Important Questions for Class 9 Maths (Ganita Manjari I) Chapter 1 Orienting Yourself: The Use of Coordinates - 2026-27

Students can use the Class 9 Maths Chapter 1 important questions PDF to revise coordinate geometry concepts and practise different question formats. The downloadable resource is intended to support preparation for class tests, periodic assessments and annual examinations. It also helps students revisit formulas, practise graph-based problems and strengthen their understanding of coordinates before exams.

Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates MCQ

Q1. Which of the following points lies in the second quadrant of the Cartesian plane?

(a) (4, 7) 

(b) (−6, 3) 

(c) (−5, −2) 

(d) (8, −4)

Answer: (b) (−6, 3)

Explanation: In the second quadrant, the x-coordinate is negative and the y-coordinate is positive. Therefore, (−6, 3) lies in Quadrant II.

Q2. A point is located 7 units to the left of the origin and 4 units below the x-axis. What are its coordinates?

(a) (7, 4) 

(b) (−7, 4) 

(c) (−7, −4) 

(d) (7, −4)

Answer: (c) (−7, −4)

Explanation: Moving left gives a negative x-coordinate, and moving downward gives a negative y-coordinate.

Q3. Which statement is correct for every point on the y-axis?

(a) Its y-coordinate is zero. 

(b) Its x-coordinate is zero. 

(c) Both coordinates are positive. 

(d) Both coordinates are negative.

Answer: (b) Its x-coordinate is zero.

Explanation: Every point on the y-axis has coordinates of the form (0, y).

Q4. What is the distance between the points A(−3, 5) and B(4, 5)?

(a) 1 unit 

(b) 7 units 

(c) 8 units 

(d) 9 units

Answer: (b) 7 units

Explanation:

Both points have the same y-coordinate, so:

$$AB = |4 - (-3)| = 7 \text{ units}$$

Q5. What happens to the coordinates of a point (x, y) when it is reflected in the y-axis?

(a) (x, −y) 

(b) (−x, y) 

(c) (−x, −y) 

(d) (y, x)

Answer: (b) (−x, y)

Explanation: Reflection in the y-axis changes the sign of the x-coordinate while keeping the y-coordinate unchanged.

Q6. The distance between P(1, 2) and Q(4, 6) is:

(a) 4 units 

(b) 5 units 

(c) 6 units 

(d) 7 units

Answer: (b) 5 units

Explanation:

Using the distance formula,

$$PQ = \sqrt{(4-1)^2 + (6-2)^2}$$

$$= \sqrt{9+16} = \sqrt{25} = 5 \text{ units}$$

Q7. Which of the following pairs of points are mirror images of each other in the x-axis?

(a) (3, 5) and (−3, 5) 

(b) (4, −6) and (4, 6) 

(c) (−2, 3) and (2, −3) 

(d) (5, 7) and (7, 5)

Answer: (b) (4, −6) and (4, 6)

Explanation: Points reflected in the x-axis have the same x-coordinate and opposite y-coordinates.

Class 9 Maths Chapter 1 Very Short Answer Questions

Q8. What is the Cartesian plane? Name its two coordinate axes.

Answer: 

The Cartesian plane is a two-dimensional plane used to locate points using ordered pairs of numbers. Its two perpendicular axes are the horizontal x-axis and the vertical y-axis.

Q9. Write the coordinates of a point situated 9 units below the origin on the y-axis.

Answer: 

(0, −9)

The x-coordinate is zero because the point lies on the y-axis, and the y-coordinate is negative because it lies below the origin.

Q10. State the signs of the x-coordinate and y-coordinate in the fourth quadrant.

Answer: 

In the fourth quadrant, the x-coordinate is positive and the y-coordinate is negative.

The sign pattern is (+, −).

Q11. What are the abscissa and ordinate of the point P(−8, 6)?

Answer:

• Abscissa (x-coordinate) = −8

• Ordinate (y-coordinate) = 6

Q12. A point P has coordinates (5, −3). What are its perpendicular distances from the x-axis and y-axis?

