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Short Intro

In this post, students can find complete step-by-step solutions for Class 9 Maths Chapter 1 End-of-Chapter Exercises – Coordinate Geometry. All questions are solved in simple and exam-oriented language based on the latest NCERT syllabus. This chapter helps students understand coordinates, quadrants, plotting points, midpoint concepts, and distance between points in the Cartesian plane.

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Quick Information Box

Item	Details
Board	NCERT / CBSE
Class	9
Subject	Maths
Chapter	Coordinate Geometry
Exercise Type	End-of-Chapter Exercise
Main Topics	Coordinates, Distance, Midpoint

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Concepts Used (Topics Covered)

• Cartesian Plane
• Coordinates of Points
• Quadrants
• Distance Formula
• Midpoint Formula
• Reflection in Axes
• Collinearity of Points
• Coordinate Geometry Applications

The chapter introduces the coordinate plane, quadrants, and methods to locate and measure distances between points.

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Important Formulas

Distance Formula

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$(x2​−x1​)2+(y2​−y1​)2​

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Midpoint Formula

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(2×1​+x2​​,2y1​+y2​​)-10-8-6-4-2246810-2246810A = (-6, -1)B = (6, 7)M = (0, 3)

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Coordinates of Origin

(0, 0)

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Questions & Step-by-step Solutions

Question 1

What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Solution

The x-axis and y-axis intersect at the origin.

Therefore:

Coordinates = (0, 0)

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Question 2

Point W has x-coordinate equal to −5. Predict coordinates of H on line parallel to y-axis.

Solution

If H lies on a line parallel to the y-axis through W, then:

x-coordinate of H = −5

Possible coordinates:

(−5, y)

H may lie in:

• Quadrant II
• Quadrant III

depending on value of y.

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Question 3

Consider points R (3, 0), A (0, −2), M (−5, −2) and P (−5, 2).

(i) Perpendicular sides

Solution

RA ⟂ AM

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(ii) Side parallel to axis

Solution

AM is parallel to x-axis

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(iii) Mirror image points

Solution

A (0, −2) and P (−5, 2)

are reflected across x-axis.

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Question 4

Plot Z (5, −6) and construct triangle.

Solution

Using origin O:

Distance OZ:

$\sqrt{5^2+(-6)^2}=\sqrt{61}$52+(−6)2​=61​

Sides:

Base = 5 units
Height = 6 units
Hypotenuse = √61 units

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Question 5

What if negative numbers did not exist?

Solution

Without negative numbers:

• Only first quadrant would exist.
• We could not locate points left or below origin.
• Complete coordinate geometry would not be possible.

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Question 6

Are M (−3, −4), A (0, 0), and G (6, 8) collinear?

Solution

Distance MA:

$\sqrt{(-3)^2+(-4)^2}=5$(−3)2+(−4)2​=5

Distance AG:

$\sqrt{6^2+8^2}=10$62+82​=10

Distance MG:

$\sqrt{9^2+12^2}=15$92+122​=15

Since:

MA + AG = MG

Therefore:

Points are collinear.

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Question 7

Check whether R (−5, −1), B (−2, −5), C (4, −12) are collinear.

Solution

Using distance formula:

RB = 5
BC = √85
RC = √130

Since:

RB + BC ≠ RC

Therefore:

Points are not collinear.

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Question 8

Plot triangles using origin.

Solution

(i) Right-angled isosceles triangle

Possible vertices:

O(0,0), A(4,0), B(0,4)

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(ii) Isosceles triangle in Quadrant III and IV

Possible vertices:

A(−4,−2), B(4,−2), O(0,0)

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Question 9

Check midpoint cases.

Solution

Midpoint formula:

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(2×1​+x2​​,2y1​+y2​​)-10-8-6-4-2246810-2246810A = (-6, -1)B = (6, 7)M = (0, 3)

Answers

S	M	T	Midpoint?
(−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

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Question 10

Find coordinates of B.

Given:

M(−7,1)
A(3,−4)

Using midpoint formula:

(3 + x)/2 = −7

x = −17

Similarly:

(−4 + y)/2 = 1

y = 6

Therefore:

B = (−17, 6)

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Question 11

Coordinates of trisection points.

Given:

A(4,7), B(16,−2)

Using section division:

P = (8,4)
Q = (12,1)

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Question 12

Circle centered at origin.

Solution

Distance OA:

$\sqrt{1^2+(-8)^2}=\sqrt{65}$12+(−8)2​=65​

Similarly:

OB = √65
OC = √65

Therefore:

All points lie on same circle.

Radius:

√65 units

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Question 13

Find coordinates of A, B and C.

Using midpoint relations:

A = (11,3)
B = (1,−1)
C = (−1,7)

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Question 14

City roads model.

Solution

(i)

Draw coordinate grid using:

1 cm = 200 m

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(ii)

(a)

Only 1 intersection can be (4,3)

(b)

Only 1 intersection can be (3,4)

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Question 15

Computer graphics circles.

Solution

Circle A:

No part lies outside screen.

Circle B:

No part lies outside screen.

Distance between centers:

$\sqrt{(250-100)^2+(230-150)^2}$(250−100)2+(230−150)2​

≈ 170 units

Since:

80 + 100 > 170

Therefore:

Circles intersect.

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Question 16

Is ABCD a square?

Solution

Using distance formula:

All sides are equal.

Diagonals are perpendicular.

Therefore:

ABCD is a square.

Area:

Side² = 10 square units

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Common Mistakes

• Writing coordinates in wrong order
• Forgetting negative signs
• Incorrect midpoint calculations
• Mistakes in distance formula
• Wrong quadrant identification

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Exam Tips

• Learn quadrant signs carefully.
• Practice graph plotting daily.
• Always simplify square roots properly.
• Use graph paper neatly.

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Practice MCQs

MCQ 1

Coordinates of origin are:

A. (1,0)
B. (0,1)
C. (0,0)
D. (1,1)

Answer:

C. (0,0)

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MCQ 2

Point (−3, 5) lies in:

A. Quadrant I
B. Quadrant II
C. Quadrant III
D. Quadrant IV

Answer:

B. Quadrant II

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MCQ 3

Distance between points is found using:

A. Midpoint Formula
B. Distance Formula
C. Area Formula
D. Ratio Formula

Answer:

B. Distance Formula

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FAQ Section

What is the Cartesian plane?

The plane formed by x-axis and y-axis is called Cartesian plane.

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What are coordinates of origin?

(0, 0)

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What is midpoint formula?

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(2×1​+x2​​,2y1​+y2​​)-10-8-6-4-2246810-2246810A = (-6, -1)B = (6, 7)M = (0, 3)

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Why is Coordinate Geometry important?

It is used in mathematics, engineering, maps, graphics, and navigation.

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