C^{2}$ , so the angle opposite AC, that is $∠ B$ , is $9 0^{\circ}$ .
$A B = B C = 5$ units, so the triangle is also isosceles.

Answer: It is right-angled at B and isosceles, with $A B = B C = 5$ units and $A C = 52$ units.

Do three points lie on one line?

The CBSE syllabus asks you to decide whether three points lie on a straight line. Two easy checks work at Class 9 level.

Distance check: find AB, BC and AC. The points lie on one line exactly when the two shorter distances add up to the longest one, for example $A B + B C = A C$ . If they do not, the points make a triangle.
Step check: going from A to B, note the shift along x and the shift along y. Do the same from B to C. If the second pair of shifts is the first pair multiplied by the same number, the direction has not changed, so the points are on one line.

The step check avoids square roots, so it is quicker when distances do not come out as whole numbers.

Worked example

Check whether $P (- 2, - 1)$ , $Q (1, 3)$ and $R (7, 11)$ lie on one line, using both checks.

Try it first, then show the solutionSolution

Distance check: $P Q = 3^{2} + 4^{2} = 5$ , $QR = 6^{2} + 8^{2} = 10$ , $P R = 9^{2} + 1 2^{2} = 15$ .
$P Q + QR = 5 + 10 = 15 = P R$ , so the points lie on one line.
Step check: P to Q shifts by $(3, 4)$ ; Q to R shifts by $(6, 8) = 2 \times (3, 4)$ .
Same direction, so the same conclusion.

Answer: Yes. P, Q and R lie on one straight line, with Q between P and R.

The midpoint of a segment

The midpoint M of the segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ is halfway along both axes, so each coordinate is the average:

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

The midpoint is in the CBSE Class 9 syllabus for this unit. In the NCERT book it appears through the end-of-chapter problems, so learn the formula and practise both directions:

given both ends, find the midpoint;
given one end A and the midpoint M, find the other end: $B = (2 m_{1} - x_{1}, 2 m_{2} - y_{1})$ , where $M = (m_{1}, m_{2})$ .

Worked example

(i) Find the midpoint of the segment joining $(- 5, 8)$ and $(9, - 2)$ . (ii) M $(5, 1)$ is the midpoint of AB, where A is $(2, - 3)$ . Find B.

Try it first, then show the solutionSolution

(i) $x = \frac{- 5 + 9}{2} = 2$ and $y = \frac{8 + ( - 2 )}{2} = 3$ , so the midpoint is $(2, 3)$ .
(ii) Let B be $(x, y)$ . Then $\frac{2 + x}{2} = 5$ , so $x = 8$ .
$\frac{- 3 + y}{2} = 1$ , so $y = 5$ .
Check: the midpoint of $(2, - 3)$ and $(8, 5)$ is $(5, 1)$ . Correct.

Answer: (i) $(2, 3)$ (ii) $B (8, 5)$

Maths in real life: Orienting Yourself: The Use of Coordinates

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Coordinates and quadrants

Suppose your town's two main roads cross at the clock tower, one running East–West and one North–South, and every street is 150 m from the next. You give directions using East and North as positive.

The library is at $(3, - 2)$ in street blocks. Describe how to reach it from the clock tower and how far you walk along the streets.

Try it first, then show the solutionSolution

$x = 3$ is positive, so go 3 blocks East.
$y = - 2$ is negative, so go 2 blocks South.
Total blocks $= 3 + 2 = 5$ , and $5 \times 150 = 750$ m.

Answer: Walk 3 blocks East and 2 blocks South, 750 m in all. The library is in Quadrant IV of the town map.

Memory hook: A pair of numbers is a full address: across first, then up or down.

Distance formula

Suppose a shop sends a parcel by drone. On the delivery map, 1 unit = 100 m, the shop is at $(- 2, 1)$ and your home is at $(4, 9)$ .

How far does the drone fly in a straight line?

Try it first, then show the solutionSolution

Shifts: $4 - (- 2) = 6$ and $9 - 1 = 8$ .
Distance $= 36 + 64 = 10$ units.
$10 \times 100 = 1000$ m.

Answer: 1000 m, that is 1 km.

Memory hook: Drones fly the hypotenuse; people walk the two legs.

Midpoint

Suppose two friends on a town map (1 unit = 50 m) are at $(1, 7)$ and $(9, 3)$ and want to meet exactly halfway.

Where should they meet, and how far does each walk in a straight line?

Try it first, then show the solutionSolution

Meeting point $= (\frac{1 + 9}{2}, \frac{7 + 3}{2}) = (5, 5)$ .
Distance from $(1, 7)$ to $(5, 5)$ $= 16 + 4 = 20 \approx 4.47$ units.
$4.47 \times 50 \approx 224$ m.

Answer: Meet at $(5, 5)$ ; each walks about 224 m.

