NCERT Solutions Class 9 Maths Chapter 1 Orienting Yourself the Use of Coordinates

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

Chapter 1: Orienting Yourself:
The Use of Coordinates

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

1 – 15 12 39 English

Solutions, notes, sample papers & more at 54 pages

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 1: Orienting Yourself: The
Use of Coordinates
This chapter turns position into arithmetic. Two perpendicular number lines fix an origin, four quadrants
and a signed address (x, y) for every point of the plane; the Baudhāyana–Pythagoras Theorem then turns any
two addresses into the distance between them.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 1 – 15

SECTIONS QUESTIONS

12 39

MEDIUM

English

In-text Questions — Page 3

Page 1 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Section 1.2 Settling In

Q1 Notice that this only shows the map of the floor. Do you see why the position of the
windows cannot be marked on this map?

Left Wall of the Room
Bathing Area

Right Wall of the Room
Left Wall of the Bathroom

Bed

Bedroom
12 ft × 10 ft

Bathroom
6 ft × 9 ft
Wardrobe
4 ft × 2 ft
Door Door

Fig. 1.1, page 3 — Shalini’s sketch of Reiaan’s room, pinned out to a scale of 1 cm : 1 foot.

Because Fig. 1.1 is a map of the floor only, and a window is not on the floor.

A point on the floor needs two numbers: how far along the room, how far across.
A window needs a third number as well: how high up the wall it is.

Why it happens: Shalini’s grid of pins is a flat, two-dimensional model. Fixing two
axes on the floor fixes exactly two independent directions — along the room and
across it. Every point of the floor gets one address (x, y), and that address is
complete. Height is a direction perpendicular to both axes, so no pair (x, y) can
record it. On the floor plan a window would collapse onto the wall line and be
indistinguishable from the wall itself.

The doors can be marked, because a doorway is a gap that reaches down to the floor. It leaves a
real gap in the wall line of the plan. A window leaves no gap at floor level.

Page 2 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Tip: To record windows you would need a third axis (height) — that is 3-D
coordinates (x, y, z). Architects get around this by drawing a separate elevation: a
second 2-D drawing whose two axes are “along the wall” and “height”. Two flat
drawings together carry the three-dimensional information.

Check it yourself: Reiaan can feel the wool along the bottom wall and find the door
gap with his fingers. Ask yourself what he would feel where a window is — an
unbroken thread, exactly like the rest of the wall.

Exercise Set 1.1 — Page 5
Section 1.3 The 2-d Cartesian Coordinate System

Q(i) If D₁R₁ represents the door to Reiaan’s room, how far is the door from the left wall
(the y-axis) of the room? How far is the door from the x-axis?

From Fig. 1.3, the door runs from D₁ (8, 0) to R₁ (11.5, 0) along the bottom wall.

Distance of the door from the left wall (y-axis) = 8 ft at its near end D₁

(the far end R₁ is 11.5 ft from the left wall)

Distance of the door from the x-axis = 0

So the doorway begins 8 ft from the left wall, and it is 0 ft from the x-axis.

Why it happens: The bottom wall of the room is the x-axis in this figure. Every point
of the doorway therefore has y-coordinate 0, and the perpendicular distance from
the x-axis of a point (x, 0) is 0. The distance from the y-axis is read off the x-
coordinate, and along the doorway that x-coordinate runs from 8 to 11.5, so “how
far from the left wall” is answered by its nearest edge, 8 ft.

Tip: Whenever a wall has been chosen as an axis, objects standing against that wall
get a zero in their coordinates. Choosing the axes well is half the work in a
coordinate problem.

Page 3 of 54

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a g l
Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
m.
What are the coordinates of D₁?
e
Q(ii)

m l as
.co a g
a s em
aislthe pin on the x-axis at the +8 mark.
D₁ g

co m
. ag
D₁ = (8, 0)
e m
g l as
a
Why it happens: D₁ lies on the bottom wall, which is the x-axis, so its y-coordinate
must be 0. Counting the unit marks from O along the x-axis gives 8, and the scale is
co m
1 cm : 1 foot, so D₁ is 8 ft to the right of the left wall.
em.
m l as
m .co a g
l a se
g
a Q(iii) If R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable

s
width for the room door? If a person in a wheelchair wants to enter the room, will
m a
.co agl
he/she be able to do so easily?

se m
g l a
a
D₁ and R₁ both lie on the x-axis, so the width is just the difference of the x-coordinates.

co m
m .
m as e
.co
Width of door = |11.5 − 8| = 3.5 ft
a g l
se m
a
3.5 ft = 3.5 × 12 in = 42 inches ≈ 1.07 m

ag l
Yes, this is a comfortable width. Ordinary bedroom doors in India are about 2.5 ft to 3 ft (750–
se m
com g l a
. a
900 mm) wide, so 3.5 ft is generously wide.
m
gl ase
Yes, a wheelchair user can enter easily. A standard wheelchair is about 24–26 in across, and

a
barrier-free design guidelines ask for a clear opening of at least 32 in, with 36 in preferred. A 42-
inch doorway clears that comfortably, and leaves room for a helper’s hands on the push
handles.
co m
m .
m as e
.co a g
Why it happens: The two points differ only in their x-coordinates, so the segmentl
se m D₁R₁ is parallel to the x-axis and its length is the plain difference |x₂ − x₁|. No square
g l a
a roots are needed — that is the whole point of choosing the wall as an axis.
c
m .
m a s e
m . co
Check it yourself: Measure the clear opening of a door at home — the gap when the
e agl
las
door is fully open, not the outside width of the frame. The door leaf and hinges eat
up an inch or two. ag

co m
m .
m ase
.co


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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q(iv) If B₁ (0, 1.5) and B₂ (0, 4) represent the ends of the bathroom door, is the bathroom
door narrower or wider than the room door?

B₁ and B₂ both lie on the y-axis, so their separation is the difference of the y-coordinates.

Width of bathroom door = |4 − 1.5| = 2.5 ft

Width of room door = 3.5 ft

3.5 − 2.5 = 1

The bathroom door is narrower, by 1 ft.

Why it happens: The bathroom door sits in the left wall, and the left wall is the y-
axis. Both ends therefore have x-coordinate 0, the segment is vertical, and its length
is |y₂ − y₁|. Notice how the same subtraction idea handles both doors — only the
axis changes.

Did you know? 2.5 ft is 30 inches, which is below the 32-inch clear width that
barrier-free guidelines ask for. So Reiaan’s room door is wheelchair friendly but his
bathroom door is not — a very common problem in Indian homes.

Think and Reflect — Page 5
After Exercise Set 1.1

Q1 What are the standard widths for a room door? Look around your home and in
school.

Common widths used in Indian homes and schools:

Page 5 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

DOOR USUAL WIDTH IN FEET

Main entrance door 900–1000 mm 3 – 3.3 ft

Bedroom / room door 750–900 mm 2.5 – 3 ft

Bathroom / toilet door 700–750 mm 2.3 – 2.5 ft

Classroom door 900–1200 mm 3 – 4 ft

Doors are usually 2.0–2.1 m (about 6.5–7 ft) tall.

Try This: Measure five doors around you and record each as an ordered pair (width,
height) in centimetres. Then treat width as x and height as y and plot the five points.
You will see the points cluster in a narrow band — doors are not made in random
sizes, and the plot shows that at a glance.

Why it happens: Measure the clear opening, not the frame: when a door is fully
open the leaf and hinges still stand in the way, so the usable gap is typically 25–50
mm less than the frame width.

Q2 Are the doors in your school suitable for people in wheelchairs?

Test each door against three things, not just one:

1. Clear width. India’s Harmonised Guidelines for Barrier Free Built Environment ask for a clear
opening of at least 900 mm. A 750 mm bathroom door fails this.
2. Threshold. A raised sill or step at the door stops a wheelchair even if the door is wide. The
floor should be level, or ramped at a gentle slope of about 1 : 12.
3. Turning space. A wheelchair needs roughly a 1500 mm × 1500 mm clear circle to turn, so a
wide door opening into a narrow corridor is still not usable.

In most school buildings the classroom doors pass the width test and the toilet doors fail it —
which is exactly the door that matters most.

Try This: Make a table of every door in your school block with columns: width,
threshold height, and pass/fail. That table is a small accessibility audit, and it is the
sort of evidence a school can actually act on.

Page 6 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

In-text Questions — Page 6
Section 1.3 The 2-d Cartesian Coordinate System

Q1 So far, we have only considered points on the two coordinate axes. What can you
say about the coordinates of points that are not on either axes?

Both coordinates are non-zero, and their two signs together decide which of the four quadrants
the point is in.

y

4
Quadrant II Quadrant I
(−, +) 2 (+, +)

x
-4 -2 O 2 4

Quadrant III -2 Quadrant IV
(−, −) (+, −)
-4

The four quadrants. The sign pair (x-sign, y-sign) is the address of a quadrant.

