Orienting Yourself: The Use of Coordinates: Class 9 Ganita Manjari · Chapter 1

Ganita Manjari Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 26 Sep 2026 · Updated 28 Sep 2026

75 min
3 simulations
28 practice questions
29 MCQs

What you will learn in Orienting Yourself: The Use of Coordinates

Explain how a coordinate system uses a fixed origin and two reference directions to name any position, and trace how the idea grew in India and beyond.
Draw and read a scale floor plan on a grid, and explain why a 2-D plan cannot show height.
Identify the axes, origin and quadrants of the Cartesian plane and use the sign of each coordinate to place a point.
Plot a point from its coordinates and read the coordinates of a plotted point, treating (x, y) as an ordered pair.
Find the distance between two points on a line parallel to an axis, and between any two points using the Baudhāyana–Pythagoras theorem.
Use distances to test shapes: equal sides, right angles, collinear points and points on a circle.
Show that reflection in an axis changes positions but keeps all lengths.

Original companion text aligned to NCERT Mathematics Class 9, Chapter 1, pp. 1–15. Not affiliated with or endorsed by NCERT.

1.1 Introduction

Suppose you leave your bicycle somewhere in a huge, empty maidan and phone a friend to fetch it. "Somewhere in the middle" is no help at all. "Walk 40 steps from the gate towards the water tank, then 15 steps to your left" works every time. Two numbers, counted from an agreed starting point along agreed directions, are enough to pin down a spot on flat ground.

That is the whole idea of a coordinate system (निर्देशांक पद्धति): a fixed starting point and fixed reference lines that let us name every location with numbers. Once places have numbers, we can calculate with them, for example to find how far apart two places are.

The idea is very old in our subcontinent. Cities of the Sindhu-Sarasvatī Civilisation, such as Mohenjo-daro, were planned with wide main streets running North–South and East–West, crossing at right angles. In a town like that, finding a place becomes simple counting: so many blocks east, then so many blocks north (Fig. 1.1).

4/4The stall's address is the pair (3, 2): blocks east first, then blocks north.

Fig. 1.1Counting blocks from a fixed crossing turns a location into a pair of numbers.

A bird's-eye map of a town whose streets form a square grid, with a few round trees in the blocks and a compass showing north at the top. In the first step the central crossing is marked 'centre' and a starred stall is marked a little way up and to the right. In the second step a walker moves 3 blocks east along the main street, shown by a dotted orange arrow. In the third step the walker turns and moves 2 blocks north, shown by a dotted blue arrow, reaching the stall. In the last step a panel shows the stall's address as the pair (3, 2), meaning east first and then north.

Around 800 BCE, the Śulba-sūtra of Baudhāyana gave exact rules for building fire altars. The builders began by stretching a cord along an East–West line. Next they fixed a North–South line at right angles to it. Every corner of the altar was then measured from these two lines. The same text states the rule we now call the Baudhāyana–Pythagoras theorem, which will give us distances in Section 1.4.

To mark a place on the round Earth, we need two angles, today called latitude and longitude. Latitude starts from the equator, but longitude has no natural zero, so someone has to choose one. Astronomers of the Indian Siddhānta texts chose the north–south line through Ujjayinī, today's Ujjain. Around 150 CE, Ptolemy of Alexandria compiled a gazetteer giving both angles for several thousand places, Ujjain among them. Āryabhaṭa (499 CE) built his astronomy on the half-chord, the idea that grew into today's sine. Tables of half-chords turned hard angle problems about stars and places into quick look-ups. The sky got coordinates too: a planet's place was given by angles measured along the ecliptic, the Sun's yearly track, and away from it.

The plane you are about to meet has 0 where its two axes cross, and negative numbers on half of each axis. That design needs zero and negative numbers to be accepted as real numbers. In 628 CE, Brahmagupta did exactly that in his Brāhmasphuṭasiddhānta. He called positive numbers fortunes and negative numbers debts. His rules include “a debt minus zero is a debt” and “the product of two debts is a fortune”.

Knowledge then travelled west. In Baghdad, around 770 CE, a Sanskrit astronomy text in Brahmagupta's tradition was translated into Arabic as the Sindhind. With it came Ujjain's meridian, which Arab map-makers renamed Arin and used as their zero of longitude. Al-Bīrūnī later spent years in India and learned Sanskrit. His Book of India (c. 1030 CE) explains Indian astronomy in detail, and his other works list the latitudes and longitudes of many cities. He also wrote a manual on the astrolabe, a brass instrument for measuring the height of the Sun or a star. The poet-mathematician ʿUmar Khayyām (1048–1131) solved cubic equations by drawing two curves and finding where they meet.

In Europe, the decisive step came in the 1630s. Pierre de Fermat, a lawyer in Toulouse, and René Descartes, a philosopher living in the Netherlands, had the same insight. A curve can be described by an equation that links two numbers. Descartes published his method in 1637, in La Géométrie. From then on, a problem about shapes could be turned into algebra, and an equation could be drawn as a picture.

Timeline

How the idea of coordinates grew

c. 2600–1900 BCE
Grid-planned cities
Main streets of cities such as Mohenjo-daro run North–South and East–West.
c. 800 BCE
Baudhāyana's Śulba-sūtra
Altars measured from an East–West line and a North–South line; the diagonal rule for rectangles.
c. 150 CE
Ptolemy's Geography
Latitude and longitude for thousands of places.
499 CE
Āryabhaṭa
Half-chord (sine) tables for the positions of stars and cities.
628 CE
Brahmagupta
Zero and negative numbers treated as numbers, with rules.
c. 1030 CE
Al-Bīrūnī
Latitudes and longitudes of many cities; a manual on the astrolabe.
c. 1070 CE
ʿUmar Khayyām
Cubic equations solved by making two curves meet.
1637
Descartes's La Géométrie
Points named by two numbers; geometry meets algebra.

In this chapter we build the two-number system step by step. We start on the floor of a room, then move to an endless flat plane that uses negative numbers too. Finally, we use it to measure the straight-line distance between any two points.

1.2 Settling In

The Grade 9 students of a school in Nashik have been handed an empty room and a challenge. Their teacher wants them to turn it into a Ganit Lab: a maths room with a work table, a shelf of puzzles and a noticeboard. One rule matters most. Their classmate Farhan uses a wheelchair, so every doorway and every path must work for him too.

Before moving any furniture, the team draws the room on squared paper. They choose a scale of 1 cm on paper for 1 m in the room, so a wall 8 m long becomes a line 8 cm long. They pick one corner of the room as their starting point and measure everything from the two walls that meet there.

A drawing like this is a floor plan: the room as a lizard on the ceiling would see it, looking straight down. Fig. 1.2 shows what survives the trip from the real room to the plan, and what gets lost.

4/4The plan shows where the window is along the wall, but not how high it is.

Fig. 1.2A floor plan keeps length and breadth but not height.NCERT Fig. 1.1 · p. 3

On the left, a room is drawn in 3-D with a back wall, a left wall and a floor; there is a window on the back wall, a door on the left wall and a small table on the floor. In the second step an arrow labelled 'top view' points to a floor plan on the right: a rectangle with a window mark on the top edge and a door gap with its swing on the left edge. In the third step grid squares appear on the plan with the note '1 square = 1 m', and the table appears as a green rectangle. In the last step an arrow on the 3-D room marks the height of the window sill, and a label 'height?' beside the plan shows that the plan cannot record it.

Look at what the plan cannot tell us. It shows the window as a mark on the back wall, so we know how far along the wall it is. It does not show how high the window sill is, or how tall the table stands. A floor plan records only two directions, length and breadth. Height is a third direction, and it would need a third number.

Compare

How many numbers fix a position?