Memory hook: Halfway is the average: add and halve.

Right-angle test

Suppose a mason is marking the corners of a room floor at $(0, 0)$ , $(12, 0)$ and $(12, 9)$ in feet, and wants to be sure the corner at $(12, 0)$ is square.

What length should the diagonal from $(0, 0)$ to $(12, 9)$ measure if the corner is a right angle?

Try it first, then show the solutionSolution

The two walls are 12 ft and 9 ft long.
For a right angle, diagonal $^{2} = 1 2^{2} + 9^{2} = 144 + 81 = 225$ .
Diagonal $= 15$ ft. By the distance formula, $(0, 0)$ to $(12, 9)$ is also $225 = 15$ ft.

Answer: 15 ft. If the tape shows 15 ft, the corner is square.

Memory hook: Masons check corners with Baudhāyana's rope: 9, 12, 15.

Point inside, on or outside a circle

Suppose a garden sprinkler at the origin waters everything within 5 m, and plants stand at $(3, 4)$ , $(- 4, 4)$ and $(1, - 5)$ (in metres).

Which plants get water?

Try it first, then show the solutionSolution

$(3, 4)$ : distance $9 + 16 = 5$ , exactly on the edge.
$(- 4, 4)$ : distance $32 \approx 5.66 > 5$ , outside.
$(1, - 5)$ : distance $26 \approx 5.10 > 5$ , outside.

Answer: Only the plant at $(3, 4)$ is reached, right at the edge. The other two need the sprinkler moved.

Memory hook: Compare distance from the centre with the radius.

Topics in this chapter

Orienting Yourself: The Use of Coordinates formula sheet

Origin: $O = (0, 0)$; Where the x-axis and y-axis cross.
Point on the x-axis: $(x, 0)$; y-coordinate is zero.
Point on the y-axis: $(0, y)$; x-coordinate is zero.
Quadrant signs: $I : (+, +), II : (-, +), III : (-, -), IV : (+, -)$; Numbered anticlockwise from the top right.
Order matters: $(x, y) = (y, x) ⟺ x = y$; Swapping the coordinates gives a different point unless they are equal.
Horizontal distance: $distance between (x_{1}, y) and (x_{2}, y) = ∣ x_{2} - x_{1} ∣$; Same y-coordinate: the segment is parallel to the x-axis.
Vertical distance: $distance between (x, y_{1}) and (x, y_{2}) = ∣ y_{2} - y_{1} ∣$; Same x-coordinate: the segment is parallel to the y-axis.
Distance formula: $A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; From the Baudhāyana–Pythagoras theorem.
Distance from the origin: $O P = x^{2} + y^{2}$; Special case with $(x_{1}, y_{1}) = (0, 0)$ .
Midpoint: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$; Average the x-coordinates and average the y-coordinates.
Other end from a midpoint: $B = (2 m_{1} - x_{1}, 2 m_{2} - y_{1})$; When A $(x_{1}, y_{1})$ and the midpoint M $(m_{1}, m_{2})$ are known.
Reflections: $in the x-axis : (x, y) \to (x, - y); in the y-axis : (x, y) \to (- x, y)$; Lengths do not change under reflection.
Right-angle test: $A B^{2} + B C^{2} = A C^{2} \Rightarrow ∠ B = 9 0^{\circ}$; Converse of the Baudhāyana–Pythagoras theorem; compare squares, not roots.
Three points on one line: $A B + B C = A C$; B lies between A and C on the segment AC.
Point and circle: $d = distance from centre; d < r inside, d = r on, d > r outside$; Compare the distance from the centre with the radius r.

How the school exam asks Orienting Yourself: The Use of Coordinates

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Quadrant and axis MCQs	A point is given and you choose its quadrant, or you are told a point lies on an axis and must pick its form, such as $(0, y)$ .	Look at the signs first. If either coordinate is 0, the point is on an axis and in no quadrant. Then use the sign table (+, +), (−, +), (−, −), (+, −).
Reading a floor plan or map	A room or town is drawn on a grid. You read off corners, find a door width, place a table or find the fourth corner of a rectangle.	Write the coordinates of every corner you need. Lengths parallel to an axis are differences of one coordinate. For a rectangle with sides along the grid, the fourth corner takes its x from one neighbour and its y from the other.
Distance between two points	Find AB, or find a perimeter, or find how far a point is from the origin.	Write the formula, then substitute with brackets around negative numbers. Simplify the squares before adding. Give the answer with units, and as a surd if it is not a perfect square.
Unknown coordinate from a distance	The distance between $(x, 4)$ and $(1, - 2)$ is 10; find x. Or find a point on an axis that is equally far from two given points.	Square both sides to remove the root. Expect two answers when you take a square root, and keep both unless the question rules one out. For the axis question, write the point as $(x, 0)$ or $(0, y)$ and set the two squared distances equal.
Proving a shape	Show that four points form a square, rhombus or rectangle, or that three points form a right-angled or isosceles triangle.	Find all sides, and for quadrilaterals both diagonals. Square: four equal sides and equal diagonals. Rhombus: four equal sides, unequal diagonals. Rectangle: opposite sides equal and equal diagonals. Right angle: use the converse of the Baudhāyana–Pythagoras theorem.
Midpoint and collinearity	Find a midpoint or a missing end; decide whether three points lie on one line.	Average the coordinates for a midpoint; double and subtract for a missing end. For collinearity, compare shifts (step check) or check whether two distances add up to the third.

