Important Questions Class 9 Maths Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates

Contents

Part A — Short Answer Questions (2–3 marks)
Part B — Long Answer Questions (5 marks)
Topic-wise Coverage Summary
Exam Tips & Common Mistakes

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Part A — Short Answer Questions (2–3 Marks)

📍 Coordinate System — Basics

Q1
What is the Cartesian plane? Name its parts.
Theory
Easy

The Cartesian plane (also called the coordinate plane or xy-plane) is a two-dimensional plane formed by two number lines — one horizontal and one vertical — intersecting at right angles.

The horizontal line is called the x-axis.
The vertical line is called the y-axis.
Their point of intersection is called the origin O, with coordinates (0, 0).
The axes divide the plane into four parts called quadrants (I, II, III, IV).

Q2
What are the coordinates of the origin? What is the x-coordinate of any point lying on the y-axis?
Theory
Easy

The coordinates of the origin are (0, 0).
Any point on the y-axis is of the form (0, y). So the x-coordinate is always 0 for any point on the y-axis.
Similarly, any point on the x-axis is of the form (x, 0), so its y-coordinate is 0.

Q3
State the sign of coordinates (x, y) in each of the four quadrants.
Theory
Easy

Quadrant I

(+, +) — both x and y are positive

Quadrant II

(−, +) — x is negative, y is positive

Quadrant III

(−, −) — both x and y are negative

Quadrant IV

(+, −) — x is positive, y is negative

Q4
In which quadrant do each of these points lie? (a) (−3, 5) (b) (4, −6) (c) (−2, −8) (d) (7, 3)
Numerical
Easy

(−3, 5) → x is negative, y is positive → Quadrant II
(4, −6) → x is positive, y is negative → Quadrant IV
(−2, −8) → both negative → Quadrant III
(7, 3) → both positive → Quadrant I

📐 Coordinates & Points

Q5
A point P has x-coordinate 4 and lies on the x-axis. Write its coordinates. Also write the coordinates of point Q which is 3 units above the y-axis on the y-axis.
Numerical
Easy

P lies on the x-axis, so y-coordinate = 0. Coordinates of P = (4, 0).
Q lies on the y-axis, so x-coordinate = 0. Since it is 3 units above origin, y = 3. Coordinates of Q = (0, 3).

Tip

Any point on the x-axis has form (x, 0). Any point on the y-axis has form (0, y).

Q6
Is (3, 5) the same as (5, 3)? Justify your answer.
Theory
Easy

No, (3, 5) and (5, 3) are not the same.

In a coordinate pair (x, y), the first value is the x-coordinate (distance from the y-axis) and the second is the y-coordinate (distance from the x-axis). Since 3 ≠ 5, swapping them gives a different point in a different location on the plane.

Rule: (x, y) = (y, x) only when x = y.

Q7
Point W has x-coordinate equal to −5. Point H lies on the line through W parallel to the y-axis. In which quadrants can H lie? Give possible coordinates.
Theory
Medium

Since W has x-coordinate −5, the vertical line through W has the equation x = −5. Point H also lies on this line, so its x-coordinate is always −5. H can be (−5, y) for any value of y.

If y > 0, H is in Quadrant II. Example: (−5, 4)
If y < 0, H is in Quadrant III. Example: (−5, −2)
If y = 0, H lies on the x-axis (not in any quadrant)

So H can lie in Quadrant II or Quadrant III.

📏 Distance Between Two Points

Q8
State the distance formula between two points (x₁, y₁) and (x₂, y₂). Name the theorem it is based on.
Theory
Easy

The distance between two points (x₁, y₁) and (x₂, y₂) is given by:

Distance = √[ (x₂ − x₁)² + (y₂ − y₁)² ]

This formula is derived from the Baudhāyana–Pythagoras Theorem. The horizontal shift (x₂ − x₁) and vertical shift (y₂ − y₁) form the two legs of a right triangle, and the distance is the hypotenuse.

Remember

It does not matter whether (x₂ − x₁) or (y₂ − y₁) are positive or negative, since we square them.