Page 7 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

QUADRANT POSITION SIGN OF X SIGN OF Y

I right of the y-axis, above the x-axis + +

II left of the y-axis, above the x-axis − +

III left of the y-axis, below the x-axis − −

IV right of the y-axis, below the x-axis + −

Why it happens: To reach a point from O you make two independent moves: a
horizontal shift and a vertical shift. Each shift has a size and a direction, and the sign
convention records the direction — right and up are positive, left and down are
negative. Since each of the two signs can be chosen in 2 ways, there are 2 × 2 = 4
possible sign patterns, and that is precisely why the axes cut the plane into four
quadrants, no more and no fewer.

Why the axes belong to no quadrant: A point on an axis is the case where one of
the two shifts is zero, and zero is neither positive nor negative. (x, 0) sits on the x-
axis, (0, y) sits on the y-axis, and O (0, 0) has both shifts zero. These are the boundary
cases between quadrants, which is why they are counted separately.

Tip: To name the quadrant of a point you never need to plot it. Just read the two
signs. (−7, 2) is (−, +), so Quadrant II. (0.5, −9) is (+, −), so Quadrant IV.

In-text Questions — Page 7

Page 8 of 54

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Section 1.3 The 2-d Cartesian Coordinate System
co m
e m.
m l as
Q1
.co
Copy Fig. 1.4 and mark S and Q in your diagram. Mark any point P in Quadrant I and
m a g
l a se
any point R in Quadrant III, and write down their coordinates.

a g

com
Quadrant II
e m . y-axis Quadrant I ag
g l as 4
a
Q (-5, 3) 3
2
co m
em.
m 1
l as
m .co O (0, 0)
a g x-axis

l a se -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
a g -1
-2
a s
.com -3 agl
m
Quadrant IIIgl ase Quadrant IV
a
-4
-5 S (3, -5)

co m
m .
m as e
.co a g l
se m
a
Fig. 1.4, page 6 — the coordinate plane and its four quadrants.

ag l
se m
com g l a
m . a
ase
agl
S (3, −5) and Q (−5, 3) are fixed by the question. P and R may be chosen freely — here P (4, 2) and
R (−4, −3).

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 9 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y
6

4
Q (−5, 3)
P (4, 2)
2

x
-6 -4 -2 2 4 6

-2

R (−4, −3)
-4

S (3, −5)
-6

S and Q are given; P and R are choices. The dashed paths show how each point is reached from O:
first the horizontal shift, then the vertical one.

S (3, −5): from O go 3 right, then 5 down → Quadrant IV

Q (−5, 3): from O go 5 left, then 3 up → Quadrant II

P = (4, 2) — both coordinates positive → Quadrant I

R = (−4, −3) — both coordinates negative → Quadrant III

Page 10 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why it happens: The order in the pair is what makes the marking unambiguous.
Look at S (3, −5) and Q (−5, 3): they use the same two numbers, 3 and −5, yet they are
different points in different quadrants. The first entry is always the horizontal shift
and the second always the vertical one — that agreement is what lets one pair of
numbers name one point, and only one.

Try This: Your P and R will not match the ones above, and that is fine — the question
says any point. Check your own choice by its signs alone: (+, +) must be Quadrant I
and (−, −) must be Quadrant III. If you picked P (4, 0) you have made a mistake: that
point lies on the x-axis, not in Quadrant I.

Think and Reflect — Page 7
After Fig. 1.4

Q1 What is the x-coordinate of a point on the y-axis?

x-coordinate of any point on the y-axis = 0

Why it happens: The x-coordinate of a point is defined as its perpendicular distance
from the y-axis, taken as positive to the right and negative to the left. A point that
lies on the y-axis is at zero distance from it, so its x-coordinate can only be 0. Every
such point therefore has the form (0, y) — for example H (0, 4) and G (0, −4.5) in Fig.
1.2.

Tip: Read the rule the other way round too. If a point’s x-coordinate is 0, it must lie
on the y-axis. The condition x = 0 is exactly the equation of the y-axis.

Q2 Is there a similar generalisation for a point on the x-axis?

Yes, with the roles of the two axes swapped.

Page 11 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y-coordinate of any point on the x-axis = 0

Every point on the x-axis has the form (x, 0)

Why it happens: The y-coordinate measures perpendicular distance from the x-axis.
A point sitting on the x-axis is at zero distance from it, so y = 0. B (4.5, 0) and E (−2.9,
0) in Fig. 1.2 are examples. The condition y = 0 is the equation of the x-axis.

Did you know? Putting the two results together: a point lying on both axes must
have x = 0 and y = 0, so it is (0, 0). That is a short proof that the two axes meet at
exactly one point — the origin.

Q3 Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.

Yes — exactly when x = y.

Two points coincide ⇔ both coordinates match

(x, y) = (y, x) ⇔ x = y and y = x

Both conditions say the same thing, so ⇔ x = y

For example P (4, 4) and Q (4, 4) are the same point. But P (4, 7) and Q (7, 4) are two different
points.

Why it happens: (x, y) is an ordered pair: the first entry is reserved for the
horizontal shift and the second for the vertical one. Swapping the entries swaps the
two shifts, which moves the point — unless the two shifts happen to be equal, in
which case swapping changes nothing. Geometrically, swapping the coordinates
reflects the point in the line through O that bisects Quadrants I and III (the line y = x),
and the only points a reflection leaves fixed are those on the mirror line itself.

Check it yourself: P (4, 7) and Q (7, 4) are both at distance √(16 + 49) = √65 from O.
Equal distance from the origin is not enough to make two points coincide — you
need both coordinates to agree.

Page 12 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q4 If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?

Yes, the claim is true. Both halves follow from one fact about ordered pairs: (a, b) = (c, d)
exactly when a = c and b = d.

If x = y: then (x, y) = (x, x) and (y, x) = (x, x), so the two are the same point.

If (x, y) = (y, x): matching first entries gives x = y. So equality forces x = y.

Together: (x, y) = (y, x) if and only if x = y.

Contrapositive: if x ≠ y, then (x, y) ≠ (y, x).

Why it happens: The second statement is an “if and only if”, so it needs both
directions proved, and the first statement is nothing but the contrapositive of one of
them. That is why the two sentences in the claim are not really two separate facts —
proving the “if and only if” proves the whole claim.

Tip: Watch the direction of an “if and only if” statement. “x = y gives (x, y) = (y, x)”
alone would not settle the question — you must also rule out any other way the two
pairs could be equal, and that is what the second direction does.

Exercise Set 1.2 — Page 8

Page 13 of 54

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates
a g l AglaSem · NCERT Solutions

Section 1.3 The 2-d Cartesian Coordinate System
co m
e m.
m l as
Q1
.co
Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11,
m (11, 7). (i) Where will the fourth foot of the table be? (ii) Is this a good spot for
9)eand
a g
a s
a gl the table? (iii) What is the width of the table? The length? Can you make out the
height of the table?

co m
m . ag
l a se
+10 agC (0, 10) B (12, 10)
R W F

co m
+9

m.
S₄ S₃

e
+8
Showering

as
r₁

m l
.co S
Area

g
+7
y-axis

a
Bed

sem
+6

Right wall
H
a
agl
+5
S₁ S₂
+4
Bathroom B₂

m a s
+3

.co agl
W₄ W₃

B₁ m
+2

s e
la
+1 Wardrobe

ag
A (12, 0)
-6 -5 -4 -3 -2 -1 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12
P O (0, 0) W₁ x-axis W₂ D₁ R₁ (11.5, 0)

co m
m .
m as e
.co l
Fig. 1.5, page 7 — Reiaan’s room and bathroom on the coordinate grid; 1 unit = 1 foot.

a g
se m
g l a
a
m

a se
com g l
(i) The fourth foot is at (8, 7).

m . a
ase
agl
(8, 9) and (11, 9) share the same y → that side is horizontal

(11, 9) and (11, 7) share the same x → that side is vertical
co m
m .
e
The fourth vertex must share x = 8 with (8, 9) and y = 7 with (11, 7)
m l as
m .co
Fourth foot = (8, 7)
a g
l a se
ag
.c
m
Why it happens: In a rectangle the opposite sides are parallel and equal. The shift

m a s e
co agl
from (11, 9) to (8, 9) is 3 units left, so the same shift applied to (11, 7) must land on

m .
e
the fourth foot: (11 − 3, 7) = (8, 7). You never need to measure — the shift does the
as
work.
a g l

co m
m .
m as e
.co


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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

(ii) Yes, it is a workable spot, and it can be improved.
The table covers 8 ≤ x ≤ 11 and 7 ≤ y ≤ 9. Check it against everything else in the room:

Bed: x from 0.5 to 6.5. Since 8 > 6.5, no clash.
Wardrobe: x from 3 to 7, y from 0 to 2. No clash.
Room door D₁R₁: x from 8 to 11.5 on the bottom wall. The table is at y ≥ 7, far from the swing
of the door.