Tricks to remember Orienting Yourself: The Use of Coordinates

Short memory aids. Say each one aloud once, then use it on a question the same day.

Walk along the corridor, then take the lift
To plot $(x, y)$ , first walk along the corridor (x, left or right), then take the lift (y, up or down).
Use it when: Stops you from swapping x and y when you plot or read a point.
All, Second, Third, Fourth signs
Quadrant I is happy (+, +). Go anticlockwise and the minus sign appears first in x (II), spreads to both (III), then leaves x and stays in y (IV).
Use it when: Rebuild the quadrant sign table in seconds.
Zero tells you the axis
Zero in the y-place means you are lying on the x-axis floor; zero in the x-place means you are standing on the y-axis wall.
Use it when: Points like $(0, - 3)$ and $(5, 0)$ in MCQs.
Shift, square, add, root
Distance formula in four words: find the shifts, square them, add them, take the root.
Use it when: Keeps you from forgetting either the squares or the square root.
Midpoint is the average
Midpoint = average marks of two tests: add and halve, separately for x and y.
Use it when: Writing the midpoint formula correctly, with a plus sign, not a minus.

Orienting Yourself: The Use of Coordinates: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Point on an axis · 1 mark · MCQExam level

The point $(- 4, 0)$ lies
(a) in Quadrant II
(b) on the positive x-axis
(c) on the negative x-axis
(d) on the negative y-axis

Try it first, then show the solutionSolution

The y-coordinate is 0, so the point is on the x-axis, not in a quadrant.
The x-coordinate is negative, so it is to the left of O.

Answer: (c) on the negative x-axis

Example 2: Quadrant signs (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): The point $(- 3, - 7)$ lies in Quadrant III.
Reason (R): In Quadrant III, both coordinates of a point are negative.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R is the sign rule for Quadrant III, so R is true.
Both coordinates of $(- 3, - 7)$ are negative, so by R the point is in Quadrant III. A is true and R explains it.

Answer: (a)

Example 3: Distance with negative coordinates · 2 marks · Very short answerExam level

Find the distance between $(- 1, - 3)$ and $(5, 5)$ .

Try it first, then show the solutionSolution

Shifts: $5 - (- 1) = 6$ along x, $5 - (- 3) = 8$ along y.
Distance $= 6^{2} + 8^{2} = 36 + 64 = 100 = 10$ units.

Answer: 10 units

Example 4: Distance from a mirror image · 2 marks · Very short answerExam level

A is the point $(4, - 3)$ and $A^{'}$ is its image in the y-axis. Find $A A^{'}$ and the distance of $A^{'}$ from the origin.

Try it first, then show the solutionSolution

Image in the y-axis: change the sign of x, so $A^{'} (- 4, - 3)$ .
A and $A^{'}$ have the same y-coordinate, so $A A^{'} = ∣4 - (- 4) ∣ = 8$ units.
$O A^{'} = (- 4)^{2} + (- 3)^{2} = 5$ units.

Answer: $A A^{'} = 8$ units and $O A^{'} = 5$ units.

Example 5: Unknown coordinate from a distance · 3 marks · Short answerHarder exam level

The distance between $P (x, 4)$ and $Q (1, - 2)$ is 10 units. Find the possible values of x.

Try it first, then show the solutionSolution

$P Q^{2} = (x - 1)^{2} + (4 - (- 2))^{2} = (x - 1)^{2} + 36$ .
$P Q = 10$ , so $(x - 1)^{2} + 36 = 100$ , giving $(x - 1)^{2} = 64$ .
$x - 1 = 8$ or $x - 1 = - 8$ , so $x = 9$ or $x = - 7$ .
Check: for $x = 9$ , $64 + 36 = 10$ ; for $x = - 7$ , $64 + 36 = 10$ . Both work.

Answer: $x = 9$ or $x = - 7$

Example 6: Point on the x-axis equidistant from two points · 3 marks · Short answerHarder exam level

Find the point on the x-axis that is equidistant from $A (2, - 5)$ and $B (- 2, 9)$ .