Q9
Find the distance between points A(3, 4) and D(7, 1).
Numerical
Easy

Using the distance formula:

AD = √[ (7 − 3)² + (1 − 4)² ] = √[ 4² + (−3)² ] = √[16 + 9] = √25 = 5 units

∴ Distance AD = 5 units

Q10
Find the distance between points D(7, 1) and M(9, 6).
Numerical
Medium

Using the distance formula:

DM = √[ (9 − 7)² + (6 − 1)² ] = √[ 2² + 5² ] = √[4 + 25] = √29 units

∴ Distance DM = √29 units ≈ 5.39 units

Q11
What is the distance between the points (x₁, y) and (x₂, y)? When would you use the simpler formula instead of the full distance formula?
Theory
Medium

When two points share the same y-coordinate, they lie on a horizontal line. Their distance is simply:

Distance = |x₂ − x₁|

Similarly, for points (x, y₁) and (x, y₂) (same x-coordinate, on a vertical line):

Distance = |y₂ − y₁|

You use these simpler formulas when the line segment is parallel to one of the axes. The full distance formula is needed only when the segment is at an angle to both axes.

🏙️ Real-Life Application & Reflection

Q12
A room has corners at O(0,0), A(12,0), B(12,10) and C(0,10). What are the dimensions (length and width) of the room?
Numerical
Medium

Length along x-axis = distance from O(0,0) to A(12,0) = |12 − 0| = 12 ft
Width along y-axis = distance from O(0,0) to C(0,10) = |10 − 0| = 10 ft

∴ The room is 12 ft long and 10 ft wide.

Q13
The ends of a door are at B₁(0, 1.5) and B₂(0, 4) on the y-axis. The ends of another door are at D(8, 0) and R(11.5, 0) on the x-axis. Which door is wider?
Numerical
Medium

Width of door B₁B₂ = |4 − 1.5| = 2.5 ft
Width of door DR = |11.5 − 8| = 3.5 ft

∴ Door DR (the room door) is wider at 3.5 ft, compared to door B₁B₂ (the bathroom door) at 2.5 ft.

Q14
Triangle ADM has A(3, 4), D(7, 1), M(9, 6). Triangle A’D’M’ is the reflection of triangle ADM in the y-axis. Write the coordinates of A’, D’ and M’.
Numerical
Medium

When a point (x, y) is reflected in the y-axis, its image is (−x, y) — the x-coordinate changes sign, y stays the same.

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

Tip

Reflection in the y-axis preserves all side lengths of the triangle.

Q15
A rectangular study table has three feet at (8, 9), (11, 9) and (11, 7). Find the coordinates of the fourth foot.
Numerical
Hard

Let the three given points be P₁(8, 9), P₂(11, 9) and P₃(11, 7). For a rectangle, opposite sides must be parallel and equal.

P₁ and P₂ share y = 9 → horizontal side of length 11 − 8 = 3 ft
P₂ and P₃ share x = 11 → vertical side of length 9 − 7 = 2 ft
The fourth corner P₄ must be at x = 8 (same as P₁) and y = 7 (same as P₃)

∴ Fourth foot of the table is at (8, 7). Table is 3 ft × 2 ft.

Common Mistake

Students often guess randomly. Always use the property: opposite vertices of a rectangle have the same x or y as adjacent vertices.

📖

Part B — Long Answer Questions (5 Marks)

Q1
Explain the 2-D Cartesian Coordinate System in detail. Define x-axis, y-axis, origin, quadrants and coordinates of a point. Give one example of a point in each quadrant.
Theory
Medium

Definition: The 2-D Cartesian coordinate system uses two number lines at right angles to locate any point in a plane.
x-axis: The horizontal number line. Positive values go to the right of origin; negative values to the left.
y-axis: The vertical number line. Positive values go upwards; negative values downwards.
Origin (O): The point of intersection of the x-axis and y-axis. Coordinates = (0, 0).
Quadrants: The two axes divide the plane into 4 regions:

Quadrant I: (+, +) | Quadrant II: (−, +) | Quadrant III: (−, −) | Quadrant IV: (+, −)
Coordinates (x, y): x = perpendicular distance from y-axis; y = perpendicular distance from x-axis.
Examples: Q I: (3, 5), Q II: (−2, 4), Q III: (−3, −6), Q IV: (5, −1)

Remember

The order in (x, y) matters — the x-coordinate always comes first (horizontal), y-coordinate second (vertical).