So nothing overlaps. But it leaves a wasted 1 ft gap on the right (x = 11 to the right wall at x =
12) and 1 ft above (y = 9 to the top wall at y = 10). Pushing the table into the corner, with feet at
(9, 8), (12, 8), (12, 10) and (9, 10), would clear more floor and — important for Reiaan — put two
of its sides against walls he can trace with his hand.
(iii) Width 2 ft, length 3 ft; the height cannot be found.

Length = |11 − 8| = 3 ft (along the x-direction)

Width = |9 − 7| = 2 ft (along the y-direction)

Height = cannot be determined

Why the height is missing: Fig. 1.5 is a plan of the floor. Its two axes fix the two
directions that lie in the floor; height is perpendicular to both and no pair (x, y) can
record it. This is the same reason the windows could not be marked on Fig. 1.1.

Q2 If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the
wardrobe? Are there any changes you would suggest if the door is made wider?

No, it will not hit the wardrobe — but the clearance is only 0.5 ft.

Hinge B₁ = (0, 1.5); door leaf B₁B₂ = |4 − 1.5| = 2.5 ft

Swinging on the hinge, the free edge traces a quarter circle of radius 2.5 about B₁

Wardrobe occupies 3 ≤ x ≤ 7, 0 ≤ y ≤ 2

Nearest point of the wardrobe to B₁ is (3, 1.5), because y = 1.5 already lies between 0 and 2

Distance = |3 − 0| = 3 ft

3 ft > 2.5 ft → the door misses the wardrobe by 0.5 ft

Page 15 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why it happens: Every point of the swinging door stays within 2.5 ft of the hinge,
because the hinge holds one end fixed. So the whole question reduces to one
comparison: is the nearest bit of the wardrobe more than 2.5 ft from B₁? Turning
“does it hit?” into “compare two distances” is exactly what coordinates are for.

If the door is made wider. A door of width w hinged at B₁ will just graze the wardrobe when w =
3 ft, and will strike it for any w > 3 ft. Since 2.5 ft (30 in) is below the 32-inch clear width that
wheelchair users need, widening the door is worth doing — so make one of these changes:

Move the wardrobe 0.5–1 ft to the right, from W₁ (3, 0)–W₂ (7, 0) to (3.5, 0)–(7.5, 0). There is
room: the right wall is at x = 12.
Hinge the door at B₂ (0, 4) instead, so it swings upward, away from the wardrobe. The
nearest wardrobe corner is then (3, 2), at distance √(3² + 2²) = √13 ≈ 3.6 ft — so a door up to
about 3.5 ft would clear.
Make it open into the bathroom, or fit a sliding door, which sweeps no floor at all.

Tip: The door swing is dead floor space — you cannot put anything there. On a plan,
sketch each door’s quarter circle before you place the furniture, not after.

Page 16 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q3 Look at Reiaan’s bathroom. (i) What are the coordinates of the four corners O, F, R,
and P of the bathroom? (ii) What is the shape of the showering area SHWR in
Reiaan’s bathroom? Write the coordinates of the four corners. (iii) Mark off a 3 ft × 2
ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates
of the corners of these spaces.

C (0, 10) B (12, 10)
+10
R W F
+9
S₄ S₃
+8
r₁ Showering
Area +7
y-axis

Bed
S +6

Right wall
H
+5
S₁ S₂
+4
Bathroom B₂
+3
W₄ W₃
+2

+1 B₁
Wardrobe
A (12, 0)
-6 -5 -4 -3 -2 -1 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12
P O (0, 0) W₁ x-axis W₂ D₁ R₁ (11.5, 0)

Fig. 1.5, page 7 — Reiaan’s room and bathroom on the coordinate grid; 1 unit = 1 foot.

(i) Reading Fig. 1.5, the bathroom lies to the left of the y-axis:

O = (0, 0), F = (0, 9), R = (−6, 9), P = (−6, 0)

Check: width = |0 − (−6)| = 6 ft, length = |9 − 0| = 9 ft → a 6 ft × 9 ft bathroom, as Fig. 1.1

states

Why the x-coordinates are negative: The bathroom is on the far side of the wall
that was chosen as the y-axis. Distances to the left of that axis are counted as
negative, so the bathroom sits in Quadrant II. This is the whole payoff of allowing
negative numbers: one origin and one pair of axes can serve both rooms.

(ii) The showering area is a trapezium (a right trapezium), not a rectangle.

Page 17 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y
10
R W F

8
SHWR

S 6
H

4

Bathroom
2

P
x
-6 -4 -2 O

The showering area SHWR inside the 6 ft × 9 ft bathroom. SH and RW are parallel; RS is perpendicular
to both.

Page 18 of 54

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
e m.
S = (−6, 6), H = (−3, 6), W = (−2, 9), R = (−6, 9)
m l as
.co
SH: from (−6, 6) to (−3, 6) — both have y = 6, so SH is horizontal, length 3 ft
m a g
l a se
g
RW: from (−6, 9) to (−2, 9) — both have y = 9, so RW is horizontal, length 4 ft
aSH ∥ RW but 3 ≠ 4 → exactly one pair of parallel sides → trapezium

co m
. ag
RS: from (−6, 9) to (−6, 6) — both have x = −6, so RS is vertical, length 3 ft, and is
e m
perpendicular to SH and RW
g l as
a
HW = √((−2 − (−3))² + (9 − 6)²) = √(1 + 9) = √10 ≈ 3.16 ft — the slanting glass edge

co m
m.
Area = ½ × (3 + 4) × 3 = 10.5 sq ft

m as e
.co a g l
a s em
Why we can tell the shape without measuring: Two segments are parallel to each

a gl other when the same coordinate is constant along both. Here y is constant along SH
and along RW, so both are horizontal; x is constant along RS, so it is vertical and

a s
com agl
therefore perpendicular to them. HW has neither coordinate constant, so it is

m .
as e
slanted — and its length needs the Baudhāyana–Pythagoras Theorem.

a g l
(iii) Answers may differ. One workable arrangement, keeping both fittings clear of the shower

m
(which occupies y ≥ 6 on the left) and leaving a clear path from the doorway on the y-axis:

. co
em
m l as
.co
FITTING SIZE CORNERS

a g
se m
l a
Washbasin 3 ft × 2 ft (−6, 0), (−3, 0), (−3, 2), (−6, 2)

ag
m
Toilet 2 ft × 3 ft (−6, 2.5), (−4, 2.5), (−4, 5.5), (−6, 5.5)

a se
.com a g l
m
ase
agl
Washbasin: 3 units wide along x (−6 to −3), 2 units along y (0 to 2) → 3 ft × 2 ft ✓

Toilet: 2 units along x (−6 to −4), 3 units along y (2.5 to 5.5) → 2 ft × 3 ft ✓

Both lie inside −6 ≤ x ≤ 0 and 0 ≤ y ≤ 9, and both stay below y = 6 → clear of the shower
co m
m .
o m l a se
m .cCheck it yourself: Whatever corners you choose, verifyagthem the same way —
l a se
ag subtract the x-coordinates to get one side, subtract the y-coordinates to get the

.c
m
other, and check that all four corners satisfy −6 ≤ x ≤ 0 and 0 ≤ y ≤ 9. Also leave the

m a s e
co agl
strip near x = 0, y = 1.5 to 4 free: that is where the doorway is.

m .
as e
a g l

co m
m .
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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q4 Other rooms in the house: (i) Reiaan’s room door leads from the dining room which
has the length 18 ft and width 15 ft. The length of the dining room extends from
point P to point A. Sketch the dining room and mark the coordinates of its corners.
(ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining
room. Write down the coordinates of the feet of the table.

C (0, 10) B (12, 10)
+10
R W F
+9
S₄ S₃
+8
r₁ Showering
Area +7
y-axis

Bed
S +6

Right wall
H
+5
S₁ S₂
+4
Bathroom B₂
+3
W₄ W₃
+2

+1 B₁
Wardrobe
A (12, 0)
-6 -5 -4 -3 -2 -1 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12
P O (0, 0) W₁ x-axis W₂ D₁ R₁ (11.5, 0)

Fig. 1.5, page 7 — Reiaan’s room and bathroom on the coordinate grid; 1 unit = 1 foot.