Answer:

Distance from the x-axis = 3 units.

Distance from the y-axis = 5 units.

Distances are always non-negative, even when coordinate values are negative.

Q13. Under what condition do the ordered pairs (a, b) and (b, a) represent the same point?

Answer: 

The ordered pairs represent the same point when a=b.

For example, (4, 4) and (4, 4) represent the same point.

Class 9 Maths Chapter 1 Short Answer Questions

Q14. Identify the positions of the following points in the Cartesian plane.

A(7, 2), B(−4, 6), C(−3, −5), D(5, −7), E(0, 8) and F(−6, 0).

Answer:

Explanation: The signs of the coordinates determine the quadrant. Points with one coordinate equal to zero lie on an axis, not inside any quadrant.

Q15. Find the distance between the points A(−5, 3) and B(2, 3). Explain why the distance formula can be simplified.

Answer:

Given:

A(−5, 3) and B(2, 3).

Since both points have the same y-coordinate, the line joining them is parallel to the x-axis.

Therefore,

$$AB = |x_2 - x_1|$$

$$= |2 - (-5)| = 7 \text{ units}$$

Final Answer: 7 units.

Q16. A point P(−4, 7) is reflected first in the x-axis and then in the y-axis. Find its coordinates after each reflection.

Answer:

Original point: P(−4, 7).

Step 1: Reflection in the x-axis

The x-coordinate remains unchanged, and the y-coordinate changes sign.

$$P' = (-4, -7)$$

Step 2: Reflection in the y-axis

The x-coordinate changes sign while the y-coordinate remains unchanged.

$$P'' = (4, -7)$$

Final Answer: After the first reflection: (−4, −7). After the second reflection: (4, −7).

Q17. Find the distance between A(−2, −3) and B(4, 5) using the Baudhāyana–Pythagoras theorem.

Answer:

Given:

A(−2, −3), B(4, 5).

The horizontal distance is:

4 - (-2) = 6

The vertical distance is:

5 - (-3) = 8

Using the Baudhāyana–Pythagoras theorem:

$$AB = \sqrt{6^2 + 8^2}$$

$$= \sqrt{36 + 64} = \sqrt{100}$$

Final Answer: AB = 10 units.

Q18. Three vertices of a rectangle are A(2, 3), B(8, 3) and C(8, 7), taken in consecutive order. Find the fourth vertex and the rectangle's perimeter.

Answer:

The points A and B have the same y-coordinate, while B and C have the same x-coordinate.

Therefore, the fourth vertex is: 

D = (2, 7)

Length:

AB = 8 - 2 = 6 units

Breadth:

BC = 7 - 3 = 4 units

Perimeter:

P = 2(l+b)

=2(6+4) = 20 units 

Final Answer: Fourth vertex D(2, 7); perimeter = 20 units.

Q19. Explain why the points P(3, 4) and Q(4, 3) are different. Calculate the distance between them.

Answer:

In an ordered pair, the first number represents the x-coordinate and the second represents the y-coordinate.

For P(3, 4), x = 3 and y = 4.

For Q(4, 3), x = 4 and y = 3.

Thus, the points represent different positions.

Their distance is:

$$PQ = \sqrt{(4-3)^2 + (3-4)^2}$$

$$= \sqrt{1+1} = \sqrt{2}$$

Class 9 Maths Chapter 1 Important Long Answer Questions

Q20. Four points A(2, 2), B(8, 2), C(8, 7) and D(2, 7) represent the corners of a rectangular garden. Find the lengths of all four sides, its perimeter and its area.