On a line (1-D)

One number
A kilometre stone on a highway
A mark on a ruler

On a flat surface (2-D)

Two numbers
A square on a chessboard
A cinema seat: row and seat number
A corner on a floor plan

In space (3-D)

Three numbers
A kite in the sky
A ceiling fan in a room
A window sill: along the wall and above the floor

Try this

Draw a plan of your own room

You’ll need

squared paper
a measuring tape
a pencil and a ruler

Predict first

How many measurements do you need to fix one corner of your bed on the plan?

onetwothreefour

Choose one corner of the room as your starting point and call it O.
Measure the two walls that meet at O, and pick a scale such as 1 cm for 50 cm.
For each piece of furniture, measure how far each corner is from both walls.
Draw the walls to scale, then mark the furniture corners and join them.
Show doors and windows as gaps or marks on the walls.

Table 1.1My measurements

Corner	From wall 1 (m)	From wall 2 (m)
Bed, nearest corner
Table, nearest corner
Door hinge

Your entries are saved automatically.

Quick check

1 of 3

A plan uses 1 cm for 50 cm. A table is drawn 3 cm long. How long is the real table, in metres?

Show answer

Answer 1.5 m

Each centimetre stands for 50 cm, so

3 \times 50 = 150

cm, which is 1.5 m.

1.3 The 2-D Cartesian Coordinate System

On a number line, one number fixes a point: 3 lies three units to the right of 0, and $- 2$ lies two units to the left. A number line is one-dimensional. To describe points on a flat surface we need a second number, and so a second number line.

Now take a second number line, stand it upright, and slide it until its 0 sits exactly on the 0 of the first (Fig. 1.3). The two lines are now perpendicular and share their zero. The horizontal line is the x-axis and the upright one is the y-axis; as a pair they are called the coordinate axes. Their shared zero point is the origin (मूलबिंदु), labelled O.

4/4Right and up are positive; left and down are negative.

Fig. 1.3Two number lines at right angles make the coordinate axes.NCERT Fig. 1.2 · p. 4

In the first step a horizontal number line from minus 5 to 5, with 0 in the middle, is labelled x-axis. In the second step a vertical number line from minus 3 to 3 crosses it at zero, with a small square marking the right angle; it is labelled y-axis. In the third step the crossing point is marked O with the label (0, 0). In the last step the right half of the x-axis and the upper half of the y-axis are highlighted in green with labels '+ right' and '+ up', and the left and lower halves are dashed in orange with labels 'minus left' and 'minus down'.

Each axis is an ordinary number line, and both use the same unit length. So the familiar number-line rule holds on each one. On the x-axis, points right of O are positive and points left of O are negative. On the y-axis, up is positive and down is negative.

Definition

Origin

The point where the x-axis and the y-axis cross. Its coordinates are $(0, 0)$ .

ExampleOn the Ganit Lab plan, the origin is the corner where the front wall meets the left wall.

Points on the axes

Start with points that sit on an axis. The point K in Fig. 1.4 is on the x-axis, three and a half units right of O. Getting there from O takes 3.5 units in the x-direction and none in the y-direction, so K is described by the pair $(3.5, 0)$ . We write $K = (3.5, 0)$ , usually shortened to $K (3.5, 0)$ . The point L, two and a half units left of O, is $L (- 2.5, 0)$ .

Points on the y-axis work the other way round: the first number is 0. The point M, 2.5 units above O, is $M (0, 2.5)$ , and N, 3 units below O, is $N (0, - 3)$ .

3/3Every point of the x-axis is $(x, 0)$ ; every point of the y-axis is $(0, y)$ .

Fig. 1.4On an axis, one of the two coordinates is always 0.

Coordinate axes with unit marks. In the first step two blue points sit on the x-axis: K at (3.5, 0) to the right of O and L at (minus 2.5, 0) to the left. In the second step two orange points appear on the y-axis: M at (0, 2.5) above O and N at (0, minus 3) below it. In the last step the x-axis is shaded blue with the label 'on x-axis: (x, 0)' and the y-axis is shaded orange with the label 'on y-axis: (0, y)'.

Back to the Ganit Lab

The students now put axes on their plan. The origin O is the corner where the front wall meets the left wall. The x-axis runs along the front wall and the y-axis along the left wall (Fig. 1.5). The room is 8 m wide and 6 m deep, and one unit stands for 1 m.

3/3A shelf, a noticeboard and a cupboard complete the plan.

Fig. 1.5The Ganit Lab with its corners, doors and furniture. One unit is 1 m.NCERT Fig. 1.3 · p. 5

A rectangular room drawn on a grid with the x-axis along its front wall and the y-axis along its left wall, marked 1 to 8 and 1 to 6. In the first step the corners are labelled O at the origin, A (8, 0), B (8, 6) and C (0, 6). In the second step two doors appear: the main door DE is a gap in the front wall between 5 and 6 with a dashed swing, and the store-room door ST is a smaller gap in the left wall between about 3.5 and 4.3. In the last step a shelf appears against the left wall above the store door, a green noticeboard along the back wall and a cupboard against the right wall.

Every corner of the room now has a name made of two numbers: $O (0, 0)$ , $A (8, 0)$ , $B (8, 6)$ and $C (0, 6)$ . The whole room lies to the right of the left wall and above the front wall, so none of its points has a negative coordinate.

Example

Reading the lab plan

On the lab plan, the noticeboard runs along the back wall from $G (1.5, 6)$ to $H (4.5, 6)$ . (a) How long is it? (b) How far is its right end from the right wall AB? (c) Where is its midpoint?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 4 questions

Practice 1.1: questions with answers

Use the Ganit Lab plan in Fig. 1.5. One unit is 1 m.

1
Short answerEasyNCERT Exercise Set 1.1 (i)–(ii)
The main door runs from D to E along the front wall.
1. How far is D from the left wall (the y-axis), in metres?
  m
  Show answer
  Answer 5 m
  D sits at the 5 m mark on the front wall.
2. How far is D from the front wall (the x-axis), in metres?
  m
  Show answer
  Answer 0 m
  D is on the front wall, so its distance from the x-axis is 0.
3. Write the coordinates of D and of E.
  Show solution
  Answer $D (5, 0)$ and $E (6, 0)$
  Both points are on the x-axis, so their y-coordinates are 0. Their x-coordinates are the marks 5 and 6, giving $D (5, 0)$ and $E (6, 0)$ .
2
Short answerEasyNCERT Exercise Set 1.1 (iii)
A doorway needs at least 0.9 m of clear width for a wheelchair.
1. How wide is the main door DE, in metres?
  m
  Show answer
  Answer 1 m
  $6 - 5 = 1$ m.
2. The main door is wide enough for Farhan's wheelchair.
  TrueFalse
  Show answer
  Answer True
  1 m is more than 0.9 m, with 10 cm to spare.
3
Short answerMediumNCERT Exercise Set 1.1 (iv)
The store-room door runs from $S (0, 3.5)$ to $T (0, 4.3)$ .
1. How wide is the store-room door, in metres?
  m
  Show answer
  Answer 0.8 m
  $4.3 - 3.5 = 0.8$ m.
2. Is it narrower or wider than the main door, and by how much?
  Show solution
  AnswerNarrower, by 0.2 m.
  The main door is 1 m wide and the store-room door 0.8 m, so the store-room door is narrower by $1 - 0.8 = 0.2$ m.
3. Can Farhan use the store-room door as it is? What would you suggest?
  Show solution
  AnswerNo: 0.8 m is less than 0.9 m.
  The doorway is 0.1 m narrower than the 0.9 m a wheelchair needs. The team could widen it to about 1 m, as the main door is, or keep items Farhan needs in the lab itself.
Quick answerEasy
Which corner of the room has coordinates $(0, 6)$ ?
OABC
Show answer
Answer (D) C
$(0, 6)$ is on the y-axis, 6 m from O: the back-left corner C.

Think about itNCERT Think and Reflect · p. 5

How wide should a door be?