(i) P = (−6, 0) and A = (12, 0) are already on the plan, and

PA = |12 − (−6)| = 18 ft — exactly the stated length ✓

So PA is the wall shared with the bedroom and bathroom, and the dining room must lie on the
other side of it — below the x-axis. Going 15 ft down from PA gives the corners:

P (−6, 0), A (12, 0), (12, −15), (−6, −15)

Page 20 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y

10
Bathroom Bedroom

5

P A
x
-5 5 10

-5
Table

-10
Dining room 18 ft × 15 ft
(−6, −15) (12, −15)
-15

The dining room (blue) lies below the x-axis, sharing the wall PA with the bedroom and bathroom.
The dining table (amber) is centred at (3, −7.5).

Page 21 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why the y-coordinates are negative: The x-axis is the wall that separates the
dining room from Reiaan’s side of the house. Distances measured downwards from
that wall are negative by the sign convention, so the dining room occupies parts of
Quadrants III and IV. That is also why the exercise told you to mark the y-axis all the
way down to (0, −15) — the graph has to be large enough to hold the room.

(ii) The centre of a rectangle is the midpoint of a diagonal.

Centre = midpoint of P (−6, 0) and (12, −15)

= ((−6 + 12)/2, (0 + (−15))/2)

= (3, −7.5)

Table 5 ft along x and 3 ft along y → go 2.5 each side in x and 1.5 each side in y:

x: 3 − 2.5 = 0.5 and 3 + 2.5 = 5.5

y: −7.5 − 1.5 = −9 and −7.5 + 1.5 = −6

Feet = (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9)

Check: the midpoint of the table’s own diagonal is ((0.5 + 5.5)/2, (−6 + (−9))/2) = (3, −7.5) ✓ — the
table really is centred.

Try This: Turn the table through a right angle, 3 ft along x and 5 ft along y. The feet
become (1.5, −5), (4.5, −5), (4.5, −10), (1.5, −10) — still centred at (3, −7.5). Both
answers are correct; the question fixes the centre, not the orientation.

In-text Questions — Page 9
Section 1.4 Distance Between Two Points in the 2-D Plane

Q1 Triangle ADM is an acute angled triangle in the first quadrant. How do we find the
lengths of its sides AD, DM and MA?

None of AD, DM, MA is parallel to an axis, so we cannot subtract coordinates directly. Instead,
make each side the hypotenuse of a right triangle whose legs are parallel to the axes.

Page 22 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y

M (9, 6)
6

A (3, 4)
4

AC = 3
2
CD = 4
C (3, 1) D (7, 1)
x
2 4 6 8 10
Dropping A straight down to C (3, 1) makes ∆ACD right-angled at C, with legs parallel to the axes.

A (3, 4), D (7, 1), M (9, 6)

AD: horizontal shift = |7 − 3| = 4, vertical shift = |4 − 1| = 3

AD = √(4² + 3²) = √25 = 5 units

DM: horizontal shift = |9 − 7| = 2, vertical shift = |6 − 1| = 5

DM = √(2² + 5²) = √29 ≈ 5.39 units

MA: horizontal shift = |9 − 3| = 6, vertical shift = |6 − 4| = 2

MA = √(6² + 2²) = √40 = 2√10 ≈ 6.32 units

Page 23 of 54

as e
a g l
Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
m.
Why it happens: Take any two points. Travel from the first to the second in two

m l a se
stages — first horizontally, then vertically. The corner where you turn is the third
o
vertex, C .(3, a g line and the
c 1) for the side AD, and since one stage is along a horizontal
m
sealong a vertical line, the angle there is a right angle. The Baudhāyana–
g l a
other
aPythagoras Theorem then gives the direct distance from the two stages. This works
for any pair of points, which is why it becomes the general distance formula
co m
√((x₂−x₁)² + (y₂−y₁)²).
e m . ag
g l as
a
Check it yourself: The book calls ∆ADM acute-angled. Verify it. The longest side is
MA, with MA² = 40. Now AD² + DM² = 25 + 29 = 54, and 54 > 40, so the angle opposite
co m
MA (the angle at D) is acute. The other two angles are opposite shorter sides, so they
em.
c o m
are acute as well. All three angles acute — the claim checks out, and it was settled by
g l as
.
m alone. a
s e
arithmetic

a gla
m a s
Think and Reflect — Page 9
m .co agl
l a se
g
Section 1.4 Distance Between Two Points in the 2-D Plane

a
m
In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis?
co
Q1
What about the distance along the y-axis?
m .
as e
. com a g l
s em

gl a
a
m
Along the x-axis: 7 − 3 = 4 units (to the right)
a se
.
Along the y-axis: 1 − 4 = −3, that is 3 units downwards
com a g l
m
ase
agl
In Fig. 1.7 these two shifts appear as the segments CD = 4 and AC = 3, where C = (3, 1).

. com
Why it happens: The x-coordinate records position along the horizontal direction

m a s em and nothing
only, so the change in x measures the horizontal part of the journey
.celse. agl not an error — it says the
o Likewise for y. The minus sign in 1 − 4 is information,
m movement was downwards. For a length we take the size, 3 units.
l a se
ag
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
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.co


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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q2 Can these distances help you find the distance AD?

Yes. The two shifts are the legs of a right triangle whose hypotenuse is AD.

Turning point C = (3, 1)

AC is vertical (x = 3 throughout) and CD is horizontal (y = 1 throughout)

→ ∡ACD = 90°

By the Baudhāyana–Pythagoras Theorem:

AD² = AC² + CD² = 3² + 4² = 9 + 16 = 25
AD = 5 units

Why it happens: A horizontal line and a vertical line always meet at a right angle —
that is how the two axes were set up in the first place. So splitting any journey into a
horizontal stage and a vertical stage automatically manufactures a right triangle,
and the theorem is always available. This single observation is what turns geometry
into algebra: AD = √((x₂ − x₁)² + (y₂ − y₁)²).

Tip: It makes no difference which point you start from. Going from D to A the shifts
are −4 and +3, and squaring turns both signs into the same positive numbers: (−4)² +
3² = 25. Distance has no direction.

In-text Questions — Page 11

Page 25 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Section 1.4 Distance Between Two Points in the 2-D Plane

Q1 What if, x₁, x₂, y₁, y₂ take negative values? In Fig. 1.9, triangle AMD is reflected in the
y-axis. What are the coordinates of the images of points A, M, and D?

M′ y-axis M (9, 6)
6

5
A′ A (3, 4)
4

3

2

1 C (3, 1) D (7, 1)
D′ C′ x-axis

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Fig. 1.9, page 11 — triangle AMD and its image after reflection in the y-axis; the image
vertices are marked A′, M′ and D′.

Reflection in the y-axis sends (x, y) to (−x, y).

A (3, 4) → A′ = (−3, 4)

M (9, 6) → M′ = (−9, 6)

D (7, 1) → D′ = (−7, 1)

(and the helper point C (3, 1) → C′ = (−3, 1))

Page 26 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y
M' M

A' A
4

D' D x
-8 -4 4 8
∆ADM in Quadrant I and its mirror image ∆A′D′M′ in Quadrant II. Each point keeps its height and
swaps sides.

Why the x-coordinate changes sign and the y-coordinate does not: The mirror is
the y-axis. Reflection keeps a point at the same perpendicular distance from the
mirror but on the opposite side. That distance is measured along the x-direction, so
its size stays the same and its sign flips. Nothing moves the point up or down, so the
y-coordinate is untouched.

The side lengths are unchanged, and the arithmetic shows why:

A′D′: shifts |−3 − (−7)| = 4 and |4 − 1| = 3 → √(16 + 9) = 5 units

D′M′: shifts |−9 − (−7)| = 2 and |6 − 1| = 5 → √(4 + 25) = √29 units

M′A′: shifts |−9 − (−3)| = 6 and |6 − 4| = 2 → √(36 + 4) = √40 units

Why negatives cause no trouble in the distance formula: Only the differences x₂ −
x₁ and y₂ − y₁ enter the formula, and they are then squared. Changing the sign of
both x-coordinates changes the sign of their difference but not its square, so the
distance is unaffected. The formula was never restricted to the first quadrant.

Think and Reflect — Page 11

Page 27 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Section 1.4 Distance Between Two Points in the 2-D Plane

Q1 What has remained the same and what has changed with this reflection?

UNCHANGED CHANGED

All three side lengths: 5, √29, √40 The sign of every x-coordinate

All three angles, so the shape is identical Position: from Quadrant I to Quadrant II

Perimeter, and the area (13 sq units) Orientation: A→D→M was anticlockwise, A′→D′→M′ is
clockwise

Every y-coordinate, so every point’s height above the Which side of the y-axis each point is on
x-axis

Each point’s distance from the y-axis (its size, not its Distances from any fixed point that is not on the y-axis
sign)

Why lengths survive but orientation does not: Reflection preserves both shifts in
size — the vertical shift exactly, the horizontal shift up to sign — and the distance
formula squares them, so every length comes through untouched. A reflection is
therefore a congruence: the image triangle can be laid exactly on the original. But it
cannot be slid onto it without being turned over, because a mirror reverses the sense
in which the vertices are read. That is the one thing a reflection always destroys.