Try it first, then show the solutionSolution

A point on the x-axis is $P (x, 0)$ .
$P A^{2} = (x - 2)^{2} + 25$ and $P B^{2} = (x + 2)^{2} + 81$ .
Set them equal: $x^{2} - 4 x + 4 + 25 = x^{2} + 4 x + 4 + 81$ .
$- 8 x = 56$ , so $x = - 7$ .
Check: $P A^{2} = 81 + 25 = 106$ and $P B^{2} = 25 + 81 = 106$ . Equal.

Answer: $P (- 7, 0)$ , which is $106$ units from both A and B.

Example 7: Other end of a diameter · 3 marks · Short answerHarder exam level

The centre of a circle is $C (2, - 3)$ and one end of a diameter is $A (- 1, 1)$ . Find the other end B of the diameter and the radius of the circle.

Try it first, then show the solutionSolution

The centre is the midpoint of the diameter AB. Let B be $(x, y)$ .
$\frac{- 1 + x}{2} = 2 \Rightarrow x = 5$ and $\frac{1 + y}{2} = - 3 \Rightarrow y = - 7$ .
Radius $= C A = (- 1 - 2)^{2} + (1 + 3)^{2} = 9 + 16 = 5$ units.
Check: $C B = 3^{2} + (- 4)^{2} = 5$ . Equal, as it must be.

Answer: $B (5, - 7)$ and the radius is 5 units.

Example 8: Proving a square · 5 marks · Long answerHarder exam level

Show that $A (2, 1)$ , $B (5, 5)$ , $C (9, 2)$ and $D (6, - 2)$ are the vertices of a square. Find its area.

Try it first, then show the solutionSolution

$A B = 3^{2} + 4^{2} = 5$ and $B C = 4^{2} + (- 3)^{2} = 5$ .
$C D = (- 3)^{2} + (- 4)^{2} = 5$ and $D A = (- 4)^{2} + 3^{2} = 5$ .
All four sides are equal, so ABCD is at least a rhombus.
Diagonals: $A C = 7^{2} + 1^{2} = 50$ and $B D = 1^{2} + (- 7)^{2} = 50$ . Equal.
Four equal sides and equal diagonals, so ABCD is a square.
Area $= 5^{2} = 25$ square units.

Answer: ABCD is a square of side 5 units, and its area is 25 square units.

Example 9: Same line or not? (step check) · 3 marks · Short answerHOTS / competency

Without plotting, decide whether $A (1, - 1)$ , $B (4, 3)$ and $C (10, 10)$ lie on one straight line. Explain why the distance check is awkward here and which check you prefer.

Try it first, then show the solutionSolution

Shifts from A to B: $(4 - 1, 3 - (- 1)) = (3, 4)$ .
Shifts from B to C: $(10 - 4, 10 - 3) = (6, 7)$ .
If the points were on one line, $(6, 7)$ would be a multiple of $(3, 4)$ . Doubling $(3, 4)$ gives $(6, 8)$ , not $(6, 7)$ , so the direction changes at B.
Distances are $A B = 5$ , $B C = 85$ , $A C = 202$ . In decimals $5 + 9.22 = 14.22$ and $202 \approx 14.21$ , so rounding nearly hides the difference. The step check gives an exact answer with no roots.

Answer: No, the points do not lie on one line. The step check is better here because the distances are surds that are very close after rounding.

Example 10: Icons on a screen · 5 marks · Long answerHOTS / competency

A screen is 800 pixels wide and 600 pixels high, with the origin at the bottom-left corner. A round icon of radius 90 pixels is centred at $A (120, 140)$ and another of radius 120 pixels is centred at $B (300, 260)$ . (i) Does any part of either icon go off the screen? (ii) Do the two icons overlap?

Try it first, then show the solutionSolution

(i) Icon A reaches from $x = 120 - 90 = 30$ to $x = 210$ , and from $y = 140 - 90 = 50$ to $y = 230$ . All values are between 0 and the screen edges, so it fits.
Icon B reaches from $x = 300 - 120 = 180$ to $x = 420$ , and from $y = 260 - 120 = 140$ to $y = 380$ . It also fits.
(ii) Distance between centres: $A B = 18 0^{2} + 12 0^{2} = 32400 + 14400 = 46800 \approx 216.3$ pixels.
Two circles overlap only if the distance between centres is less than the sum of the radii, $90 + 120 = 210$ .
$216.3 > 210$ , so there is a gap of about 6 pixels between them.

Answer: (i) No, both icons are fully on the screen. (ii) No, they do not overlap; the centres are about 216.3 pixels apart, more than 210.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Planning a classroom reading corner

A Class 9 section draws its classroom floor on graph paper with 1 unit = 1 ft. The corners of the room are $O (0, 0)$ , $A (24, 0)$ , $B (24, 18)$ and $C (0, 18)$ . The door is on wall AB, from $(24, 5)$ to $(24, 8.5)$ . The teacher's desk is at $T (4, 15)$ and a student bench is at $S (16, 6)$ . The class wants to place a water cooler exactly halfway between the desk and the bench.