Measure the main door and the bathroom door at home, and a classroom door at school. Which is the narrowest?
Accessibility guidelines in India ask for doorways at least about 90 cm wide, so that a wheelchair can pass. Which of your doors pass?
Walk around your school. Where would Farhan get stuck: steps, narrow doors, heavy doors, or something else?

Table 1.2Door survey

Door	Width (cm)	Wide enough?
Main door at home
Bathroom door at home
Classroom door
School washroom door

Your entries are saved automatically.

Quick check

1 of 4

Which point lies on the x-axis?

(0, - 2)

(2, 2)

(- 2, - 2)

(- 2, 0)

Show answer

Answer (D) $(- 2, 0)$

On the x-axis the y-coordinate is 0, so

(- 2, 0)

is the one.

(0, - 2)

has its x-coordinate 0, which puts it on the y-axis instead.

Quadrants

Most points, of course, sit on neither axis. The whole flat surface on which the axes are drawn has three common names: the Cartesian plane, the coordinate plane and the xy-plane. Its two axes cross like two fences in a field, splitting it into four regions called quadrants (चतुर्थांश). We number them I, II, III and IV, starting at the top right and going anticlockwise (Fig. 1.6).

3/3P, Q, R and S each sit in a different quadrant.

Fig. 1.6The four quadrants, numbered anticlockwise from the top right.NCERT Fig. 1.4 · p. 6

A grid with the x-axis and y-axis crossing in the middle. In the first step the four regions are shaded in different colours and labelled with large Roman numerals: I top right, II top left, III bottom left and IV bottom right, with a curved arrow around the origin showing that the numbering goes anticlockwise. In the second step each quadrant shows its sign pattern: (+, +) in I, (minus, +) in II, (minus, minus) in III and (+, minus) in IV. In the last step four points appear: P (5, 4) in Quadrant I, Q (minus 6, 1) in Quadrant II, R (minus 3, minus 4) in Quadrant III and S (6, minus 3) in Quadrant IV.

Table 1.3Signs of the coordinates in each quadrant

Quadrant	x-coordinate	y-coordinate	Example
I (top right)	positive	positive	$(5, 4)$
II (top left)	negative	positive	$(- 6, 1)$
III (bottom left)	negative	negative	$(- 3, - 4)$
IV (bottom right)	positive	negative	$(6, - 3)$

Definition

Quadrant

One of the four regions into which the coordinate axes divide the plane. Points on the axes belong to no quadrant.

Example $(- 3, 0)$ lies on the x-axis, so it is in no quadrant, even though one of its numbers is negative.

What the two numbers mean

Look at the point P in Fig. 1.7. To reach P from O, go 4 units to the right and then 3 units up. The 4 is P's distance from the y-axis: the length of the horizontal dashed line. The 3 is P's distance from the x-axis: the length of the vertical dashed line. Signs show the side. A negative first number puts a point left of the y-axis, and a negative second number puts it below the x-axis.

The pair $(4, 3)$ is called the coordinates (निर्देशांक) of P. The first number is the x-coordinate and the second is the y-coordinate. Because the order carries meaning, $(x, y)$ is an ordered pair: $(4, 3)$ and $(3, 4)$ are different points.

3/4P is 3 units from the x-axis, so its y-coordinate is 3. So P = (4, 3).

Fig. 1.7The x-coordinate is the distance from the y-axis; the y-coordinate is the distance from the x-axis.

A first-quadrant grid numbered 1 to 6 along x and 1 to 5 along y, with a panel on the right. In the first step only the point P is marked and the panel asks 'where is P?'. In the second step a dashed blue line runs from P across to the y-axis, labelled 4 units, and the panel shows x = 4. In the third step a dashed orange line runs from P down to the x-axis, labelled 3 units, and the panel adds y = 3 and P (4, 3). In the last step a purple point (3, 4) appears up and to the left of P with faint guide lines, and the panel notes that (3, 4) is not equal to (4, 3).

Definition

Coordinates of a point

The ordered pair $(x, y)$ in which $x$ is the point's signed distance from the y-axis and $y$ is its signed distance from the x-axis.

Example $P (4, 3)$ is 4 units to the right of the y-axis and 3 units above the x-axis.

Step by step

Plotting a point in four moves

Step 1 of 4

Start at the origin

Put your pencil on O. To plot $B (- 3, 4)$ , read the two numbers in order.

1 / 4

Example

Signs decide the quadrant

Where does each point lie: (a) $(- 2.5, 7)$ (b) $(6, - 0.5)$ (c) $(0, - 4)$ (d) $(- 3, - 3)$ ? (e) If $a < 0$ and $b > 0$ , in which quadrants are $(a, b)$ and $(b, a)$ ?

Try it yourself first. Then reveal the solution one step at a time.

SimulationExplore the four quadrants

A coordinate plane from minus 10 to 10 across and minus 8 to 8 up, with the four quadrants shaded. Four labelled points A, B, C and D start in different quadrants; drag any point to a grid crossing and its coordinates and quadrant update as you move it.

Fig. 1.8Drag the points and watch their coordinates and quadrants change.

Try this

Drag A into Quadrant II. Which of its coordinates changed sign?
Put B exactly on the y-axis. What is its x-coordinate now?
Can you place C where both coordinates are equal? In which quadrants is that possible?

Beyond the lab walls

The lab is not alone on its floor. To its left, beyond the y-axis, lie a store room and a small wash area. In front of it, beyond the x-axis, runs a corridor (Fig. 1.9). Keep the same origin and axes, and every point in these spaces gets a negative x-coordinate, a negative y-coordinate, or both. One floor now spreads over all four quadrants.

4/4One floor now spreads over all four quadrants.

Fig. 1.9With one origin, the rooms on this floor lie in all four quadrants.NCERT Fig. 1.5 · p. 7

A floor plan on a grid numbered minus 3 to 8 along the bottom and minus 2 to 6 up the left side, with dashed orange axes crossing at the lab's front-left corner. In the first step only the Ganit Lab is shown, to the right of the y-axis and above the x-axis. In the second step a store room from y = 2 to 6 and a wash area from y = 0 to 2 appear to the left of the y-axis. In the third step a corridor 2 units deep appears below the x-axis, running the full width. In the last step the four quadrants are labelled with their sign patterns: I (+, +) in the lab, II (minus, +) in the store, and III (minus, minus) and IV (+, minus) in the corridor.

Example

The missing corner

On a map of the school garden, with 1 unit for 1 m, three corners of a rectangular flower bed are $K (- 3, 2)$ , $L (5, 2)$ and $M (5, - 4)$ . Find the fourth corner N, the length and breadth of the bed, and the quadrant of each corner.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 4 questions

Practice 1.2: questions with answers

Use the whole-floor plan in Fig. 1.9, with 1 unit for 1 m. On squared paper, draw the x-axis from $- 4$ to 9 and the y-axis from $- 3$ to 7 (1 cm for each unit) and copy the plan.