Check it yourself: Going from A (3, 4), the shift to D is (4, −3) and to M is (6, 2). In the
image, the shift from A′ to D′ is (−4, −3) and to M′ is (−6, 2) — the horizontal parts
have reversed while the vertical parts have not. That mismatch is exactly what flips
the triangle over.

Q2 Would these observations be the same if ΔADM is reflected in the x-axis (instead of
the y-axis)?

Yes — every observation carries over, with the roles of x and y interchanged.

Page 28 of 54

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
e m.
Reflection in the x-axis sends (x, y) to (x, −y)
m l as
.co
A (3, 4) → A″ (3, −4) D (7, 1) → D″ (7, −1) M (9, 6) → M″ (9, −6)
m a g
l a se
g
aA″D″: shifts 4 and |−4 − (−1)| = 3 → √(16 + 9) = 5 units

co m
. ag
D″M″: shifts 2 and |−6 − (−1)| = 5 → √29 units
e m
M″A″: shifts 6 and |−6 − (−4)| = 2 → √40 units
g l as
a
m
The triangle moves from Quadrant I to Quadrant IV. Lengths, angles and area are preserved;
co
m.
the x-coordinates are now the ones left alone and the y-coordinates change sign; the

m as e
.co l
orientation is reversed once again.
a g
a s emthe answer had to be the same: The distance formula treats the two
gl
Why
a coordinates symmetrically — both differences are squared and added. So whichever

m
axis is used as the mirror, one difference keeps its sign, the other reverses, and
a s
m .co
squaring erases the difference between the two cases. Any reflection in a straight
agl
l a se
line preserves distance; the choice of mirror only decides where the image lands, not
what shape it is. a g

co m
m .
e
Try This: Reflect ∆ADM in the y-axis and then reflect the result in the x-axis. You land
m l as
.co g
on (−3, −4), (−7, −1), (−9, −6) in Quadrant III. Two reflections have restored the

em a
s
anticlockwise order — the combined effect is a half turn about the origin, not a
a
l mirror image.
ag
se m
com g l a
m 12–14 . a
End-of-Chapter Exercises —sePages
a
agl
Chapter 1 Orienting Yourself: The Use of Coordinates

co m
m
What are the x-coordinate and y-coordinate of the point of intersection of the two.
as e
Q1

m l
.co g
axes?

emANSWER a
a s
agl
.c
m
The two axes meet at the origin.

m a s e
x-coordinate = 0, y-coordinateem
.co agl
as
=0

The point is O (0, 0) a g l

co m
m .
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.co


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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why it happens: The point of intersection lies on the y-axis, so its perpendicular
distance from the y-axis is 0, giving x = 0. It also lies on the x-axis, so y = 0. And these
two conditions together pin down a single point: only (0, 0) satisfies both. That is
why two intersecting lines can share exactly one point.

Q2 Point W has x-coordinate equal to − 5. Can you predict the coordinates of point H
which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

H = (−5, k), where k can be any real number

Why the x-coordinate is forced: A line parallel to the y-axis is a vertical line. Every
point on it is the same perpendicular distance from the y-axis and on the same side
of it, so every point on it has the same x-coordinate. W has x = −5, so the whole line is
x = −5, and H must share that value. Its y-coordinate, however, is completely free —
moving up or down the line changes nothing about the distance from the y-axis.

Quadrants: since x = −5 is negative, H lies to the left of the y-axis. So:

VALUE OF K POINT WHERE IT LIES

k>0 (−5, k) Quadrant II

k=0 (−5, 0) on the x-axis, in no quadrant

k<0 (−5, k) Quadrant III

So H can lie only in Quadrant II or Quadrant III — never in I or IV, because those need a
positive x-coordinate.

Tip: x = −5 is the equation of that vertical line. In Grade 10 you will meet such
equations everywhere; the idea starts right here, with one coordinate held fixed
while the other roams.

Page 30 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q3 Consider the points R (3, 0), A (0, − 2), M (− 5, − 2) and P (− 5, 2). If they are joined in
the same order, predict: (i) Two sides of RAMP that are perpendicular to each other.
(ii) One side of RAMP that is parallel to one of the axes. (iii) Two points that are
mirror images of each other in one axis. Which axis will this be? Now plot the points
and verify your predictions.

Predict first, by reading the coordinates only.
(i) AM and MP are perpendicular.

A (0, −2) and M (−5, −2) share y = −2 → AM is horizontal

M (−5, −2) and P (−5, 2) share x = −5 → MP is vertical
Horizontal ⊥ vertical → AM ⊥ MP, the right angle being at M
(ii) AM is parallel to the x-axis (both ends have y = −2). MP is parallel to the y-axis, so either
answer is acceptable; the other two sides, RA and PR, are parallel to neither.
(iii) M (−5, −2) and P (−5, 2) are mirror images in the x-axis.

Same x-coordinate, opposite y-coordinates: −2 and +2

Reflection in the x-axis sends (x, y) to (x, −y) → M ↔ P ✓

Page 31 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y
4

P (−5, 2)
2

R (3, 0)
x
-6 -4 -2 2 4

-2
M (−5, −2) A (0, −2)

-4

Quadrilateral RAMP. AM is horizontal, MP is vertical, so the angle at M is a right angle; M and P are
mirror images in the x-axis.

Verification by calculation:

RA = √((0 − 3)² + (−2 − 0)²) = √(9 + 4) = √13

AM = |−5 − 0| = 5

MP = |2 − (−2)| = 4

PR = √((3 − (−5))² + (0 − 2)²) = √(64 + 4) = √68 = 2√17

Right angle at M: the shift M→A is (5, 0), the shift M→P is (0, 4). One is purely horizontal

and the other purely vertical, so they meet at 90°.

Page 32 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why the predictions could be made without a diagram: A shared y-coordinate is
precisely the condition for a segment to be horizontal; a shared x-coordinate is the
condition for it to be vertical. Equal x with opposite y is precisely reflection in the x-
axis. Each geometric property has an exact arithmetic signature — that is what
coordinate geometry buys you.

Did you know? RAMP is not a rectangle, though it has a right angle: RA = √13 and PR
= 2√17 are unequal and neither is parallel to an axis. One right angle is not enough.

Q4 Plot point Z (5, − 6) on the Cartesian plane. Construct a right-angled triangle IZN
and find the lengths of the three sides. (Comment: Answers may differ from person
to person.)

Z (5, −6) is in Quadrant IV: 5 to the right of O, then 6 down. The easiest way to guarantee a right
angle at Z is to take one other vertex directly above it and one directly beside it.

Choose I = (5, −2) — same x as Z, so ZI is vertical

Choose N = (8, −6) — same y as Z, so ZN is horizontal

→ ∡IZN = 90°

Page 33 of 54

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
y em.
com g l as
m . a
as e
a g l
x
2 4 om 6 8 10
e m .c ag
ag las I (5, −2)
-2
co m
s em.
.co
m
4 5 agla
a s em
-4
ag l

a s
. com agl
-6
a s em
aZgl(5, −6)
3 N (8, −6)

co m
-8 m .
ase
. com a g l
se m
l a
ag A right-angled triangle IZN with the right angle at Z. The legs are chosen parallel to the axes, so their
lengths are simple differences.
se m
com g l a
m . a
ase
agl
ZI = |−2 − (−6)| = 4 units

ZN = |8 − 5| = 3 units

co m
IN = √((8 − 5)² + (−6 − (−2))²) = √(9 + 16) = √25 = 5 units
m .
m as e
.co a g l
se m Check: ZN² + ZI² = 9 + 16 = 25 = IN² ✓
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why this construction always works: Making one leg vertical and the other
horizontal guarantees the right angle, because the two axis directions are
perpendicular by definition. It also makes both leg lengths plain subtractions, so
only the hypotenuse needs the theorem. Deliberately choosing legs of 3 and 4 turns
the hypotenuse into a whole number, 5.

Try This: Your I and N will very likely differ from these — the question says so. Test
your own triangle the same way: verify that the sum of the squares of the two
shorter sides equals the square of the longest. If it does not, your triangle is not
right-angled.

Q5 What would a system of coordinates be like if we did not have negative numbers?
Would this system allow us to locate all the points on a 2-D plane?

No. Without negative numbers you could label only one quarter of the plane.

