🎯 What You Will Learn
After completing these Class 9 Maths Chapter 1 End Exercise Solutions, you will be able to confidently solve coordinate geometry questions asked in CBSE examinations.
📑 Chapter 1 End Exercise
Quick Navigation to All Questions
Learn Before You Solve
Before solving the Class 9 Maths Chapter 1 End Exercise, quickly revise these important concepts and formulas. Spending one minute here will help you solve the questions faster and avoid common mistakes.
📍 Origin
The origin is the point where the x-axis and
y-axis intersect.
Origin = (0, 0)
↔ Parallel Lines
Parallel to x-axis → y = constant
Parallel to y-axis → x = constant
📏 Distance Formula
√[(x₂ − x₁)² + (y₂ − y₁)²]
📌 Midpoint Formula
((x₁ + x₂)/2 , (y₁ + y₂)/2)
🎯 Collinear Points
Three or more points are collinear if they lie on the same straight line.
🪞 Reflection Rules
Across x-axis → (x, −y)
Across y-axis → (−x, y)
💡 Quick Exam Tip
Revise these concepts before attempting the exercise. Most mistakes in coordinate geometry occur because students forget the formulas for distance, midpoint, or the rules for parallel lines.
📝 Class 9 Maths Chapter 1 End Exercise Solutions
Find complete step-by-step NCERT solutions for all questions of the Class 9 Maths Chapter 1 End Exercise. Each solution follows the latest NCERT Ganita Manjari (2026) and CBSE guidelines using a student-friendly notebook writing style.
Q1 Intersection of Two Axes
Given: x-axis and y-axis
To Find: Coordinates of their point of intersection
Solution:
All points on x-axis have y = 0
All points on y-axis have x = 0
At intersection, both conditions are satisfied:
x = 0, y = 0
Q2 Point on Line Parallel to y-axis
Given: Point W has x-coordinate = −5
To Find: Coordinates of point H and possible quadrants
Solution:
A line parallel to the y-axis is vertical, so x-coordinate remains constant.
Therefore, coordinates of H will be:
H = (−5, y)
If y > 0 → Quadrant II
If y < 0 → Quadrant III
Q3 Quadrilateral RAMP
Given: R(3,0), A(0,−2), M(−5,−2), P(−5,2)
To Find:
(i) Perpendicular sides
(ii) Side parallel to axis
(iii) Mirror image points
Solution:
(i) AM has same y-coordinate → horizontal
MP has same x-coordinate → vertical
Therefore, AM ⟂ MP
(ii) AM is parallel to x-axis
(iii) M(−5,−2) and P(−5,2) have same x but opposite y
Hence, they are mirror images about x-axis
(i) AM ⟂ MP
(ii) AM ∥ x-axis
(iii) M and P are mirror images about x-axis
Q4 Right-Angled Triangle IZN
Given: Z(5, −6)
To Find: Lengths of sides of triangle IZN
Solution:
Choose points on axes:
I(5,0) and N(0,−6)
Using distance formula:
IZ = √[(5−5)² + (−6−0)²] = √36 = 6
ZN = √[(5−0)² + (−6+6)²] = √25 = 5
IN = √[(5−0)² + (0+6)²] = √61
Q5 Importance of Negative Numbers
Given: Coordinate system without negative numbers
To Find: Whether all points can be located
Solution:
Without negative numbers, only positive values of x and y exist.
This represents only the first quadrant.
Other quadrants cannot be represented.
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Q6 Collinearity (Distance Method)
Given: M(−3,−4), A(0,0), G(6,8)
To Find: Whether points are collinear
Solution:
Distance MA:
√[(0 + 3)² + (0 + 4)²] = √(9 + 16) = 5
Distance AG:
√[(6 − 0)² + (8 − 0)²] = √(36 + 64) = 10
Distance MG:
√[(6 + 3)² + (8 + 4)²] = √(81 + 144) = 15
MA + AG = 5 + 10 = 15 = MG
Q7 Collinearity Check (Distance Method)
Given: R(−5,−1), B(−2,−5), C(4,−12)
To Find: Whether points are collinear
Solution:
RB:
√[(−2 + 5)² + (−5 + 1)²] = √(9 + 16) = 5
BC:
√[(4 + 2)² + (−12 + 5)²] = √(36 + 49) = √85
RC:
√[(4 + 5)² + (−12 + 1)²] = √(81 + 121) = √202
RB + BC ≠ RC
Q8 Triangles
To Find: Coordinates of required triangles
Solution:
(i) Right-angled isosceles triangle:
O(0,0), A(2,0), B(0,2)
(ii) Isosceles triangle:
A(0,0), B(−2,−2), C(2,−2)
Q9 Midpoint Verification
To Find: Whether M is midpoint of ST
Solution:
| S | M | T | Result |
|---|---|---|---|
| (−3,0) | (0,0) | (3,0) | Yes |
| (2,3) | (3,4) | (4,5) | Yes |
| (0,0) | (0,5) | (0,−10) | No |
| (−8,7) | (0,−2) | (6,−3) | No |
Q10 Find Coordinates of B (Using Midpoint Formula)
Given: M(−7,1), A(3,−4), B(x,y)
Step 1: Use midpoint formula
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
Step 2: Substitute values
−7 = (3 + x)/2
1 = (−4 + y)/2
Step 3: Solve
Multiply both sides by 2:
−14 = 3 + x → x = −17
2 = −4 + y → y = 6
Q11 Trisection of Line Segment
Given: A(4,7), B(16,−2)
Diagram: Points P and Q divide AB into 3 equal parts
Step 1: Find Total Change in Coordinates
Change in x-coordinate = 16 − 4 = 12
Change in y-coordinate = −2 − 7 = −9
Step 2: Divide Changes into 3 Equal Parts
One-third change in x-coordinate = 12/3 = 4
One-third change in y-coordinate = −9/3 = −3
Step 3: Find Coordinates of Point P
Point P is one-third distance from A toward B.