1
Short answerMediumNCERT Exercise Set 1.2 Q1
The team places a rectangular work table with three of its feet at $(1, 4)$ , $(3.5, 4)$ and $(3.5, 5.2)$ .
1. Where must the fourth foot go?
  Show solution
  Answer $(1, 5.2)$
  The first two feet share $y = 4$ and the last two share $x = 3.5$ . The fourth foot must be level with $(3.5, 5.2)$ and directly above $(1, 4)$ , so it is at $(1, 5.2)$ .
2. How long and how wide is the table?
  Show solution
  Answer2.5 m by 1.2 m
  Length: $3.5 - 1 = 2.5$ m. Width: $5.2 - 4 = 1.2$ m.
3. The shelf stands against the left wall from $(0, 4.5)$ to $(0.4, 5.5)$ . Farhan needs a clear gap of about 0.9 m to reach it. Is the table well placed?
  Show solution
  AnswerNo: the gap is only 0.6 m.
  The front of the shelf is at $x = 0.4$ and the table's left edge is at $x = 1$ , and they face each other between $y = 4.5$ and $y = 5.2$ . The gap is $1 - 0.4 = 0.6$ m, which is 0.3 m too narrow. Moving the table at least 0.3 m to the right, for example to feet at $(1.3, 4)$ , $(3.8, 4)$ , $(3.8, 5.2)$ and $(1.3, 5.2)$ , fixes it.
4. Can you tell the height of the table from the plan?
  Show solution
  AnswerNo.
  A floor plan is two-dimensional: it records length and breadth only. Height would need a third number.
2
Short answerHardNCERT Exercise Set 1.2 Q2
The store-room door is hinged at $S (0, 3.5)$ and swings into the lab. The shelf stands along the left wall from $(0, 4.5)$ to $(0.4, 5.5)$ .
1. The door is 0.8 m wide. Will it hit the shelf as it opens?
  Show solution
  AnswerNo.
  Every point of the door stays within 0.8 m of the hinge S. The nearest point of the shelf, $(0, 4.5)$ , is $4.5 - 3.5 = 1$ m above S, so the door clears the shelf with 0.2 m to spare.
2. To let Farhan in, the door is widened to 1.2 m, still hinged at S. What must change?
  Show solution
  AnswerThe shelf must move up by at least 0.2 m.
  The doorway would now run from $y = 3.5$ to $y = 3.5 + 1.2 = 4.7$ , but the shelf starts at $y = 4.5$ , inside the doorway. The shelf must move up by at least 0.2 m (a little more in practice, for clearance), so that it starts at $y = 4.7$ or higher.
  The swinging door then clears the shelf too: its lower front corner $(0.4, 4.7)$ is about 1.26 m from S, more than the 1.2 m reach of the door. You will learn to compute such distances in Section 1.4.
3
Short answerMediumNCERT Exercise Set 1.2 Q3
Look at the store room and the wash area.
1. Write the coordinates of the four corners of the store room.
  Show solution
  Answer $(- 3, 2)$ , $(0, 2)$ , $(0, 6)$ , $(- 3, 6)$
  The store room runs from $x = - 3$ to $x = 0$ and from $y = 2$ to $y = 6$ . Its corners are $(- 3, 2)$ , $(0, 2)$ , $(0, 6)$ and $(- 3, 6)$ .
2. What shape is the wash area? Give its corners, length and breadth.
  Show solution
  AnswerA rectangle 3 m by 2 m.
  Its corners are $(- 3, 0)$ , $(0, 0)$ , $(0, 2)$ and $(- 3, 2)$ . It is a rectangle $0 - (- 3) = 3$ m long and $2 - 0 = 2$ m wide.
3. Mark a 1.5 m by 1 m space for a water cooler in one corner of the wash area. Write the coordinates of its corners.
  Show solution
  AnswerFor example $(- 3, 0)$ , $(- 1.5, 0)$ , $(- 1.5, 1)$ , $(- 3, 1)$ .
  Put one corner at $(- 3, 0)$ . Going 1.5 m to the right and 1 m up gives $(- 3, 0)$ , $(- 1.5, 0)$ , $(- 1.5, 1)$ and $(- 3, 1)$ . Other corners of the wash area give other correct answers, such as $(- 1.5, 1)$ , $(0, 1)$ , $(0, 2)$ , $(- 1.5, 2)$ .
4
Short answerMediumNCERT Exercise Set 1.2 Q4
The corridor is 11 m long and 2 m wide. It runs along the front of the wash area and the lab.
1. Write the coordinates of the corridor's four corners.
  Show solution
  Answer $(- 3, - 2)$ , $(8, - 2)$ , $(8, 0)$ , $(- 3, 0)$
  It runs from $x = - 3$ to $x = 8$ , a length of $8 - (- 3) = 11$ m, and from $y = - 2$ to $y = 0$ . So the corners are $(- 3, - 2)$ , $(8, - 2)$ , $(8, 0)$ and $(- 3, 0)$ .
2. A bench 2 m long and 1 m wide is placed exactly at the centre of the corridor, parallel to its length. Find the coordinates of its corners.
  Show solution
  Answer $(1.5, - 1.5)$ , $(3.5, - 1.5)$ , $(3.5, - 0.5)$ , $(1.5, - 0.5)$
  The centre is halfway from $- 3$ to 8 across, which is $2.5$ , and halfway from $- 2$ to 0 up, which is $- 1$ : the point $(2.5, - 1)$ . The bench reaches 1 m to each side along x and 0.5 m up and down along y, giving corners $(1.5, - 1.5)$ , $(3.5, - 1.5)$ , $(3.5, - 0.5)$ and $(1.5, - 0.5)$ .
3. In which quadrant does the whole bench lie?
  Quadrant IQuadrant IIQuadrant IIIQuadrant IV
  Show answer
  Answer (D) Quadrant IV
  Every corner has a positive x-coordinate and a negative y-coordinate: $(+, -)$ , which is Quadrant IV.

Let's explore

Coordinate hide-and-seek

You’ll need

two sheets of squared paper
a pencil each
a partner

Predict first

Your partner says all their ships are where both coordinates are negative. Where should you search?

Quadrant IQuadrant IIQuadrant IIIQuadrant IV

Each player secretly draws axes from $- 5$ to 5 in both directions.
Hide three ships by marking three points with whole-number coordinates.
Take turns calling out a point, such as $(- 2, 3)$ . Your partner answers "hit" if a ship is there, "near" if a ship is exactly one unit away along a grid line, and "miss" otherwise.
Record every guess and answer on your own sheet.
The first player to find all three ships wins.

SimulationPlot it!

A challenge on a coordinate plane from minus 8 to 8 across and minus 6 to 6 up. Each round names a point by its coordinates; tap the grid crossing where it belongs to score, and a wrong tap shows a hint before the next try.

Fig. 1.10Eight rounds: tap where each point belongs.

Try this

Before you tap, say which quadrant the point must be in.
Play again for a new set of points and try to score 8 out of 8.

1.4 Distance Between Two Points in the 2-D Plane

Farhan wants to know how far his wheelchair must roll from the door to the shelf. The team wants to know how long a cable must be to run straight across the lab from one corner to the opposite one. Both are questions about the distance between two points. When the two points share a coordinate, one subtraction answers the question. When they do not, the Baudhāyana–Pythagoras theorem does the work.

Along a line parallel to an axis

The points $A (- 4, 2)$ and $B (3, 2)$ have the same y-coordinate, so the segment AB is parallel to the x-axis. Counting squares from A to B gives 7 (Fig. 1.11). Subtracting the x-coordinates gives the same answer: $3 - (- 4) = 7$ . We take the absolute value, because a distance is never negative; subtracting the other way gives $∣ (- 4) - 3 ∣ = ∣ - 7 ∣ = 7$ too.

3/3C and D share an x-coordinate, so subtract their y-coordinates: 5.

Fig. 1.11Subtracting the coordinates that change gives the same answer as counting squares.

A grid with both axes numbered and a strip for working below it. In the first step a blue segment joins A (minus 4, 2) and B (3, 2), and seven small dashed hops above it are numbered 1 to 7, counting the unit steps. In the second step the working AB = |3 minus (minus 4)| = 7 appears in the strip below the grid, showing that subtraction gives the same count. In the last step an orange vertical segment joins C (5, minus 2) and D (5, 3) with five hops beside it, and the working CD = |3 minus (minus 2)| = 5 appears in the strip next to the first result.

Same y-coordinate

d = ∣ x_{2} - x_{1} ∣

where

$(x_{1}, y), (x_{2}, y)$: two points on a line parallel to the x-axis

Same x-coordinate

d = ∣ y_{2} - y_{1} ∣

where

$(x, y_{1}), (x, y_{2})$: two points on a line parallel to the y-axis

In the same way, $C (5, - 2)$ and $D (5, 3)$ share the x-coordinate 5, so CD is parallel to the y-axis and $C D = ∣ 3 - (- 2) ∣ = 5$ . When one point is above the x-axis and the other below it, the distance is the two parts added together: 2 below and 3 above.