(i) · 1 mark
How wide is the door?
Answer
Answer: 3.5 ft
Both ends have x = 24, so the width is $∣8.5 - 5∣ = 3.5$ ft.
(ii) · 1 mark
Find the straight-line distance between the teacher's desk and the bench.
Answer
Answer: 15 ft
$T S = (16 - 4)^{2} + (6 - 15)^{2} = 144 + 81 = 225 = 15$ ft.
(iii) · 2 marks
Find the position of the water cooler and its distance from corner O.
Answer
Answer: $(10, 10.5)$ , 14.5 ft from O
Midpoint of TS $= (\frac{4 + 16}{2}, \frac{15 + 6}{2}) = (10, 10.5)$ (1 mark). Distance from O $= 1 0^{2} + 10. 5^{2} = 100 + 110.25 = 210.25 = 14.5$ ft (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the distance between $P (- 6, 7)$ and $Q (- 1, - 5)$ . (2 marks)

Step	Marks
$P Q = (- 1 - (- 6))^{2} + (- 5 - 7)^{2}$ , correct substitution in the distance formula	1
$= 25 + 144 = 169 = 13$ units	1

Find a if the point $(a, 2)$ is equidistant from $(3, - 4)$ and $(- 1, 6)$ . (3 marks)

Step	Marks
Equal distances, so equal squares: $(a - 3)^{2} + (2 + 4)^{2} = (a + 1)^{2} + (2 - 6)^{2}$	1
$a^{2} - 6 a + 9 + 36 = a^{2} + 2 a + 1 + 16$ , so $- 8 a = - 28$	1
$a = 3.5$ . Check: both squared distances are $36.25$	1

Show that $A (- 2, - 1)$ , $B (1, 3)$ , $C (6, 3)$ and $D (3, - 1)$ are the vertices of a rhombus but not of a square. Also show that its diagonals bisect each other. (5 marks)

Step	Marks
$A B = 3^{2} + 4^{2} = 5$ , $B C = 5^{2} + 0^{2} = 5$	1
$C D = (- 3)^{2} + (- 4)^{2} = 5$ , $D A = (- 5)^{2} + 0^{2} = 5$ ; all four sides equal	1
Diagonals: $A C = 8^{2} + 4^{2} = 80$ , $B D = 2^{2} + (- 4)^{2} = 20$ ; unequal, so not a square	1
Conclusion: four equal sides with unequal diagonals, so ABCD is a rhombus and not a square	1
Midpoint of AC $= (2, 1)$ and midpoint of BD $= (2, 1)$ ; same point, so the diagonals bisect each other	1

Where students lose marks in Orienting Yourself: The Use of Coordinates

The slips examiners see most often, and what to write instead.

Mistake: Swapping x and y
Students read the vertical scale first or write the coordinates in the order they noticed them.
Wrong working
The point 2 units right and 5 units down is written $(- 5, 2)$ .
Right working
x first (2 right, so $+ 2$ ), then y (5 down, so $- 5$ ): $(2, - 5)$ .
Do this: Always say 'across, then up or down' before you write a point.
Usually costs: The whole 1-mark answer, and every later step that uses that point.
Mistake: Sign errors when subtracting a negative
Writing $3 - - 5$ quickly as $- 2$ .
Wrong working
$A B = (3 - 5)^{2} + \dots$ for $A (- 5, 1)$ and $B (3, 7)$
Right working
$A B = (3 - (- 5))^{2} + (7 - 1)^{2} = 64 + 36 = 10$
Do this: Put every negative coordinate in brackets when you substitute.
Usually costs: Usually 1 mark for the wrong value and the final answer mark.
Mistake: Adding the shifts instead of using the formula
Walking distance along streets is confused with straight-line distance.
Wrong working
Distance from $(0, 0)$ to $(3, 4)$ $= 3 + 4 = 7$
Right working
$3^{2} + 4^{2} = 25 = 5$
Do this: Add the shifts only when the question asks for a route along the grid lines. For the straight-line distance, square, add and take the root.
Usually costs: All the marks for that distance.
Mistake: Calling a rhombus a square
Four equal sides look like enough.
Wrong working
All four sides are 5 units, so ABCD is a square.
Right working
All four sides are 5 units, and the diagonals are $80$ and $20$ , which are unequal, so ABCD is a rhombus, not a square.
Do this: For a square, check the diagonals too: they must be equal.
Usually costs: 1 to 2 marks in a 5-mark proof.
Mistake: Finding the other end of a segment by averaging
The midpoint formula is applied the wrong way round.
Wrong working
A $(2, - 3)$ , M $(5, 1)$ , so B $= (\frac{2 + 5}{2}, \frac{- 3 + 1}{2})$
Right working
$\frac{2 + x}{2} = 5 \Rightarrow x = 8$ ; $\frac{- 3 + y}{2} = 1 \Rightarrow y = 5$ ; B $(8, 5)$
Do this: Write the midpoint equation with the unknown end and solve it. Check by finding the midpoint again.
Usually costs: The whole 2-mark answer.
Mistake: Dropping one of the two answers
Taking the square root of $(x - 1)^{2} = 64$ as only $x - 1 = 8$ .
Wrong working
$(x - 1)^{2} = 64 \Rightarrow x = 9$
Right working
$x - 1 = 8$ or $x - 1 = - 8$ , so $x = 9$ or $x = - 7$
Do this: Write ' $\pm$ ' every time you take a square root, then check each value.
Usually costs: 1 mark in a 3-mark question.

