Class 9 Maths Chapter 1 End Exercise Solutions (NCERT 2026):
- Origin = (0, 0)
- Parallel to y-axis → x = constant
- Collinear points → equal slopes or distance check
- Distance formula → √[(x₂−x₁)² + (y₂−y₁)²]
- Midpoint → ((x₁+x₂)/2, (y₁+y₂)/2)
These Class 9 Maths Chapter 1 End Exercise Solutions are designed to help students understand coordinate geometry concepts clearly. Each question is solved step-by-step according to CBSE guidelines and the latest NCERT Ganita Manjari 2026 book. See the diagram below for better understanding of coordinate geometry concepts
👉 Check complete chapter: Class 9 Maths Chapter 1 Full Solutions
📌 Chapter 1 End Exercise Solution Content
- Q1 Intersection of Axes
- Q2 Parallel Line
- Q3 Points RAMP
- Q4 Triangle
- Q5 Negative Numbers
- Q6 Collinearity
- Q7 Collinearity Check
- Q8 Triangles
- Q9 Midpoint
- Q10 Find Coordinates of B
- Q11 Trisection of Line
- Q12 Circle Verification
- Q13 Triangle Coordinates
- Q14 City Mapping
- Q15 Circles Position
- Q16 Square Verification
Class 9 Maths Chapter 1 End Exercise Solutions – Step-by-Step Explanation
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.
🎯 Need help in Maths? Join Class 9 Maths Coaching in Ankur Vihar for step-by-step guidance and test practice.
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: Midpoint M
M = ((4 + 16)/2 , (7 + (−2))/2)
M = (20/2 , 5/2)
M = (10 , 2.5)
Step 2: Point P (between A and M)
P = ((4 + 10)/2 , (7 + 2.5)/2)
P = (14/2 , 9.5/2)
P = (7 , 4.75)
Step 3: Point Q (between M and B)
Q = ((10 + 16)/2 , (2.5 + (−2))/2)
Q = (26/2 , 0.5/2)
Q = (13 , 0.25)
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)
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
(y₂ + y₃)/2 = 1 → y₂ + y₃ = 2
E = midpoint of CA → x₃ + x₁ = 12
y₃ + y₁ = 10
F = midpoint of AB → x₁ + x₂ = 0
y₁ + y₂ = 6
Step 2: Solve equations
From x₁ + x₂ = 0 → x₁ = −x₂
Substitute in x₃ + x₁ = 12 → x₃ − x₂ = 12
Using x₂ + x₃ = 10 → solve → x₂ = 0, x₃ = 10
So x₁ = −0 = 0
Similarly for y-values:
y₁ + y₂ = 6
y₂ + y₃ = 2
y₃ + y₁ = 10
Solving → y₁ = −3, y₂ = 5, y₃ = 7
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
Also solve previous exercises: Exercise 1.1 Solutions and Exercise 1.2 Solutions.
Need guidance? Join our Class 9 Maths Coaching in Ankur Vihar.
❓ Frequently Asked Questions
Q1. How to check if points are collinear?
Use distance formula or slope method. If sum of two distances equals third, points are collinear.
Q2. What is the distance formula?
Distance = √[(x₂−x₁)² + (y₂−y₁)²]
Q3. Why are coordinates important?
They help represent real-life positions and are important for higher classes.
Q4. Which method is best for exams?
Distance method is most reliable and preferred in CBSE exams.
📘 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
🚀 Improve Your Maths with Expert Guidance
Join Maths Gurukulam – Ankur Vihar
✔ 18+ Years Experience | ✔ CBSE Experts | ✔ Small Batches
📞 Call: 8447002272