Only x ≥ 0 and y ≥ 0 could be written → only Quadrant I survives

The axes shrink from full lines to two rays leaving O

Quadrants II, III and IV would have no addresses at all

Why it fails: Each coordinate has to record two things about a shift — how far, and
in which of the two opposite directions. A number without a sign records only the
size. So (3, 2) and (−3, 2) would collapse to the same label “3, 2”, and the same
address would name two different points. The correspondence between points and
pairs of numbers would stop being one-to-one, and that correspondence is the
whole idea of the coordinate system. Signs are what let a single number carry a
direction as well as a magnitude.

You could patch it by writing “3 left, 2 up” in words — but then “left” is doing exactly the job of
the minus sign, and the neat algebra is lost. The distance formula, for instance, works because
x2 − x1 is a signed number that gets squared.

Page 35 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Did you know? This is the connection the chapter opens with. Brahmagupta (c. 628
CE) was the first to treat zero and negative numbers as genuine algebraic quantities.
Without his work, the four-quadrant plane you are drawing in this chapter could not
exist — Descartes himself first worked mostly in what we would call the first
quadrant.

Q6 Are the points M (− 3, − 4), A (0, 0) and G (6, 8) on the same straight line? Suggest a
method to check this without plotting and joining the points.

Yes, M, A and G are collinear, with A lying between M and G.
Method 1 — compare the shifts. Read off how far you move horizontally and vertically along
each step:

M → A: shift = (0 − (−3), 0 − (−4)) = (3, 4)

A → G: shift = (6 − 0, 8 − 0) = (6, 8) = 2 × (3, 4)

Vertical shift ÷ horizontal shift: 4/3 for the first step, 8/6 = 4/3 for the second → equal

→ the direction never changes → the three points lie on one line

Method 2 — use distances. Three points are collinear exactly when the longest of the three
distances equals the sum of the other two:

MA = √(3² + 4²) = √25 = 5
AG = √(6² + 8²) = √100 = 10

MG = √((6 − (−3))² + (8 − (−4))²) = √(81 + 144) = √225 = 15

MA + AG = 5 + 10 = 15 = MG ✓ → collinear

Why the distance test works: If M, A, G were the vertices of a genuine triangle, the
triangle inequality would give MA + AG > MG strictly. Equality can happen only when
the triangle has collapsed — that is, when A lies on the segment MG. So checking MA
+ AG = MG is a complete test, and it needs no drawing.

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why the shift test works: Walking along a straight line, the ratio of vertical rise to
horizontal run is fixed — that fixed ratio is what “straight” means in coordinates. If
the ratio changes between one step and the next, the path has turned. Method 1 is
usually quicker because it avoids square roots altogether, and it also handles the
case where the middle point is not between the other two.

Tip: With Method 1, cross-multiply instead of dividing, to keep everything in whole
numbers: 4 × 6 = 24 and 3 × 8 = 24, equal, so collinear. This also avoids trouble when
a horizontal shift is 0.

Q7 Use your method (from Problem 6) to check if the points R (− 5, − 1), B (− 2, − 5) and C
(4, − 12) are on the same straight line. Now plot both sets of points and check your
answers.

No — R, B and C are not collinear, although they come remarkably close.

R → B: shift = (−2 − (−5), −5 − (−1)) = (3, −4)

B → C: shift = (4 − (−2), −12 − (−5)) = (6, −7)

Cross-multiply to compare 3 : −4 with 6 : −7 —

3 × (−7) = −21 and (−4) × 6 = −24

−21 ≠ −24 → the direction changes at B → not collinear

The distance test agrees, but only just:

RB = √(3² + 4²) = √25 = 5

BC = √(6² + 7²) = √85 ≈ 9.2195

RC = √(9² + 11²) = √(81 + 121) = √202 ≈ 14.2127

RB + BC ≈ 14.2195, while RC ≈ 14.2127

Difference ≈ 0.007 units — small, but not zero → not collinear

Page 37 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why plotting cannot settle this one: At the scale 1 cm = 1 unit, the bend at B is
about 0.07 mm — thinner than your pencil line. Draw the three points and they will
look exactly like a straight line. That is the real lesson of this pair of questions: in Q6
the algebra confirmed what a drawing would suggest, but here a drawing would
have misled you. An exact test beats a picture.

Check it yourself: If C had been (4, −13) instead, the second shift would be (6, −8) =
2 × (3, −4) and the three points would be collinear. One unit of difference in a single
coordinate decides the matter — and the eye cannot see it.

Q8 Using the origin as one vertex, plot the vertices of: (i) A right-angled isosceles
triangle. (ii) An isosceles triangle with one vertex in Quadrant III and the other in
Quadrant IV.

Answers may differ. Two convenient choices:

Page 38 of 54

as e
a g l
Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
y e m.
com g l as
m . a
ase
agl 4
(0, 4)

com
e m . ag
g l as right isosceles
a 2
m
m .co
m l
(4,a se0)
. co O
ag
m x
ase
agl -4 -2 2 4

om a s
. c
em -2 agl
a s
a gl isosceles

co m
-4 m .
e
.co
m (−3, −4) las
(3,ag−4)
se m
g l a
a
se m
com g l a
.
Two triangles with a vertex at O. In blue, a right-angled isosceles triangle with its legs on the axes; in
m a
ase
green, an isosceles triangle straddling Quadrants III and IV.

agl
(i) Right-angled isosceles triangle: O (0, 0), (4, 0), (0, 4).

com
m .
m as e
o l
One leg on the x-axis: length |4 − 0| = 4
. c a g
a s em One leg on the y-axis: length |4 − 0| = 4 → equal → isosceles
agl The axes are perpendicular → the angle at O is 90° → right-angled
c
m .
Hypotenuse = √(4² + 4²) = √32 = 4√2 ≈ 5.66 units
m a s e
e m . co agl
g l as
(ii) Isosceles triangle with one vertex in Quadrant III and one in Quadrant IV: O (0, 0), U (−3,
−4), V (3, −4). a

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m .
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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

U (−3, −4) has signs (−, −) → Quadrant III ✓

V (3, −4) has signs (+, −) → Quadrant IV ✓

OU = √((−3)² + (−4)²) = √25 = 5
OV = √(3² + (−4)²) = √25 = 5 → OU = OV → isosceles

UV = |3 − (−3)| = 6

Why picking U and V as a −/+ pair guarantees the isosceles property: U and V
have opposite x-coordinates and the same y-coordinate, so they are mirror images
of each other in the y-axis. The origin lies on that mirror line, so it is equidistant from
any pair of mirror-image points. You get OU = OV for free, without computing either
— the symmetry does the proof.

Try This: Try to make (ii) also right-angled at O. You need OU ⊥ OV, and with U (−a,
−b), V (a, −b) that happens only when a = b — for example O, (−3, −3), (3, −3). Check
that the two legs are then √18 each and the base is 6, and that 18 + 18 = 36 ✓.

Q9 The following table shows the coordinates of points S, M and T. In each case, state
whether M is the midpoint of segment ST. Justify your answer.

S M T IS M THE MID-POINT OF ST? REASON FOR YOUR
YES OR NO ANSWER

(−3, (0, 0) (3, 0)
0)

(2, 3) (3, 4) (4, 5)

(0, 0) (0, 5) (0,
−10)

(−8, (0, (6, −3)
7) −2)

When M is the mid-point of ST, can you find any connection between the
coordinates of M, S and T?

Test each row by averaging the coordinates of S and T and comparing with M.

Page 40 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

S M T IS M THE MIDPOINT OF REASON
ST?

(−3, (0, 0) (3, 0) Yes ((−3 + 3)/2, (0 + 0)/2) = (0, 0) = M
0)

(2, 3) (3, 4) (4, 5) Yes ((2 + 4)/2, (3 + 5)/2) = (3, 4) = M

(0, 0) (0, 5) (0, No ((0 + 0)/2, (0 + (−10))/2) = (0, −5), not (0,
−10) 5)

(−8, (0, (6, −3) No ((−8 + 6)/2, (7 + (−3))/2) = (−1, 2), not (0,
7) −2) −2)

The connection: each coordinate of M is the average of the corresponding coordinates of S
and T.

If M is the midpoint of ST, then

xM = (xS + xT)/2 and yM = (yS + yT)/2

Why the averaging rule is true: M is the midpoint exactly when the journey S → M
repeats itself as M → T. Compare the two journeys one coordinate at a time. The
horizontal shifts must be equal, so xM − xS = xT − xM, which rearranges to 2xM = xS +
xT. The same argument on the vertical shifts gives 2yM = yS + yT. Since each
coordinate can be handled separately, the midpoint of a slanted segment is no
harder than the midpoint of a horizontal one.

Check it yourself: Equal distances alone are not enough. In row 3, M (0, 5) is 5 units
from S and 15 units from T — it is not even on the segment, which runs downwards
from (0, 0) to (0, −10). A midpoint must lie on the segment, and the averaging rule
guarantees that automatically.