Q2
Derive the distance formula between two points (x₁, y₁) and (x₂, y₂) in a coordinate plane using the Baudhāyana–Pythagoras Theorem.
Proof
Hard

Let A(x₁, y₁) and D(x₂, y₂) be any two points in the coordinate plane.
Draw a horizontal line from A and a vertical line from D. They meet at point F(x₂, y₁).
Triangle AFD is a right-angled triangle with the right angle at F.
Horizontal leg AF: AF = |x₂ − x₁| (distance along x-axis)
Vertical leg FD: FD = |y₂ − y₁| (distance along y-axis)
By Baudhāyana–Pythagoras Theorem: AD² = AF² + FD²
Substituting: AD² = (x₂ − x₁)² + (y₂ − y₁)²

∴ AD = √[ (x₂ − x₁)² + (y₂ − y₁)² ]

Tip

Squaring removes the need for absolute value signs — whether (x₂ − x₁) is positive or negative, its square is always positive.

Q3
For triangle ADM with A(3, 4), D(7, 1) and M(9, 6), find the lengths of all three sides. Then verify whether reflection in the y-axis preserves side lengths by computing D’M’ and M’A’ where A'(−3, 4), D'(−7, 1), M'(−9, 6).
Numerical
Hard

Step 1 — Find AD:

AD = √[(7−3)² + (1−4)²] = √[16 + 9] = √25 = 5 units

Step 2 — Find DM:

DM = √[(9−7)² + (6−1)²] = √[4 + 25] = √29 units

Step 3 — Find MA:

MA = √[(3−9)² + (4−6)²] = √[36 + 4] = √40 units

Step 4 — Check reflected triangle A’D’M’:

D’M’ = √[(−9−(−7))² + (6−1)²] = √[(−2)² + 5²] = √29 units ✓

M’A’ = √[(−3−(−9))² + (4−6)²] = √[36 + 4] = √40 units ✓

∴ All side lengths are preserved after reflection: AD = A’D’ = 5, DM = D’M’ = √29, MA = M’A’ = √40 units.

Q4
Reiaan’s room has a rectangular bedroom (12 ft × 10 ft) with one corner at origin O(0,0). A wardrobe (4 ft × 2 ft) is placed with one corner at W₁(3, 0). A study table has three feet at (8, 9), (11, 9) and (11, 7). (i) Write the coordinates of all four corners of the bedroom. (ii) Find the coordinates of all four corners of the wardrobe. (iii) Find the fourth foot of the study table. (iv) Find the length of the diagonal of the bedroom.
Application
Hard

Bedroom corners: O(0, 0), A(12, 0), B(12, 10), C(0, 10) — a 12 ft × 10 ft rectangle.
Wardrobe corners: Starting from W₁(3, 0), the wardrobe is 4 ft wide (along x) and 2 ft deep (along y):

W₁(3, 0), W₂(7, 0), W₃(7, 2), W₄(3, 2)
Fourth foot of table: Three feet at (8,9), (11,9), (11,7). The fourth foot is at (8, 7).
Diagonal of bedroom: From O(0, 0) to B(12, 10):
OB = √[12² + 10²] = √[144 + 100] = √244 ≈ 15.62 ft

Bedroom diagonal = √244 ≈ 15.62 ft

Q5
Explain why a window’s position cannot be shown on a floor plan (map of the floor of a room). What limitation of the 2-D coordinate system does this illustrate? How would you extend the system to describe positions in 3-D space?
Theory
Medium

A floor plan is a 2-D representation showing only the floor area. It records positions in terms of two coordinates: East–West (x) and North–South (y).
A window is not on the floor — it is positioned on a wall, at some height above the floor. Height is the third dimension (z).
Since the floor plan only records x and y coordinates, there is no way to represent the height of the window. This is the limitation of 2-D coordinates.
Extension to 3-D: A 3-D coordinate system adds a third axis — the z-axis — perpendicular to both x and y. A point in 3-D space is written as (x, y, z), where z represents height.
In such a system, a window at height 4 ft on the north wall could be described by its x position along the wall, y = 0 (on the north wall), and z = 4 (height).