Quick check

1 of 3

How far apart are

(- 6, 3)

and

(2, 3)

units

Show answer

Answer 8 units

They share

y = 3

, so the distance is

∣ 2 - (- 6) ∣ = 8

Any two points

The grid walk has two legs at right angles: 8 units across and 6 units up. The straight path PQ is the third side of the right-angled triangle PRQ, whose corner is $R (7, 2)$ (Fig. 1.12).

TheoremBaudhāyana–Pythagoras theorem

In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides. If the legs are $a$ and $b$ and the hypotenuse is $c$ , then $a^{2} + b^{2} = c^{2}$ .

Place four copies of the triangle inside a square of side $a + b$ , one in each corner, so that their hypotenuses enclose a tilted square of side $c$ . It really is a square: at each of its corners, two acute angles of the triangle and the corner angle together make a straight angle. The two acute angles add up to $90^{\circ}$ , so each corner angle is $90^{\circ}$ .

Count the area of the big square in two ways: $(a + b)^{2} = 4 \times \frac{1}{2} a b + c^{2}$ . So $a^{2} + 2 a b + b^{2} = 2 a b + c^{2}$ , which gives $a^{2} + b^{2} = c^{2}$ .

Baudhāyana stated it for a rectangle: the square on the diagonal equals the squares on the length and the breadth together.

The converse is also true: if the sides of a triangle satisfy $a^{2} + b^{2} = c^{2}$ , then the angle opposite the side $c$ is a right angle. We use it to test whether a triangle is right-angled.

4/4 $P Q^{2} = 8^{2} + 6^{2} = 100$ , so $P Q = 10$ units.

Fig. 1.12The legs are coordinate differences; the hypotenuse is the distance.NCERT Fig. 1.6–1.7 · p. 9

A grid with axes, the x-axis from minus 2 to 8 and the y-axis from 0 to 8, and a working panel on the right. In the first step points P (minus 1, 2) and Q (7, 8) are joined by a grey dashed line with a question mark, and the panel asks PQ = ?. In the second step a blue horizontal leg runs from P to the right, labelled 8, and the panel shows PR = 8. In the third step an orange vertical leg rises from R (7, 2) to Q, labelled 6, with a right-angle mark at R, and the panel adds RQ = 6. In the last step the slanted side PQ turns vermillion and is labelled 10, and the panel shows PQ squared = 64 + 36 and PQ = 10.

3/3The hypotenuse AB gives the distance formula.

Fig. 1.13The same right triangle works for any two points.NCERT Fig. 1.8 · p. 10

Coordinate axes with two points A and B, dashed lines dropping from each point to both axes and the labels x1, x2 on the x-axis and y1, y2 on the y-axis. In the first step only A (x1, y1) and B (x2, y2) are marked, joined by a grey dashed line. In the second step a blue horizontal leg from A and an orange vertical leg down from B meet at C (x2, y1) with a right-angle mark; the legs are labelled x2 minus x1 and y2 minus y1. In the last step the side AB turns vermillion and the formula AB equals the square root of (x2 minus x1) squared plus (y2 minus y1) squared appears at the top.

DerivationThe distance formula

Let the points be $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , as in Fig. 1.13. Draw a line through A parallel to the x-axis and a line through B parallel to the y-axis. They meet at $C (x_{2}, y_{1})$ , and the angle at C is a right angle.

A and C share the y-coordinate $y_{1}$ , so $A C = ∣ x_{2} - x_{1} ∣$ . C and B share the x-coordinate $x_{2}$ , so $C B = ∣ y_{2} - y_{1} ∣$ .

By the Baudhāyana–Pythagoras theorem, $A B^{2} = A C^{2} + C B^{2}$ . Squaring removes the absolute-value bars, because $∣ a ∣^{2} = a^{2}$ for every number $a$ . Taking the positive square root gives the distance formula (दूरी सूत्र):

Distance formula

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

It does not matter which point you call A and which B, because $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ . Negative coordinates cause no trouble either, since a square is never negative. In particular, the distance of a point $P (x, y)$ from the origin comes from putting $(x_{1}, y_{1}) = (0, 0)$ :

Distance from the origin

O P = \sqrt{x^{2} + y^{2}}

Example

Distance with negative coordinates

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

Try it yourself first. Then reveal the solution one step at a time.

Example

When the answer is not a whole number

On a food-delivery map, 1 unit stands for 1 km. A kitchen is at $K (- 1, - 2)$ and a customer at $C (3, 5)$ . Find the straight-line distance KC to the nearest 0.1 km. Why will the rider's route probably be longer?

Try it yourself first. Then reveal the solution one step at a time.

Example

What kind of triangle?

Show that $P (1, 2)$ , $Q (4, 6)$ and $R (8, 3)$ are the vertices of a right-angled isosceles triangle.

Try it yourself first. Then reveal the solution one step at a time.

Example

Equally far from two points

Find the point on the x-axis that is equally far from $A (2, 3)$ and $B (6, 5)$ .

Try it yourself first. Then reveal the solution one step at a time.

Distances also tell us whether three points are collinear, that is, whether they lie on one straight line. Suppose B lies on the segment AC. Then the trip from A to B and on to C is exactly as long as the straight trip from A to C, so $A B + B C = A C$ . If B is off that line, the trip is a detour, so $A B + B C$ is greater than AC. You will use this test in the exercises.

Reflections keep lengths

Fold a graph sheet along the y-axis and press hard. Every point $(x, y)$ lands on the point $(- x, y)$ : at the same height and the same distance from the fold, but on the other side. The new point is the mirror image, or reflection (परावर्तन), of the first point in the y-axis (Fig. 1.14).

3/3The image has the same side lengths: 5, 5 and $\sqrt{50}$ .

Fig. 1.14Reflection in the y-axis changes the sign of every x-coordinate but keeps every length.NCERT Fig. 1.9 · p. 11

A grid above the x-axis from minus 8 to 8 with the y-axis in the middle. In the first step a blue triangle TUV is drawn to the right of the y-axis with T (1, 1), U (4, 5) and V (8, 2), and its sides are labelled 5, 5 and root 50. In the second step the y-axis is highlighted as a mirror, dashed lines run straight across from each vertex, and a purple dashed triangle T′U′V′ appears at (minus 1, 1), (minus 4, 5) and (minus 8, 2). In the last step the image's sides are labelled 5, 5 and root 50 as well, with a note saying 'same lengths'.

Check one side with the distance formula: $T U = \sqrt{(4 - 1)^{2} + (5 - 1)^{2}} = \sqrt{9 + 16} = 5$ and $T^{'} U^{'} = \sqrt{(- 4 - (- 1))^{2} + (5 - 1)^{2}} = \sqrt{9 + 16} = 5$ . The x-difference changed sign, from 3 to $- 3$ , but its square did not. The same thing happens for every pair of points, so a reflection keeps all lengths.

What does change? The positions change, and so does the quadrant: TUV is in Quadrant I and its image is in Quadrant II. The sense of turning flips too. Going $T \to U \to V$ is clockwise on the original but anticlockwise on the image. Shape and size stay exactly the same. Reflection in the x-axis works the same way, sending $(x, y)$ to $(x, - y)$ .

SimulationReflect and measure

A coordinate plane from minus 10 to 10 in both directions with the right-angled triangle A (2, 1), B (8, 9), C (8, 1). Its sides AB = 10, BC = 8 and AC = 6 are measured with the distance formula, and the mirror image of each vertex in the y-axis is shown as a hollow point. Drag the vertices to reshape the triangle, and switch the mirror to the x-axis or the origin to compare where the images land.

Fig. 1.15The sides of triangle ABC are measured live; the mirror images follow every drag.

Try this

Count squares between the images: are the image's two shorter sides still 6 and 8 units long?
Switch the mirror to the x-axis. Which coordinate changes sign now, and in which quadrant do the images land?
Drag B somewhere new and count again. Does the image always keep the triangle's side lengths?