NCERT exercise guide: Orienting Yourself: The Use of Coordinates

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise Set 1.1

What it tests: Reading Reiaan's room plan: the distance of a door from each wall, coordinates of door ends, door widths, and whether a door is wide enough for a wheelchair.

How to approach it: Read the coordinates of each end from the plan. If two ends lie on the same axis or on a line parallel to it, subtract the coordinate that changes. Compare widths by subtraction and comment in a sentence.

Like the door-width questions in Exercise Set 1.1

On a plan where 1 unit = 1 ft, a room door runs along the x-axis from $(6.5, 0)$ to $(9.5, 0)$ , and a balcony door runs along the y-axis from $(0, 2)$ to $(0, 4.5)$ . (i) How far is the room door from the y-axis? (ii) Find both widths. (iii) Which door is narrower?

Show the solutionSolution

(i) The end nearest the y-axis is $(6.5, 0)$ , so the door starts 6.5 ft from the y-axis wall.
(ii) Room door: $∣9.5 - 6.5∣ = 3$ ft.
Balcony door: $∣4.5 - 2∣ = 2.5$ ft.
(iii) $2.5 < 3$ , so the balcony door is narrower by $0.5$ ft.

Answer: (i) 6.5 ft (ii) 3 ft and 2.5 ft (iii) the balcony door, by 0.5 ft

Exercise Set 1.2

What it tests: Placing furniture on a full-size grid: finding the fourth foot of a rectangular table, its length and width, marking spaces of given size, and placing a table exactly in the centre of a room.

How to approach it: For a rectangle with sides along the grid, the missing corner takes the x-coordinate of one neighbour and the y-coordinate of the other. To centre an object, find the centre of the room (average of opposite corners), then go half the object's length and half its width either side.

Like the table-placing questions in Exercise Set 1.2

A room has corners $(0, 0)$ , $(16, 0)$ , $(16, 12)$ and $(0, 12)$ , with 1 unit = 1 ft. (i) Three feet of a study table are at $(3, 2)$ , $(8, 2)$ and $(8, 5)$ . Find the fourth foot and the table's length and width. (ii) A $4$ ft $\times 2$ ft rug is to be placed exactly in the centre of the room, with its longer side along the x-direction. Write the coordinates of its corners.

Show the solutionSolution

(i) $(3, 2)$ and $(8, 2)$ share y = 2; $(8, 2)$ and $(8, 5)$ share x = 8. The fourth corner shares x with $(3, 2)$ and y with $(8, 5)$ : $(3, 5)$ .
Length $= ∣8 - 3∣ = 5$ ft and width $= ∣5 - 2∣ = 3$ ft.
(ii) Centre of the room $= (\frac{0 + 16}{2}, \frac{0 + 12}{2}) = (8, 6)$ .
Half the length is 2 and half the width is 1, so x runs from 6 to 10 and y from 5 to 7.
Corners: $(6, 5)$ , $(10, 5)$ , $(10, 7)$ , $(6, 7)$ .

Answer: (i) Fourth foot $(3, 5)$ ; table 5 ft long and 3 ft wide. (ii) Rug corners $(6, 5)$ , $(10, 5)$ , $(10, 7)$ and $(6, 7)$ .

End-of-chapter Exercises

What it tests: Mixed problems: the origin, points on a line parallel to an axis, mirror images in an axis, right-angled triangles on the grid, distances, a city street grid and a test of whether four points form a square. Problems 6 to 13 are starred (enrichment, not assessed by NCERT), but midpoints and collinearity are in the CBSE syllabus, so practise the basic versions.

How to approach it: Sketch the points first; a rough sketch often shows the answer. Then confirm with distances. For a square, find four sides and two diagonals; the area is the square of a side.

Like the square-checking problem at the end of the chapter

Plot $A (3, 1)$ , $B (- 1, 3)$ , $C (- 3, - 1)$ and $D (1, - 3)$ . Is ABCD a square? Find its area.