Exam Tip

This question tests your understanding of the real-world meaning of coordinates and the dimensions involved. Always connect theory to the practical example from the chapter.

Q6
Plot points A(2, 1), B(−1, 2), C(−2, −1) and D(1, −2) in the coordinate plane. Calculate all four side lengths. Is ABCD a square? Find its area.
Numerical
Hard

Find AB: AB = √[(−1−2)² + (2−1)²] = √[9+1] = √10
Find BC: BC = √[(−2−(−1))² + (−1−2)²] = √[1+9] = √10
Find CD: CD = √[(1−(−2))² + (−2−(−1))²] = √[9+1] = √10
Find DA: DA = √[(2−1)² + (1−(−2))²] = √[1+9] = √10
All four sides are equal: AB = BC = CD = DA = √10. This is necessary for a rhombus or square.
Check diagonals: AC = √[(−2−2)² + (−1−1)²] = √[16+4] = √20; BD = √[(1−(−1))² + (−2−2)²] = √[4+16] = √20. Equal diagonals confirm it is a square.
Area: Side = √10, so Area = (√10)² = 10 square units.

∴ ABCD is a square with side √10 and area = 10 sq units.

Q7
Given that M(−7, 1) is the midpoint of segment AB where A has coordinates (3, −4), find the coordinates of B. Show all working steps.
Numerical
Hard

Let B = (x, y). The midpoint formula states that the midpoint M of segment AB is:
M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )
We know M = (−7, 1) and A = (3, −4). So:

(3 + x)/2 = −7 and (−4 + y)/2 = 1
Solve for x:

3 + x = −14 → x = −17
Solve for y:

−4 + y = 2 → y = 6
Verify: Midpoint of A(3,−4) and B(−17, 6) = ((3+(−17))/2, (−4+6)/2) = (−14/2, 2/2) = (−7, 1) ✓

∴ B = (−17, 6)

Q8
A city has two main roads intersecting at the centre, running North–South and East–West. All other streets are parallel to these roads and 200 m apart. There are 10 streets in each direction. Using a coordinate system with the centre as origin: (i) How many street intersections are there in total? (ii) How many intersections have the address (4, 3)? (iii) How many have the address (3, 4)? Are these the same?
Application
Medium

Setting up the grid: With 10 streets in each direction, there are 10 N–S streets and 10 E–W streets.
Total intersections: Each of the 10 N–S streets crosses each of the 10 E–W streets.

Total intersections = 10 × 10 = 100
Address (4, 3): This refers to the 4th N–S street and 3rd E–W street. Each combination is unique, so there is exactly 1 intersection with address (4, 3).
Address (3, 4): This refers to the 3rd N–S street and 4th E–W street. This is also exactly 1 intersection.
Are (4, 3) and (3, 4) the same? No — they are different streets, at different physical locations in the city, just as coordinates (4, 3) ≠ (3, 4) in the plane.

Total intersections = 100 | (4,3) gives 1 unique location | (3,4) gives a different unique location

Remember

This beautifully shows the real-world meaning of ordered pairs: the order matters, just like street addresses.