Recap

Orienting Yourself: The Use of Coordinates: key points and summary

A coordinate system fixes positions with numbers measured from a chosen origin along agreed directions. The idea grew from grid-planned cities, altar geometry and astronomy (Section 1.1).
A floor plan is a 2-D scale drawing: it keeps length and breadth but loses height (Section 1.2).
The horizontal x-axis and the vertical y-axis cross at right angles at the origin $O (0, 0)$ (Section 1.3).
Right of O and above O are positive; left of O and below O are negative (Fig. 1.3).
Points on the x-axis have the form $(x, 0)$ and points on the y-axis the form $(0, y)$ (Points on the axes).
The axes divide the plane into four quadrants with sign patterns $(+, +)$ , $(-, +)$ , $(-, -)$ and $(+, -)$ for I to IV. Points on the axes lie in no quadrant (Quadrants).
In $(x, y)$ , $x$ is the signed distance from the y-axis and $y$ the signed distance from the x-axis. $(x, y) = (y, x)$ only when $x = y$ (What the two numbers mean).
Along a line parallel to an axis, the distance is $∣ x_{2} - x_{1} ∣$ or $∣ y_{2} - y_{1} ∣$ (Distance along a line parallel to an axis).
Between any two points, the distance is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ , from the Baudhāyana–Pythagoras theorem (the distance formula).
Distances test shapes: equal sides, right angles (by the converse of the theorem) and collinear points (Any two points).
Reflection in the y-axis sends $(x, y)$ to $(- x, y)$ , and in the x-axis to $(x, - y)$ . Reflections keep every length (Reflections keep lengths).

Flash cards · 8

Key terms

1 / 8

Practice · 20 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

Questions 1 to 16 follow the textbook's end-of-chapter exercises; 17 to 20 are quick extra practice. Draw a sketch whenever you can.