Show the solutionSolution

$A B^{2} = (- 4)^{2} + 2^{2} = 20$ , $B C^{2} = (- 2)^{2} + (- 4)^{2} = 20$ .
$C D^{2} = 4^{2} + (- 2)^{2} = 20$ , $D A^{2} = 2^{2} + 4^{2} = 20$ . All four sides are $20$ .
Diagonals: $A C^{2} = (- 6)^{2} + (- 2)^{2} = 40$ and $B D^{2} = 2^{2} + (- 6)^{2} = 40$ , so the diagonals are equal.
Four equal sides and equal diagonals, so ABCD is a square.
Area $= (side)^{2} = 20$ square units.

Answer: Yes, ABCD is a square with side $20 = 25$ units and area 20 square units.

Practice questions on Orienting Yourself: The Use of Coordinates

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of these points lies on the y-axis?
(a) $(2, 0)$
(b) $(0, - 3)$
(c) $(- 3, - 3)$
(d) $(1, 1)$
Hint
On the y-axis the x-coordinate is 0.
Answer
(b) $(0, - 3)$
Q2 · 1 mark · MCQExam level
The point $(- 7, 2.5)$ lies in
(a) Quadrant I
(b) Quadrant II
(c) Quadrant III
(d) Quadrant IV
Hint
Negative x, positive y.
Answer
(b) Quadrant II
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The distance between $(3, 4)$ and $(- 3, 4)$ is 6 units.
Reason (R): The distance between two points with the same y-coordinate is the difference of their x-coordinates, taken as positive.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Both points have y = 4.
Answer
(a). $∣3 - (- 3) ∣ = 6$ , which is exactly what R says.
Q4 · 2 marks · Very short answerExam level
Find the distance of the point $(- 8, 15)$ from the origin.
Hint
Use $x^{2} + y^{2}$ .
Answer
$64 + 225 = 289 = 17$ units.
Q5 · 3 marks · Short answerHarder exam level
Find y if the distance between $(2, y)$ and $(- 4, 3)$ is 10 units.
Hint
Square both sides; expect two values.
Answer
$36 + (y - 3)^{2} = 100$ , so $(y - 3)^{2} = 64$ and $y = 11$ or $y = - 5$ .
Q6 · 3 marks · Short answerHarder exam level
Show that $A (- 3, 2)$ , $B (1, - 2)$ and $C (5, 2)$ form a right-angled isosceles triangle.
Hint
Compare $A B^{2} + B C^{2}$ with $A C^{2}$ .
Answer
$A B^{2} = 16 + 16 = 32$ , $B C^{2} = 16 + 16 = 32$ , $A C^{2} = 64$ . Since $32 + 32 = 64$ , $∠ B = 9 0^{\circ}$ , and $A B = B C = 42$ , so it is isosceles too.
Q7 · 2 marks · Very short answerHarder exam level
$M (4, - 1)$ is the midpoint of AB, where A is $(- 2, 5)$ . Find B.
Hint
Set up $\frac{- 2 + x}{2} = 4$ and $\frac{5 + y}{2} = - 1$ .
Answer
$x = 10$ and $y = - 7$ , so $B (10, - 7)$ .
Q8 · 3 marks · Short answerHarder exam level
Find the point on the y-axis that is equidistant from $(- 5, - 2)$ and $(3, 2)$ .
Hint
Take the point as $(0, y)$ .
Answer
$25 + (y + 2)^{2} = 9 + (y - 2)^{2}$ gives $8 y = - 16$ , so $y = - 2$ . The point is $(0, - 2)$ , 5 units from each.
Q9 · 5 marks · Long answerHarder exam level
Show that $A (- 1, - 2)$ , $B (1, 0)$ , $C (- 1, 2)$ and $D (- 3, 0)$ form a square and find its area.
Hint
Four sides and two diagonals.
Answer
Each side is $8$ (shifts of 2 and 2). Diagonals $A C = 4$ and $B D = 4$ are equal. So ABCD is a square, and its area is $(8)^{2} = 8$ square units.
Q10 · 3 marks · Short answerHarder exam level
A triangle has vertices $O (0, 0)$ , $P (6, 0)$ and $Q (3, 5)$ . Is it equilateral? Find its perimeter, leaving surds as they are.
Hint
Find all three sides.
Answer
$O P = 6$ , $O Q = 9 + 25 = 34$ , $P Q = 9 + 25 = 34$ . Two sides are equal but $34 \neq = 6$ , so it is isosceles, not equilateral. Perimeter $= 6 + 234$ units.
Q11 · 3 marks · Short answerHOTS / competency
Do $A (- 4, - 7)$ , $B (- 1, - 3)$ and $C (5, 5)$ lie on one line? Use two different methods.
Hint
Compare the shifts A to B and B to C, then check distances.
Answer
Yes. Shifts are $(3, 4)$ and $(6, 8) = 2 \times (3, 4)$ . Also $A B = 5$ , $B C = 10$ , $A C = 15$ and $5 + 10 = 15$ .
Q12 · 4 marks · Case-basedHOTS / competency
In a city grid, streets are 200 m apart and the main crossing is the origin. Riya is at the crossing $(2, 5)$ and her friend at $(7, 1)$ . How far must Riya walk along the streets to reach her friend, and how far would a drone fly in a straight line? Which is shorter, and by how much (to the nearest metre)?
Hint
Walking adds the shifts; flying uses the distance formula. Convert units to metres at the end.
Answer
Shifts are 5 and 4 units, so walking is $9 \times 200 = 1800$ m. Flying is $25 + 16 = 41 \approx 6.403$ units $\approx 1281$ m. The drone route is shorter by about 519 m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What are the coordinates of the origin?Show answer
$(0, 0)$
2.What is the y-coordinate of every point on the x-axis?Show answer
0. Points on the x-axis have the form $(x, 0)$ .
3.In which quadrant are both coordinates negative?Show answer
Quadrant III.
4.What does the x-coordinate of a point measure?Show answer
Its perpendicular distance from the y-axis, with a + sign to the right and a − sign to the left.
5.Write the distance formula.Show answer
$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
6.Which theorem gives the distance formula?Show answer
The Baudhāyana–Pythagoras theorem.
7.What is the midpoint of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ?Show answer
$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
8.What happens to $(x, y)$ when it is reflected in the y-axis?Show answer
It becomes $(- x, y)$ . Lengths stay the same.
9.When is $(x, y)$ the same point as $(y, x)$ ?Show answer
Only when $x = y$ .
10.How do you tell a rhombus from a square using coordinates?Show answer
Both have four equal sides; a square also has equal diagonals.