P = (4 + 4 , 7 + (−3))
P = (8 , 4)
Step 4: Find Coordinates of Point Q
Point Q is two-third distance from A toward B.
Q = (4 + 8 , 7 + (−6))
Q = (12 , 1)
Verification by Distance Formula
AP = √[(8−4)2 + (4−7)2]
= √[42 + (−3)2]
= √(16 + 9)
= √25 = 5
PQ = √[(12−8)2 + (1−4)2]
= √[42 + (−3)2]
= √25 = 5
QB = √[(16−12)2 + (−2−1)2]
= √[42 + (−3)2]
= √25 = 5
P = (8, 4)
Q = (12, 1)
Q11 Trisection of Line Segment Using Section Formula
Given: A(4,7), B(16,−2)
Diagram: Points P and Q trisect the line segment AB
Concept Used:
Point P divides AB internally in the ratio 1 : 2
Point Q divides AB internally in the ratio 2 : 1
Section Formula:
(x , y) = ( mx2 + nx1 m + n , my2 + ny1 m + n )
Step 1: Find Coordinates of Point P
Point P divides AB internally in the ratio 1 : 2
P = ( (1 × 16) + (2 × 4) 1 + 2 , (1 × (−2)) + (2 × 7) 1 + 2 )
P = ( 16 + 8 3 , −2 + 14 3 )
P = ( 24 3 , 12 3 )
P = (8 , 4)
Step 2: Find Coordinates of Point Q
Point Q divides AB internally in the ratio 2 : 1
Q = ( (2 × 16) + (1 × 4) 2 + 1 , (2 × (−2)) + (1 × 7) 2 + 1 )
Q = ( 32 + 4 3 , −4 + 7 3 )
Q = ( 36 3 , 3 3 )
Q = (12 , 1)
P = (8, 4)
Q = (12, 1)
Q12 Circle Verification Using Distance Formula
Given: A(1,−8), B(−4,7), C(−7,−4), Centre O(0,0)
Diagram: Points A, B, C lie on circle with centre O
Step 1: Distance from origin
OA = √(1² + (−8)²) = √65
OB = √((−4)² + 7²) = √65
OC = √((−7)² + (−4)²) = √65
👉 All distances equal → points lie on same circle
Radius = √65
(ii) Check D and E
OD = √61 < √65 → Inside
OE = 9 > √65 → Outside
Q13 Find Coordinates of Triangle (Step-by-Step)
Given: D(5,1), E(6,5), F(0,3)
To Find: Coordinates of triangle ABC
Step 1: Let
A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)
Using midpoint formula:
D = midpoint of BC
(x₂ + x₃)/2 = 5 → x₂ + x₃ = 10 …(1)
(y₂ + y₃)/2 = 1 → y₂ + y₃ = 2 …(2)
E = midpoint of CA
(x₃ + x₁)/2 = 6 → x₃ + x₁ = 12 …(3)
(y₃ + y₁)/2 = 5 → y₃ + y₁ = 10 …(4)
F = midpoint of AB
(x₁ + x₂)/2 = 0 → x₁ + x₂ = 0 …(5)
(y₁ + y₂)/2 = 3 → y₁ + y₂ = 6 …(6)
Step 2: Solve x-equations
From (5): x₁ + x₂ = 0 → x₁ = −x₂
Substitute in (3):
x₃ + (−x₂) = 12 → x₃ − x₂ = 12 …(7)
Now from (1): x₂ + x₃ = 10 …(1)
Add (1) and (7):
(x₂ + x₃) + (x₃ − x₂) = 10 + 12
2x₃ = 22 → x₃ = 11
Put x₃ = 11 in (1):
x₂ + 11 = 10 → x₂ = −1
Then x₁ = −x₂ = 1
Step 3: Solve y-equations
From (6): y₁ + y₂ = 6 …(6)
From (2): y₂ + y₃ = 2 …(2)
From (4): y₃ + y₁ = 10 …(4)
Add (6) and (2):
(y₁ + y₂) + (y₂ + y₃) = 6 + 2
y₁ + 2y₂ + y₃ = 8 …(8)
Now subtract (4) from (8):
(y₁ + 2y₂ + y₃) − (y₃ + y₁) = 8 − 10
2y₂ = −2 → y₂ = −1
Put y₂ = −1 in (6):
y₁ − 1 = 6 → y₁ = 7
Put y₁ = 7 in (4):
y₃ + 7 = 10 → y₃ = 3
A(1,7), B(−1,−1), C(11,3)
Q14 City Intersection
Each coordinate (x,y) represents intersection of streets
(4,3) → unique intersection → 1
(3,4) → unique intersection → 1
Q15 Circles (Distance Method)
Given:
A(100,150), r₁=80
B(250,230), r₂=100
Step 1: Distance between centres
d = √[(250−100)² + (230−150)²]
= √(150² + 80²)
= √(22500 + 6400)
= √28900 = 170
Step 2: Compare radii
- r₁ + r₂ = 180
- |r₁ − r₂| = 20
👉 Since 20 < 170 < 180 → circles intersect
Screen check:
- Screen size: 800×600
- Circle A: fully inside
- Circle B: fully inside
Q16 Square Verification (Distance Method)
Given: A(2,1), B(−1,2), C(−2,−1), D(1,−2)
Step 1: Find all 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 sides equal
Step 2: Check diagonals
AC = √[(−2−2)² + (−1−1)²] = √(16 + 4) = √20
BD = √[(1+1)² + (−2−2)²] = √(4 + 16) = √20
👉 Diagonals equal
Area:
Side² = (√10)² = 10 sq units
📚 Exercise Quick Revision Sheet