Q9
Consider the points R(3, 0), A(0, −2), M(−5, −2) and P(−5, 2). (i) Predict two sides of RAMP that are perpendicular to each other. (ii) Identify a side parallel to one of the axes. (iii) Find two mirror-image points and name the axis of reflection.
Numerical
Hard

Side RA: R(3,0) to A(0,−2) — slopes at an angle.
Side AM: A(0,−2) to M(−5,−2) — both have y = −2, so AM is parallel to the x-axis. Length AM = |−5−0| = 5 units.
Side MP: M(−5,−2) to P(−5,2) — both have x = −5, so MP is parallel to the y-axis. Length MP = |2−(−2)| = 4 units.
Perpendicular sides: AM is horizontal and MP is vertical, so AM ⊥ MP.
Mirror images: A(0,−2) and P(−5,2) are not mirrors. But M(−5,−2) and P(−5,2): same x-coordinate (−5), y-coordinates are −2 and +2 (opposite signs). So M and P are mirror images in the x-axis.

AM ⊥ MP | AM is parallel to x-axis | M and P are mirror images in the x-axis

Q10
A computer screen is 800 pixels wide and 600 pixels high, with origin at the bottom-left. Circle A has centre (100, 150) and radius 80 px. Circle B has centre (250, 230) and radius 100 px. (i) Does any part of Circle A lie outside the screen? (ii) Does any part of Circle B lie outside the screen? (iii) Do the two circles intersect?
Application
Hard

Check Circle A (centre (100,150), radius 80):

Left boundary: 100 − 80 = 20 > 0 ✓

Right boundary: 100 + 80 = 180 < 800 ✓

Bottom boundary: 150 − 80 = 70 > 0 ✓

Top boundary: 150 + 80 = 230 < 600 ✓

→ Circle A lies fully inside the screen.
Check Circle B (centre (250,230), radius 100):

Left: 250 − 100 = 150 > 0 ✓

Right: 250 + 100 = 350 < 800 ✓

Bottom: 230 − 100 = 130 > 0 ✓

Top: 230 + 100 = 330 < 600 ✓

→ Circle B also lies fully inside the screen.
Check if circles intersect: Find distance between centres A(100,150) and B(250,230):
d = √[(250−100)² + (230−150)²] = √[150² + 80²] = √[22500 + 6400] = √28900 = 170 px
Sum of radii = 80 + 100 = 180 px. Difference of radii = |100 − 80| = 20 px.
Since 20 < 170 < 180, the circles do intersect (the distance between centres is less than the sum of radii).

Neither circle goes outside the screen. The two circles intersect each other (d = 170 px < 180 px sum of radii).

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Topic-wise Coverage Summary

Basics of Coordinate System

Q1–Q4 (Short) + Q1 (Long) — 5 Questions

Points & Coordinates

Q5–Q7 (Short) — 3 Questions

Distance Formula

Q8–Q11 (Short) + Q2, Q3 (Long) — 6 Questions

Room / Floor Plan Application

Q12–Q13 (Short) + Q4 (Long) — 3 Questions

Reflection & Symmetry

Q14 (Short) + Q3 (Long) — 2 Questions

Real-Life Applications

Q15 + Q5–Q10 (Long) — 7 Questions

🎯

Exam Tips & Common Mistakes

Tip 1

Always write coordinates in the order (x, y) — x-coordinate first. Never swap them.

Tip 2

When finding distance, subtract corresponding coordinates: (x₂ − x₁) and (y₂ − y₁). Squaring makes the sign irrelevant.

Tip 3

For rectangle/square problems, use the property that opposite sides are equal and all angles are 90°. Missing corners can be found from shared x or y values.

Common Mistake

Students write (y, x) instead of (x, y). Remember: x always comes first, just like alphabetical order.

Common Mistake

Forgetting to take the square root at the end of the distance formula. AD² ≠ AD.

Marking Scheme

In 5-mark long answers: 1 mark for formula/setup, 2 marks for working steps, 1 mark for correct answer, 1 mark for proper conclusion/statement.

“The coordinate system you are learning today has roots going back thousands of years — from the grid cities of the Sindhu-Sarasvatī Civilisation to Brahmagupta’s formalisation of zero and negative numbers. You are thinking in the same framework that guided astronomers, navigators and mathematicians across centuries!” 🌟

Orienting Yourself: The Use of Coordinates

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