Best order to study it

Read the history in 1.1 once so you can say why two numbers and negative numbers are needed.
Practise plotting and reading points, including points on the axes, until you never swap x and y.
Learn the quadrant sign table and test yourself with ten random points.
Find lengths parallel to the axes on a floor plan (Exercise Sets 1.1 and 1.2).
Learn the distance formula and do at least ten distance questions with negative coordinates.
Finish with the midpoint, right-angle and collinearity checks, then the shape proofs (square, rhombus).

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw the axes, mark the origin and label the four quadrants without looking?
Can you say the sign pattern of each quadrant?
Can you write the form of a point on each axis?
Can you plot $(- 3.5, 2)$ and $(2, - 3.5)$ and explain why they are different?
Can you find the width of a door whose ends share a y-coordinate?
Can you write the distance formula and say where it comes from?
Can you find an unknown coordinate from a given distance, keeping both answers?
Can you find a midpoint, and the other end from a midpoint?
Can you prove a triangle is right-angled using squared distances?
Can you separate a square from a rhombus using the diagonals?

Orienting Yourself: The Use of Coordinates: frequently asked questions

Is Coordinate Geometry in the CBSE Class 9 syllabus for 2026-27?

Yes. It is Unit III of the CBSE Class 9 Mathematics syllabus for 2026-27 and is taught through this chapter of Ganita Manjari Part I. The unit covers the Cartesian system, the distance between two points and the midpoint of a segment.

The starred problems in the NCERT chapter are on midpoints and collinearity. Do I need them?

NCERT says starred (*) problems are for enrichment and are not to be assessed. But the CBSE syllabus for this unit itself lists the midpoint, checking whether three points lie on a line and checking for a right angle. So learn those skills at the level shown on this page; leave the harder starred puzzles (like trisection) for extra practice.

Is the section formula or the area of a triangle in Class 9?

No. Dividing a segment in a given ratio (section formula) is a Class 10 topic. Class 9 needs only the midpoint. Area of a triangle from coordinates is not in the syllabus.

Do I have to rationalise or simplify a surd answer like $\sqrt{80}$?

Writing \sqrt{80} is correct. If you can, simplify it to 4\sqrt{5}, because it is neater and easier to compare with other lengths. Do not round to a decimal unless the question asks.

What is the difference between 'distance along the streets' and 'distance' between two points?

Distance between two points means the straight-line distance, from the distance formula. Distance along the streets (or grid lines) is the sum of the horizontal and vertical shifts. A question will tell you which one it wants.

Why does the textbook call it the Baudhāyana–Pythagoras theorem?

The result was stated in Baudhāyana's Śulba-sūtra, which is older than Pythagoras. NCERT uses both names to give credit to both traditions. It is the same theorem you learnt in Grade 8.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part I, Mathematics Textbook for Grade 9 (First Edition 2026)