Q10 Use the connection you found to find the coordinates of B given that M (−7, 1) is
the midpoint of A (3, − 4) and B (x, y).

Apply the averaging rule one coordinate at a time and solve for the unknown.

Page 41 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

(3 + x)/2 = −7 → 3 + x = −14 → x = −17

(−4 + y)/2 = 1 → −4 + y = 2 → y = 6

B = (−17, 6)

Check: midpoint of A (3, −4) and B (−17, 6) = ((3 − 17)/2, (−4 + 6)/2) = (−7, 1) = M ✓

Why the answer is so far from A: M is not between the numbers 3 and −7 by a little
— going from A to M is a shift of (−10, 5), and B must be a second, identical shift
beyond M: (−7 − 10, 1 + 5) = (−17, 6). Thinking in shifts gives the answer in one line
and is a useful check on the algebra.

Tip: A pair of coordinates gives two independent equations, one for x and one for y.
They never mix, so you always solve them separately — never as a simultaneous pair.

Q11 Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using
your knowledge of how to find the coordinates of the midpoint of a segment, how
would you find the coordinates of P and Q? Do this for the case when the points
are A (4, 7) and B (16, −2).

The method. The midpoint rule works because you take half of the shift from A to B. Trisection
points need one-third and two-thirds of the same shift.

Shift from A to B = (xB − xA, yB − yA)

Midpoint = A + ½ of the shift

P = A + ⅓ of the shift → P = ((2xA + xB)/3, (2yA + yB)/3)

Q = A + ⅔ of the shift → Q = ((xA + 2xB)/3, (yA + 2yB)/3)

For A (4, 7) and B (16, −2):

Page 42 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Shift from A to B = (16 − 4, −2 − 7) = (12, −9)

One-third of the shift = (4, −3)

P = (4 + 4, 7 − 3) = (8, 4)

Q = (8 + 4, 4 − 3) = (12, 1)

and B = (12 + 4, 1 − 3) = (16, −2) ✓

Verification using midpoints only, as the question asks:

Midpoint of A (4, 7) and Q (12, 1) = ((4 + 12)/2, (7 + 1)/2) = (8, 4) = P ✓

Midpoint of P (8, 4) and B (16, −2) = ((8 + 16)/2, (4 − 2)/2) = (12, 1) = Q ✓

Why the midpoint idea extends: The midpoint formula is really a statement about
splitting each shift in a fixed ratio, and the coordinates handle that splitting
independently. If P cuts AB in the ratio 1 : 2, then the horizontal shift 12 is cut as 4
and 8, and the vertical shift −9 as −3 and −6 — the same ratio in both. That is why a
single fraction of the shift moves you to the right point.

Check it yourself: AP = √(4² + 3²) = 5, PQ = √(4² + 3²) = 5, QB = √(4² + 3²) = 5, and AB =
√(12² + 9²) = √225 = 15 = 3 × 5. The three pieces really are equal, so P and Q are
genuine points of trisection.

Q12 (i) Given the points A (1, − 8), B (− 4, 7) and C (−7, − 4), show that they lie on a circle K
whose center is the origin O (0, 0). What is the radius of circle K? (ii) Given the
points D (− 5, 6) and E (0, 9), check whether D and E lie within the circle, on the
circle, or outside the circle K.

(i) A circle with centre O is the set of all points at one fixed distance from O. So compute OA, OB,
OC and see whether they agree.

Page 43 of 54

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

co m
e m.
OA = √(1² + (−8)²) = √(1 + 64) = √65
m l as
.co
OB = √((−4)² + 7²) = √(16 + 49) = √65
m a g
l a se
g
OC = √((−7)² + (−4)²) = √(49 + 16) = √65
a

com
. ag
OA = OB = OC = √65 → all three lie on one circle centred at O
e m
Radius of K = √65 ≈ 8.06 units
g l as
a

. c om
Why this is a complete proof: “Lies on the circle with centre O and radius r” means

s e
exactly “is at distance r from O” — that is the definition of a circle, not ma property of
. com the three distances are equal is showing the three
it. So showing
a glapoints are concyclic
a s em O. Notice also why the three answers came out equal: 1² + 8², 4² + 7² and 7² +
gl 4² all total 65, even though the points sit in three different quadrants. Squaring
about
a
wipes out the signs.

om a s
e
. c
m It is cleaner to compare the squares and avoid roots agl
s
(ii) Compare each distance with the radius.
a
agl
altogether:

co m
.
r² = 65
e m
m l as
.co
m = (−5)² + 6² = 25 + 36 = 61 → 61 < 65 → D lies inside the circle a g
s e
la
OD²

ag OE² = 0² + 9² = 81 → 81 > 65 → E lies outside the circle

se m
com
In lengths: OD = √61 ≈ 7.81 < 8.06, and .OE g l a
m
= 9 > 8.06.
a
l a se
Why comparing squares ag is safe: Distances are never negative, and squaring is
increasing on non-negative numbers. So OD < r exactly when OD² < r². Working with

. com
61, 65 and 81 keeps everything in whole numbers and removes any doubt caused by
rounding — √61 ≈ 7.81 and √65 ≈ 8.06 are close enough that eam
m a s careless decimal could

m
o
.cmislead you. agl
l a se
ag
.c
m
Tip: This gives you a three-way test for any point P and any circle of centre C, radius

m a s e
co agl
r: compare CP² with r² — less means inside, equal means on, greater means outside.

m .
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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Q13 The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the
coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the
coordinates of A, B and C.

Take D as the midpoint of BC, E as the midpoint of CA and F as the midpoint of AB.

Midpoint rule, written as sums:

B + C = 2D = (10, 2) …(1)

C + A = 2E = (12, 10) …(2)

A + B = 2F = (0, 6) …(3)

Add all three: 2(A + B + C) = (22, 18) → A + B + C = (11, 9)

A = (A + B + C) − (B + C) = (11, 9) − (10, 2) = (1, 7)

B = (A + B + C) − (C + A) = (11, 9) − (12, 10) = (−1, −1)

C = (A + B + C) − (A + B) = (11, 9) − (0, 6) = (11, 3)

Check all three midpoints:

Midpoint of BC = ((−1 + 11)/2, (−1 + 3)/2) = (5, 1) = D ✓

Midpoint of CA = ((11 + 1)/2, (3 + 7)/2) = (6, 5) = E ✓

Midpoint of AB = ((1 − 1)/2, (7 − 1)/2) = (0, 3) = F ✓

Why adding the three equations is the key step: Each equation involves two of
the three unknowns, so no single one can be solved on its own. Adding them makes
every vertex appear exactly twice, which produces A + B + C. Subtracting any one of
the original equations from that total then isolates a single vertex. This trick — find
the total first, then peel off — is worth remembering; it works because each
coordinate can be handled separately, so what looks like one vector equation is
really two ordinary linear equations.

Page 45 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Did you know? ∆DEF is called the medial triangle of ∆ABC. The three equations
always have exactly one solution, so a triangle is completely determined by the
midpoints of its sides. Pairing the letters differently (say D with CA instead of BC)
only renames A, B and C — the triangle you get is the same one.

Q14 A city has two main roads which cross each other at the centre of the city. These
two roads are along the North–South (N–S) direction and East–West (E–W)
direction. All the other streets of the city run parallel to these roads and are 200 m
apart. There are 10 streets in each direction. (i) Using 1 cm = 200 m, draw a model
of the city in your notebook. Represent the roads/streets by single lines. (ii) There
are street intersections in the model. Each street intersection is formed by two
streets — one running in the N–S direction and another in the E–W direction. Each
street intersection is referred to in the following manner: If the second street
running in the N–S direction and 5th street in the E–W direction meet at some
crossing, then we call this street intersection (2, 5). Using this convention, find: (a)
how many street intersections can be referred to as (4, 3). (b) how many street
intersections can be referred to as (3, 4).

(i) Draw 10 vertical lines (the N–S streets) and 10 horizontal lines (the E–W streets), each set
spaced 1 cm apart, since 200 m at the scale 1 cm = 200 m is exactly 1 cm. Number the N–S
streets 1 to 10 from left to right and the E–W streets 1 to 10 from bottom to top. The grid has 10
× 10 = 100 crossings.

Page 46 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

10
9
E–W street number

8
7
6
5
(3, 4)
4
3 (4, 3)
2
1
1 2 3 4 5 6 7 8 9 10
N–S street number
The city model: 10 N–S streets and 10 E–W streets, 1 cm apart. (4, 3) and (3, 4) are two different
crossings.

(ii) (a) Exactly one. (b) Exactly one — but a different one.