These Class 9 Maths Chapter 1 End Exercise Solutions are based on the latest NCERT Ganita Manjari (2026) textbook and follow the CBSE exam pattern with step-by-step solutions. Before your test, quickly revise these important concepts from Class 9 Maths Chapter 1 End Exercise Solutions to improve your speed and accuracy.
📍 Origin
Origin is the point (0, 0), where the x-axis and y-axis intersect.
📏 Distance Formula
√[(x₂ − x₁)² + (y₂ − y₁)²]
📌 Midpoint Formula
((x₁+x₂)/2 , (y₁+y₂)/2)
↔ Parallel Lines
Parallel to x-axis → y = constant
Parallel to y-axis → x = constant
🎯 Collinear Points
If the sum of two smaller distances equals the third distance, the points are collinear.
📝 CBSE Exam Tip
Write every calculation step clearly. Always mention the formula before substituting values to score full marks.
🧠 Memory Trick
D-M-C Rule
D → Distance Formula
M → Midpoint Formula
C → Collinear Points
Revise these three concepts before every exam. Most questions in
Class 9 Maths Chapter 1 End Exercise are based on them.
✅ You have successfully completed the Class 9 Maths Chapter 1 End Exercise Solutions. Revise this sheet once before your exam to strengthen your understanding of Coordinate Geometry and improve your performance in the CBSE Class 9 Mathematics Examination.
❓ Frequently Asked Questions (FAQs)
Find answers to the most common questions about Class 9 Maths Chapter 1 End Exercise Solutions, coordinate geometry concepts, and CBSE exam preparation.
Q1. How do I solve Class 9 Maths Chapter 1 End Exercise questions easily?
First revise the important concepts such as the Cartesian plane, distance formula, midpoint formula, and collinear points. Then solve each question step by step without skipping calculations.
Q2. What is the distance formula in coordinate geometry?
The distance between two points (x₁, y₁) and (x₂, y₂) is:
√[(x₂ − x₁)² + (y₂ − y₁)²]
Q3. How can I check whether three points are collinear?
You can use the distance formula. If the sum of the two smaller distances is equal to the largest distance, the three points are collinear.
Q4. Are these Chapter 1 End Exercise solutions based on the latest NCERT book?
Yes. These solutions follow the latest NCERT Ganita Manjari (2026) textbook and are prepared according to the latest CBSE guidelines.
Q5. Is the distance method better than the slope method in CBSE exams?
For Class 9, the distance method is simple, reliable, and commonly used in CBSE examinations. Follow the method suggested in your NCERT textbook unless the question specifically asks otherwise.
Q6. Why should I practise the Chapter 1 End Exercise before exams?
The End Exercise covers all the important concepts of the chapter in one place. Practising every question helps strengthen conceptual understanding and improves confidence for school tests and CBSE examinations.
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📘 Download Class 9 Maths NCERT Book (Ganita Manjari 2026)
For better understanding of concepts, students should always refer to the original NCERT textbook. You can download the official Class 9 Maths book directly from the NCERT website.
This will help you practice questions exactly as per CBSE exam pattern and improve conceptual clarity.
📥 Download NCERT Class 9 Maths BookSource: Official NCERT Website
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These Class 9 Maths Chapter 1 End Exercise Solutions are carefully prepared according to the latest NCERT Ganita Manjari (2026) and CBSE Guidelines. Every solution is explained in a simple, step-by-step notebook style so that students can easily understand the concepts, improve problem-solving skills, and write answers confidently in the examination.
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