Quick answerEasyNCERT End-of-chapter Q1
The two axes cross at a point called the . Its x-coordinate is and its y-coordinate is .
Show answer
Answer (1) origin, (2) 0, (3) 0
The axes cross at the origin O, whose coordinates are $(0, 0)$ .
2
Short answerMediumNCERT End-of-chapter Q2
Point K has y-coordinate 3. The line $ℓ$ passes through K and is parallel to the x-axis.
1. Write the coordinates of a general point H on $ℓ$ .
  Show solution
  Answer $H (h, 3)$ , where $h$ can be any number.
  Moving along a line parallel to the x-axis changes only the x-coordinate. So every point of $ℓ$ is 3 units above the x-axis, and $H = (h, 3)$ for some number $h$ .
2. (b)
  In which quadrants can H lie?
  Quadrant IQuadrant IIQuadrant IIIQuadrant IV
  Select all that apply
  Show answer
  Answer
  (A) Quadrant I
  (B) Quadrant II
  The y-coordinate is always $+ 3$ . If $h > 0$ , H is in Quadrant I; if $h < 0$ , it is in Quadrant II. It can never be below the x-axis.
3. Can H lie on an axis? Where?
  Show solution
  AnswerOnly on the y-axis, at $(0, 3)$ .
  When $h = 0$ , H is $(0, 3)$ , on the y-axis. H can never be on the x-axis, because its y-coordinate is 3, not 0.
3
Short answerMediumNCERT End-of-chapter Q3
The points $S (4, 2)$ , $H (- 4, 2)$ , $I (- 4, - 3)$ and $P (2, - 3)$ are joined in this order to make the shape SHIP. Before plotting, predict the answers below. Then plot the points and check.
1. Name two sides of SHIP that are perpendicular to each other.
  Show solution
  AnswerSH and HI (also HI and IP).
  S and H share $y = 2$ , so SH is horizontal. H and I share $x = - 4$ , so HI is vertical. A horizontal side and a vertical side are perpendicular, so $S H ⊥ H I$ . In the same way $H I ⊥ I P$ .
2. Name a side parallel to the x-axis or the y-axis.
  Show solution
  AnswerSH (or IP) is parallel to the x-axis; HI is parallel to the y-axis.
  SH and IP each join points with equal y-coordinates, so they are parallel to the x-axis. HI joins points with equal x-coordinates, so it is parallel to the y-axis. Only PS is slanted.
3. Which two points are mirror images of each other in an axis, and in which axis?
  Show solution
  AnswerS and H, in the y-axis.
  $S (4, 2)$ and $H (- 4, 2)$ differ only in the sign of the x-coordinate, so each is the reflection of the other in the y-axis. Since SH and IP are both horizontal, SHIP is a trapezium.
4
NumericalMediumNCERT End-of-chapter Q4
Plot $R (- 6, 8)$ . Choose a point S on the x-axis so that triangle ORS has a right angle at S, where O is the origin.
1. Where must S be?
  Show solution
  Answer $S (- 6, 0)$
  A right angle at S between OS (along the x-axis) and SR means SR is vertical. So S has the same x-coordinate as R and lies on the x-axis: $S (- 6, 0)$ .
2. Find the lengths OS, SR and OR.
  Show solution
  Answer6, 8 and 10 units
  $O S = ∣ - 6 - 0 ∣ = 6$ and $S R = ∣ 8 - 0 ∣ = 8$ . By the Baudhāyana–Pythagoras theorem, $O R = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$ . The distance formula gives the same: $\sqrt{(- 6)^{2} + 8^{2}} = 10$ .
5
Long answerMediumNCERT End-of-chapter Q5
Imagine a map whose coordinates use only zero and positive numbers, measured from a fixed origin.
1. Which part of the plane can this map describe?
  Show solution
  AnswerOnly Quadrant I, with the positive halves of the axes and the origin.
  With no negative numbers, no point can be left of the y-axis or below the x-axis. The map covers only the quarter-plane of Quadrant I and its edges.
2. Could it locate every point of an endless plane? Explain.
  Show solution
  AnswerNo.
  Points to the left of the origin or below it would have no name. Moving the origin further away helps for a while, but an endless plane always has points beyond any chosen corner.
3. Screens use only non-negative pixel coordinates. Why does that work for them?
  Show solution
  A screen is a bounded rectangle. With the origin at one corner, every pixel is on the same side of both axes, so its coordinates are never negative. Negative numbers are needed only when points can lie on both sides of the origin.
Short answerHardNCERT End-of-chapter Q6
Are the points $P (- 2, 3)$ , $Q (2, 0)$ and $R (10, - 6)$ on one straight line? Describe a method that does not need plotting.
Show solution
AnswerYes: $P Q + Q R = 5 + 10 = 15 = P R$ .
$P Q = \sqrt{4^{2} + (- 3)^{2}} = 5, Q R = \sqrt{8^{2} + (- 6)^{2}} = 10, P R = \sqrt{12^{2} + (- 9)^{2}} = 15$
Since $P Q + Q R = P R$ , the point Q lies on the segment PR, so the three points are collinear.
Another check: going from P to Q moves $(+ 4, - 3)$ , and going from Q to R moves $(+ 8, - 6)$ , which is exactly twice as much in the same direction.
Short answerHardNCERT End-of-chapter Q7
Use your method from Question 6 to check whether $D (- 1, 2)$ , $E (3, 5)$ and $F (11, 10)$ lie on one straight line. Then plot them. Why might a drawing mislead you?
Show solution
AnswerNo: $D E + E F \approx 14.434$ but $D F \approx 14.422$ .
$D E = \sqrt{4^{2} + 3^{2}} = 5, E F = \sqrt{8^{2} + 5^{2}} = \sqrt{89} \approx 9.434, D F = \sqrt{12^{2} + 8^{2}} = \sqrt{208} \approx 14.422$
$D E + E F \approx 14.434$ , which is more than $D F \approx 14.422$ . So E is not on the segment DF, and the points are not collinear.
The steps tell the same story: D to E moves $(+ 4, + 3)$ , while E to F moves $(+ 8, + 5)$ rather than $(+ 8, + 6)$ . On a drawing, E misses the line DF by so little that it looks straight, which is why calculation is more trustworthy than a sketch.
8
DiagramMediumNCERT End-of-chapter Q8
Using the origin O as one vertex, give the other two vertices of each triangle and plot it.
1. A right-angled isosceles triangle with the right angle at O and one vertex on the negative x-axis.
  Show solution
  AnswerFor example $(- 5, 0)$ and $(0, 5)$ .
  Put the legs along the axes: $A (- 5, 0)$ and $B (0, 5)$ . Then $O A = O B = 5$ and the angle at O between the axes is a right angle. Any $(- k, 0)$ with $(0, k)$ or $(0, - k)$ , for $k > 0$ , also works.
2. An isosceles triangle with O as the vertex between the equal sides, one other vertex in Quadrant I and the third in Quadrant II.
  Show solution
  AnswerFor example $(4, 3)$ and $(- 4, 3)$ .
  Take mirror images in the y-axis: $A (4, 3)$ in Quadrant I and $B (- 4, 3)$ in Quadrant II. Then $O A = O B = \sqrt{16 + 9} = 5$ . In general $(a, b)$ and $(- a, b)$ with $a, b > 0$ work.
9
Short answerHardNCERT End-of-chapter Q9
For each row, decide whether M is the midpoint of the segment ST. Then look for a rule connecting the coordinates of a midpoint with those of the two ends.
Table 1.4Is M the midpoint of ST?
Row S M T
a $(- 6, 1)$ $(0, 1)$ $(6, 1)$
b $(2, - 3)$ $(4, 0)$ $(6, 4)$
c $(0, 3)$ $(0, 0)$ $(0, - 6)$
d $(- 5, 4)$ $(- 1, 1)$ $(3, - 2)$
1. Row a: M is the midpoint of ST.
  TrueFalse
  Show answer
  Answer True
  $S M = 6$ and $M T = 6$ , and all three points lie on the line $y = 1$ .
2. Row b: M is the midpoint of ST.
  TrueFalse
  Show answer
  Answer False
  $S M = \sqrt{4 + 9} = \sqrt{13}$ but $M T = \sqrt{4 + 16} = \sqrt{20}$ . The midpoint would be $(4, 0.5)$ .
3. Row c: M is the midpoint of ST.
  TrueFalse
  Show answer
  Answer False
  All three are on the y-axis, but $S M = 3$ and $M T = 6$ . The midpoint would be $(0, - 1.5)$ .
4. Row d: M is the midpoint of ST.
  TrueFalse
  Show answer
  Answer True
  $S M = \sqrt{16 + 9} = 5$ and $M T = \sqrt{16 + 9} = 5$ , and $S \to M$ and $M \to T$ are the same move $(+ 4, - 3)$ .
5. What rule links the midpoint's coordinates with those of S and T?
  Show solution
  AnswerEach coordinate of the midpoint is the average of the ends' coordinates.
  In rows a and d, each coordinate of M is exactly halfway between those of S and T. For $S (x_{1}, y_{1})$ and $T (x_{2}, y_{2})$ the midpoint is
  $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
  Checking a midpoint needs both halves to be equal and the three points to be in line; the averages test does both at once.
10
NumericalMediumNCERT End-of-chapter Q10
$M (2, - 3)$ is the midpoint of $A (- 4, 5)$ and $B (x, y)$ . Find B.
1. Find $x$ .
  Show answer
  Answer 8
  $\frac{- 4 + x}{2} = 2$ , so $- 4 + x = 4$ and $x = 8$ .
2. Find $y$ .
  Show answer
  Answer -11
  $\frac{5 + y}{2} = - 3$ , so $5 + y = - 6$ and $y = - 11$ .
NumericalHardNCERT End-of-chapter Q11
P and Q divide the segment from $A (- 5, 9)$ to $B (10, - 3)$ into three equal parts, with P nearer to A. Use midpoints to find P and Q.
Show solution
Answer $P (0, 5)$ and $Q (5, 1)$
The whole move from A to B is $(10 - (- 5), - 3 - 9) = (15, - 12)$ . One third of it is $(5, - 4)$ .
So $P = (- 5 + 5, 9 - 4) = (0, 5)$ , which happens to lie on the y-axis, and $Q = (0 + 5, 5 - 4) = (5, 1)$ .
Check with midpoints: the midpoint of A and Q is $(\frac{- 5 + 5}{2}, \frac{9 + 1}{2}) = (0, 5) = P$ , and the midpoint of P and B is $(\frac{0 + 10}{2}, \frac{5 - 3}{2}) = (5, 1) = Q$ .
12
NumericalHardNCERT End-of-chapter Q12
Circle K has its centre at the origin.
1. $A (2, 9)$ , $B (- 6, 7)$ and $C (- 7, - 6)$ all lie on K. Find the radius of K, correct to 2 decimal places.
  Show answer
  Answer 9.22
  $O A^{2} = 4 + 81 = 85$ , $O B^{2} = 36 + 49 = 85$ , $O C^{2} = 49 + 36 = 85$ . All three are equally far from O, so they lie on one circle of radius $\sqrt{85} \approx 9.22$ .
2. (b)
  Match each point with where it lies.
  Tap an item on the left, then its match on the right.
  Show answer
  Answer
  $D (- 8, 5)$ : outside K
  $E (0, - 9)$ : inside K
  $F (9, 2)$ : on K
  Compare squared distances with 85: $64 + 25 = 89 > 85$ (outside), $0 + 81 = 81 < 85$ (inside), $81 + 4 = 85$ (on the circle).
NumericalHardNCERT End-of-chapter Q13
The midpoints of the sides BC, CA and AB of triangle ABC are $D (3, 2)$ , $E (- 1, 4)$ and $F (1, - 2)$ . Find A, B and C.
Show solution
Answer $A (- 3, 0)$ , $B (5, - 4)$ , $C (1, 8)$
Twice a midpoint equals the sum of the two ends, one coordinate at a time. Write $B + C$ for “add the x-coordinates of B and C, and add their y-coordinates”. Doubling each midpoint then gives $B + C = (6, 4)$ , $C + A = (- 2, 8)$ and $A + B = (2, - 4)$ .
Adding all three: $2 (A + B + C) = (6, 8)$ , so $A + B + C = (3, 4)$ .
Subtract each pair: $A = (3, 4) - (6, 4) = (- 3, 0)$ , $B = (3, 4) - (- 2, 8) = (5, - 4)$ and $C = (3, 4) - (2, - 4) = (1, 8)$ .
Check: the midpoint of B and C is $(\frac{5 + 1}{2}, \frac{- 4 + 8}{2}) = (3, 2) = D$ .
14
Short answerMediumNCERT End-of-chapter Q14
A mela ground has 8 lanes running North–South and 6 lanes running East–West, 50 m apart, with stalls at the crossings. The crossing $(a, b)$ is where the $a$ th North–South lane (counted from the west) meets the $b$ th East–West lane (counted from the south).
1. Using 1 cm for 50 m, how long and how wide is a model of the lane grid, from the first lane to the last in each direction?
  Show solution
  Answer7 cm by 5 cm
  8 lanes have 7 gaps of 50 m each, so the grid is 350 m across: 7 cm on the model. 6 lanes have 5 gaps: 250 m, or 5 cm.
2. How many crossings are there in all?
  Show answer
  Answer 48
  Each of the 8 North–South lanes meets each of the 6 East–West lanes: $8 \times 6 = 48$ .
3. How many crossings are called $(5, 2)$ , and how many $(2, 5)$ ? Are they the same crossing?
  Show solution
  AnswerOne each, and they are different crossings.
  $(5, 2)$ is the 5th North–South lane with the 2nd East–West lane; $(2, 5)$ is the 2nd North–South lane with the 5th East–West lane. Each name fits exactly one crossing, and the two crossings are in different places.
4. Which of these names a real crossing on this ground?
  $(7, 6)$ $(6, 7)$ bothneither
  Show answer
  Answer (A) $(7, 6)$
  $(7, 6)$ uses the 7th of 8 North–South lanes and the 6th of 6 East–West lanes: it exists. $(6, 7)$ needs a 7th East–West lane, which does not exist. Order matters.
15
NumericalHardNCERT End-of-chapter Q15
A game screen is 720 pixels wide and 480 pixels high, with the origin at its bottom-left corner. Icon 1 is a circle of radius 100 pixels centred at $(90, 120)$ . Icon 2 is a circle of radius 120 pixels centred at $(330, 190)$ .
1. Does any part of either icon fall off the screen?
  Show solution
  AnswerIcon 1 is cut off on the left; icon 2 fits.
  Icon 1 reaches from $x = 90 - 100 = - 10$ to $x = 190$ and from $y = 20$ to $y = 220$ . Its leftmost 10 pixels fall off the screen.
  Icon 2 reaches from $x = 210$ to $x = 450$ and from $y = 70$ to $y = 310$ , well inside $0 \leq x \leq 720$ and $0 \leq y \leq 480$ .
2. Do the two icons overlap?
  Show solution
  AnswerNo: the centres are 250 pixels apart, more than $100 + 120 = 220$ .
  $\sqrt{(330 - 90)^{2} + (190 - 120)^{2}} = \sqrt{57 600 + 4 900} = \sqrt{62 500} = 250$
  The circles would touch only if the centres were $100 + 120 = 220$ pixels apart or closer. At 250 pixels there is a 30-pixel gap.
NumericalMediumNCERT End-of-chapter Q16
Plot $A (1, - 2)$ , $B (5, 1)$ , $C (2, 5)$ and $D (- 2, 2)$ . Is ABCD a square? Explain, and find its area.
Show solution
AnswerYes; area 25 square units.
$A B^{2} = 4^{2} + 3^{2} = 25, B C^{2} = (- 3)^{2} + 4^{2} = 25, C D^{2} = (- 4)^{2} + (- 3)^{2} = 25, D A^{2} = 3^{2} + (- 4)^{2} = 25$
$A C^{2} = 1^{2} + 7^{2} = 50, B D^{2} = (- 7)^{2} + 1^{2} = 50$
All four sides are 5 units and the two diagonals are equal, so ABCD is a square. Also $A B^{2} + B C^{2} = 50 = A C^{2}$ , so the angle at B is a right angle.
Area $= 5^{2} = 25$ square units. None of its sides is parallel to an axis, yet the distance formula measures every one of them.
Quick answerEasy
Which point is farthest from the origin?
$(5, 0)$ $(- 3, 4)$ $(- 4, - 4)$ $(0, - 6)$
Show answer
Answer (D) $(0, - 6)$
Squared distances: 25, 25, 32 and 36. $(0, - 6)$ is 6 units away; $(- 4, - 4)$ is only $\sqrt{32} \approx 5.66$ .
Quick answerMedium
Find the perimeter of the triangle with vertices $(0, 0)$ , $(6, 0)$ and $(0, 8)$ .
units
Show answer
Answer 24 units
The sides are 6, 8 and $\sqrt{36 + 64} = 10$ , so the perimeter is 24.
19
Quick answerMedium
Match each reflection of the point $(2, - 5)$ with its image.
Tap an item on the left, then its match on the right.
Show answer
Answer
in the x-axis : $(2, 5)$
in the y-axis : $(- 2, - 5)$
in the origin : $(- 2, 5)$
The x-axis flips y, the y-axis flips x, and the origin flips both.
20
Quick answerMedium
Arrange these points by distance from the origin, nearest first.
1. $(1, 1)$
2. $(- 2, 0)$
3. $(3, 3)$
4. $(0, - 3)$
Show answer
Answer
$(1, 1)$
$(- 2, 0)$
$(0, - 3)$
$(3, 3)$
The distances are $\sqrt{2} \approx 1.41$ , 2, 3 and $\sqrt{18} \approx 4.24$ .