Answer:

$$\begin{aligned} &\textbf{Step 1: Find the horizontal sides.}\\[6pt] &AB = |8 - 2| = 6 \text{ units}\\[6pt] &CD = |8 - 2| = 6 \text{ units}\\[12pt] &\textbf{Step 2: Find the vertical sides.}\\[6pt] &BC = |7 - 2| = 5 \text{ units}\\[6pt] &DA = |7 - 2| = 5 \text{ units}\\[12pt] &\textbf{Step 3: Calculate perimeter.}\\[6pt] &P = 2(6 + 5) = 22 \text{ units}\\[12pt] &\textbf{Step 4: Calculate area.}\\[6pt] &A = l \times b = 6 \times 5\\[6pt] &A = 30 \text{ square units} \end{aligned}$$

Final Answer:

• AB = CD = 6 units

• BC = DA = 5 units

• Perimeter = 22 units

• Area = 30 square units

Q21. Three vertices of a triangle are A(1, 1), B(7, 1) and C(4, 5). Find the lengths of all three sides. Determine whether the triangle is isosceles.

Answer:

$$\begin{aligned} &\text{Using the distance formula:}\\[6pt] &d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\[12pt] &\textbf{Step 1: Calculate AB.}\\[6pt] &AB = \sqrt{(7-1)^2+(1-1)^2}=6\\[12pt] &\textbf{Step 2: Calculate BC.}\\[6pt] &BC = \sqrt{(4-7)^2+(5-1)^2}\\[6pt] &= \sqrt{9+16}=5\\[12pt] &\textbf{Step 3: Calculate CA.}\\[6pt] &CA = \sqrt{(1-4)^2+(1-5)^2}\\[6pt] &= \sqrt{9+16}=5 \end{aligned}$$

Since BC = CA, the triangle has two equal sides.

Final Answer: AB = 6 units, BC = 5 units and CA = 5 units. Therefore, triangle ABC is an isosceles triangle.

Q22. A triangular figure has vertices P(2, 3), Q(6, 3) and R(2, 6). Reflect it in the y-axis. Find the new coordinates and verify that the side lengths remain unchanged.

Answer:

$$\begin{aligned} &\textbf{Step 1: Find reflected coordinates.}\\[6pt] &\text{For reflection in the y-axis:}\\[6pt] &(x,y)\rightarrow(-x,y)\\[12pt] &\text{Therefore:}\\[6pt] &P'=(-2,3)\\[6pt] &Q'=(-6,3)\\[6pt] &R'=(-2,6)\\[12pt] &\textbf{Step 2: Calculate the original side lengths.}\\[6pt] &PQ=|6-2|=4\\[6pt] &PR=|6-3|=3\\[6pt] &QR=\sqrt{(2-6)^2+(6-3)^2}\\[6pt] &=\sqrt{16+9}=5\\[12pt] &\textbf{Step 3: Calculate the reflected side lengths.}\\[6pt] &P'Q'=|-6-(-2)|=4\\[6pt] &P'R'=|6-3|=3\\[6pt] &Q'R'=\sqrt{(-2+6)^2+(6-3)^2}\\[6pt] &=\sqrt{16+9}=5 \end{aligned}$$

Final Answer: The reflected triangle has vertices P′(−2, 3), Q′(−6, 3) and R′(−2, 6). Both triangles have side lengths 3, 4 and 5 units.

Hence, reflection in the y-axis preserves distances.

Q23. A student starts at point O(0, 0), walks 6 units east, then 8 units north, and finally 3 units west. Find the student's final position, the total distance travelled and the shortest distance from the starting point.

Answer:

$$\begin{aligned} &\textbf{Step 1: Find the final coordinates.}\\[6pt] &\text{Starting point:}\\[6pt] &O=(0,0)\\[12pt] &\text{After moving 6 units east:}\\[6pt] &A=(6,0)\\[12pt] &\text{After moving 8 units north:}\\[6pt] &B=(6,8)\\[12pt] &\text{After moving 3 units west:}\\[6pt] &C=(3,8)\\[12pt] &\textbf{Step 2: Calculate the total distance travelled.}\\[6pt] &6+8+3=17\text{ units}\\[12pt] &\textbf{Step 3: Calculate the shortest distance from O to C.}\\[6pt] &OC=\sqrt{(3-0)^2+(8-0)^2}\\[6pt] &=\sqrt{9+64}=\sqrt{73}\\[6pt] &OC\approx 8.54\text{ units}\\[12pt] &\textbf{Final Answer:}\\[6pt] &\bullet\ \text{Final position: }(3,8)\\[6pt] &\bullet\ \text{Total distance travelled: }17\text{ units}\\[6pt] &\bullet\ \text{Shortest distance from the origin: }\sqrt{73}\text{ units,}\\ &\qquad\text{or approximately }8.54\text{ units.} \end{aligned}$$

Final Answer:

• Final position: (3, 8)

• Total distance travelled: 17 units

• Shortest distance from the origin: 73 units, or approximately 8.54 units.