(4, 3) = crossing of the 4th N–S street with the 3rd E–W street → 1 intersection

(3, 4) = crossing of the 3rd N–S street with the 4th E–W street → 1 intersection

(4, 3) ≠ (3, 4)

Page 47 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why the count is always one: An N–S street and an E–W street are not parallel, so
they meet, and two straight lines that are not parallel meet at exactly one point —
they cannot meet twice. Naming one street from each family therefore names one
crossing and only one. This is the same reason a point of the plane has only one
coordinate pair.

Why (4, 3) and (3, 4) are different places: The convention fixes an order: the first
number always names the N–S street, the second always the E–W street. Reading
them the other way round sends you 200 m along one road and 200 m back along
the other — a different crossing altogether. A postal address works the same way; so
does (x, y).

Did you know? The chapter opens with exactly this idea in practice. The Sindhu–
Sarasvatī cities laid their streets out N–S and E–W about 10 m apart, so a merchant
could reach a warehouse by counting streets in the two directions — a working
coordinate system thousands of years before it was written down as algebra.

Q15 A computer graphics program displays images on a rectangular screen whose
coordinate system has the origin at the bottom-left corner. The screen is 800 pixels
wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre
at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its
centre at the point B (250, 230). Determine: (i) whether any part of either circle lies
outside the screen. (ii) whether the two circles intersect each other.

With the origin at the bottom-left corner, the visible screen is exactly the set of points with 0 ≤ x
≤ 800 and 0 ≤ y ≤ 600.

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

. com
600 px s m
ehigh
m a
e m . co agl
g l as
a

co m
e m . ag
g l as
a

co m
em.
as
B (250, 230)
.com a g l
m
ase
170
agl
A (100,
m a s
.co agl
150)
se m
g l a
a
O (0, 0) 800 px wide
com
m .
o m l a se
m .c The two icons on the 800 × 600 screen. Both fit inside; their
ag 170 pixels apart.
centres are

l a se
ag (i) No — both circles lie entirely on the screen. A circle stays inside the rectangle exactly when

se m
a
the distance from its centre to every edge is at least the radius.

com
.TO BOTTOM a g l
e m
as
agl
ICON TO LEFT TO RIGHT TO TOP RADIUS VERDICT
EDGE EDGE EDGE EDGE

com 100 → inside
A (100, 100 700 150 450 80 smallest gap 80 ≤
150)
m .
m as100e
. co agl
B (250, 250 550 230 370 smallest gap 100

e m
las
230) ≤ 230 → inside

ag c
m .
m
Icon A spans x from 100 − 80 = 20 to 100 + 80 = 180, and y from 70 to 230 — all within the
a s e
e m . co agl
as
screen

a g l
Icon B spans x from 150 to 350, and y from 130 to 330 — all within the screen

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m .
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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Why checking the four edge distances is enough: The point of a circle nearest to a
given straight edge lies on the perpendicular from the centre, at distance (centre-to-
edge) − r. If that is still non-negative for all four edges, no part of the circle can cross
any edge. Distance to a vertical edge is a difference of x-coordinates only; to a
horizontal edge, of y-coordinates only — which is why the screen was set up with its
edges along the axes.

(ii) Yes, the two circles intersect — they cut each other at two points.

AB = √((250 − 100)² + (230 − 150)²)

= √(150² + 80²) = √(22500 + 6400) = √28900 = 170 pixels

Sum of radii = 80 + 100 = 180

Difference of radii = 100 − 80 = 20

20 < 170 < 180, that is, |r₁ − r₂| < AB < r₁ + r₂ → the circles meet at two points

Why the two comparisons decide it: If the centres were farther apart than 180 the
circles would be completely separate, and if they touched at exactly 180 they would
meet at a single point. If the centres were closer than 20 the smaller circle would sit
wholly inside the larger one, meeting it nowhere. Only when the centre distance lies
strictly between the difference and the sum of the radii can the two boundaries cross
— and 170 is in that range, with just 10 pixels to spare before they would separate.

Did you know? Games and drawing programs test for collisions in exactly this way.
Comparing AB² = 28900 with (r₁ + r₂)² = 32400 avoids the square root entirely, which
is why it is fast enough to run for hundreds of objects in every frame.

Q16 Plot the points A (2, 1), B (−1, 2), C (−2, −1), and D (1, −2) in the coordinate plane. Is
ABCD a square? Can you explain why? What is the area of this square?

Yes, ABCD is a square, and its area is 10 square units.

Page 50 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

y
3

B (−1, 2)
2
√10
A (2, 1)
1

x
-3 -2 -1 1 2 3

-1
C (−2, −1)
-2
D (1, −2)

-3

ABCD with its diagonals. All four sides are √10 and both diagonals are √20 — a tilted square.

Page 51 of 54

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

The four sides:

AB = √((−1 − 2)² + (2 − 1)²) = √(9 + 1) = √10

BC = √((−2 + 1)² + (−1 − 2)²) = √(1 + 9) = √10
CD = √((1 + 2)² + (−2 + 1)²) = √(9 + 1) = √10

DA = √((2 − 1)² + (1 + 2)²) = √(1 + 9) = √10

→ all four sides equal → ABCD is a rhombus

The two diagonals:

AC = √((−2 − 2)² + (−1 − 1)²) = √(16 + 4) = √20

BD = √((1 + 1)² + (−2 − 2)²) = √(4 + 16) = √20

→ the diagonals are equal → the rhombus is a square

Why equal sides alone are not enough: Four equal sides make a rhombus, and a
rhombus can lean over as far as you like without any side changing length. What
fixes the angles is the diagonals: in a rhombus the diagonals are equal only when
the angles are right angles. You can also see it directly with the converse of the
Baudhāyana–Pythagoras Theorem — AB² + BC² = 10 + 10 = 20 = AC², so the angle at
B is 90°, and a rhombus with one right angle is a square.

Area, two ways:

side² = (√10)² = 10 square units

½ × d₁ × d₂ = ½ × √20 × √20 = ½ × 20 = 10 square units ✓

Did you know? This square is tilted, so its sides are not whole numbers — yet its
area is the whole number 10. That is possible because area only needs the square of
the side, and the squaring in the distance formula undoes the square root. The
centre of the square is the midpoint of AC, which is ((2 − 2)/2, (1 − 1)/2) = (0, 0) — the
square is centred on the origin, which is why the four points came in ± pairs.

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

Chapter at a glance
Two perpendicular number lines — the horizontal x-axis and the vertical y-axis — meet at
the origin O (0, 0). Together they are the coordinate axes, and the plane they lie in is the
Cartesian plane (also: coordinate plane, xy-plane).
For a point P (x, y), x is the perpendicular distance of P from the y-axis measured along the
x-axis, and y is its perpendicular distance from the x-axis measured along the y-axis. Right
and up are positive; left and down are negative.
Because each of the two displacements carries its own sign, there are 2 × 2 = 4 sign
patterns, and so exactly four quadrants: I (+, +), II (−, +), III (−, −), IV (+, −). A point with a
zero displacement sits on an axis and belongs to no quadrant: (x, 0) on the x-axis, (0, y) on
the y-axis.
The pair is ordered. (x, y) = (y, x) only when x = y; otherwise the two are different points,
mirror images of each other in the line through O that bisects Quadrants I and III.
Distances along a grid line are differences: |x2 − x1| for a horizontal segment, |y2 − y1| for
a vertical one.
For any two points, the horizontal and vertical shifts are the legs of a right triangle, so by
the Baudhāyana–Pythagoras Theorem the distance is √((x2 − x1)² + (y2 − y1)²). Squaring
removes the signs, so it does not matter which point is called first.

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Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates AglaSem · NCERT Solutions

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Quick revision
m l a se
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Origin The point where the two axes cross — the (0, 0) O (0, 0)

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zero of both number lines

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x-coordinate first entry of (x, y) For D (7, 1), x = 7
axis, counted along theg
Signed perpendicular distance from the y-

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y-coordinate second entry of (x,
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For D (7, 1), y = 1
Signed perpendicular distance from the x-

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axis
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aPoint on the y- Zero distance from the y-axis (0, y) H (0, 4), G (0, −4.5)
axis
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Quadrant I Right of the y-axis and above the x-axis (+, +) A (3, 4)
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Quadrant II aand
Left of the y-axis (−, +) Q (−5, 3)

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Quadrant III Left of the y-axis and below the x-axis (−, −) M (−3, −4)

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Quadrant IV
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Right of the y-axis and below the x-axis (+, −)
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Distance along a Only one coordinate changes |x2 − x1| or |y2 − C (3, 1) to D (7, 1) is 4
y1| units

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Distance, any two
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The two shifts are the legs of a right √((x2−x1)² + A (3, 4) to D (7, 1) is 5
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points triangle (y2−y1)²) units

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Each coordinate is the average of the two ((x1+x2)/2, Midpoint of (−3, 0) and
end coordinates (y1+y2)/2) (3, 0) is (0, 0)

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