Row	S	M	T
a	$(- 6, 1)$	$(0, 1)$	$(6, 1)$
b	$(2, - 3)$	$(4, 0)$	$(6, 4)$
c	$(0, 3)$	$(0, 0)$	$(0, - 6)$
d	$(- 5, 4)$	$(- 1, 1)$	$(3, - 2)$

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 1.1 (i)–(ii)	Practice 1.1 · Q1
NCERT Exercise Set 1.1 (iii)	Practice 1.1 · Q2
NCERT Exercise Set 1.1 (iv)	Practice 1.1 · Q3
NCERT Exercise Set 1.2 Q1	Practice 1.2 · Q1
NCERT Exercise Set 1.2 Q2	Practice 1.2 · Q2
NCERT Exercise Set 1.2 Q3	Practice 1.2 · Q3
NCERT Exercise Set 1.2 Q4	Practice 1.2 · Q4
NCERT End-of-chapter Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Q16	End-of-chapter exercises · Q16

Quick check: MCQs on Orienting Yourself: The Use of Coordinates

1.1 Introduction4 questions
1.2 Settling In3 questions
Back to the Ganit Lab4 questions
What the two numbers mean4 questions
Beyond the lab walls3 questions
Along a line parallel to an axis3 questions
Any two points4 questions
Reflections keep lengths4 questions

Glossary · 14

Key terms (English–Hindi)

Cartesian planeकार्तीय तलkaarteey tal: The flat surface on which the coordinate axes are drawn; every point on it has coordinates.First seen in the chapter →
collinearसंरेखsanrekh: Lying on one straight line. A, B, C are collinear with B between them when $A B + B C = A C$ .First seen in the chapter →
coordinate systemनिर्देशांक पद्धतिnirdeshank paddhati: A fixed starting point and reference lines that let us name every location with numbers.First seen in the chapter →
coordinatesनिर्देशांकnirdeshank: The ordered pair $(x, y)$ that fixes a point: its signed distances from the y-axis and from the x-axis.First seen in the chapter →

distance formulaदूरी सूत्रdoori sootra: The distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .First seen in the chapter →

midpointमध्य-बिंदुmadhya-bindu: The point halfway along a segment; its coordinates are the averages of the ends' coordinates.First seen in the chapter →

ordered pairक्रमित युग्मkramit yugm: A pair of numbers in which order matters: $(a, b) = (b, a)$ only when $a = b$ .First seen in the chapter →
originमूलबिंदुmoolbindu: The point $O (0, 0)$ where the two axes cross.First seen in the chapter →

quadrantचतुर्थांशchaturthansh: One of the four regions made by the axes, numbered I to IV anticlockwise from the top right.First seen in the chapter →

reflectionपरावर्तनparaavartan: The mirror image of a point or shape in a line. In the y-axis, $(x, y)$ goes to $(- x, y)$ .First seen in the chapter →

x-axisx-अक्षx-aksh: The horizontal number line of the coordinate plane. Its points have the form $(x, 0)$ .First seen in the chapter →
x-coordinateभुजbhuj: The first number of $(x, y)$ : how far the point lies to the right (+) or left (−) of the y-axis.First seen in the chapter →

y-axisy-अक्षy-aksh: The vertical number line of the coordinate plane. Its points have the form $(0, y)$ .First seen in the chapter →
y-coordinateकोटिkoti: The second number of $(x, y)$ : how far the point lies above (+) or below (−) the x-axis.First seen in the chapter →

1.1 Introduction

1.2 Settling In

1.3 The 2-D Cartesian Coordinate System

Points on the axes

Back to the Ganit Lab

Quadrants

What the two numbers mean

Beyond the lab walls

1.4 Distance Between Two Points in the 2-D Plane

Along a line parallel to an axis

Any two points

Reflections keep lengths

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Quick check: MCQs on Orienting Yourself: The Use of Coordinates

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