Q24. A school plans a rectangular playground on a coordinate grid with vertices A(−4, −2), B(6, −2), C(6, 4) and D(−4, 4). A water station is located at P(1, 1). Find the playground's dimensions, its area, the distance from P to each corner and explain the significance of P.

Answer:

$$\begin{aligned} &\textbf{Step 1: Calculate playground dimensions.}\\[6pt] &\text{Horizontal length:}\\[6pt] &AB=|6-(-4)|=10\text{ units}\\[12pt] &\text{Vertical breadth:}\\[6pt] &BC=|4-(-2)|=6\text{ units}\\[12pt] &\textbf{Step 2: Calculate the area.}\\[6pt] &A=10\times6=60\text{ square units}\\[12pt] &\textbf{Step 3: Calculate the distance from P to A.}\\[6pt] &PA=\sqrt{(-4-1)^2+(-2-1)^2}\\[6pt] &=\sqrt{25+9}=\sqrt{34}\\[12pt] &\textbf{Step 4: Calculate the distance from P to B.}\\[6pt] &PB=\sqrt{(6-1)^2+(-2-1)^2}\\[6pt] &=\sqrt{25+9}=\sqrt{34}\\[12pt] &\textbf{Step 5: Calculate the distance from P to C.}\\[6pt] &PC=\sqrt{(6-1)^2+(4-1)^2}\\[6pt] &=\sqrt{25+9}=\sqrt{34}\\[12pt] &\textbf{Step 6: Calculate the distance from P to D.}\\[6pt] &PD=\sqrt{(-4-1)^2+(4-1)^2}\\[6pt] &=\sqrt{25+9}=\sqrt{34}\\[12pt] &\text{Thus,}\\[6pt] &PA=PB=PC=PD=\sqrt{34}\\[12pt] &\textbf{Step 7: Explain the position of P.}\\[6pt] &\text{The horizontal midpoint of the playground is:}\\[6pt] &\dfrac{-4+6}{2}=1\\[12pt] &\text{The vertical midpoint is:}\\[6pt] &\dfrac{-2+4}{2}=1\\[12pt] &\text{Therefore, P(1,1) is the centre of the rectangular playground.} \end{aligned}$$

Final Answer:

• Length = 10 units

• Breadth = 6 units

• Area = 60 square units

• Distance from P to each corner = $$\sqrt{34}$$ units, approximately 5.83 units

• P is the centre of the rectangle and is equally distant from all four corners.

Important Concepts and Formulas for Class 9 Maths Chapter 1 Exam Preparation

Before attempting Chapter 1 maths Class 9 Important Questions, students should understand how coordinates represent the position of points and how formulas are used to solve geometrical problems.

How Vedantu Helps Students Prepare for Class 9 Maths Chapter 1 Important Questions

Vedantu's Class 9 Maths Chapter 1 important questions provide students with a focused way to practise coordinates and their applications. The questions are organised by answer type, allowing students to understand basic definitions before moving on to calculations and reasoning-based problems.

• Concept-Based Practice: Understand coordinates, quadrants, axes and the position of points through focused questions.

• Step-by-Step Learning: Use explained solutions to understand the mathematical steps involved in distance and midpoint calculations.

• Different Question Formats: Practise MCQs, very short, short and long answer questions for different school assessment formats.

• Application-Based Preparation: Strengthen problem-solving skills through real-life situations involving coordinate grids and spatial reasoning.

• Quick Concept Review: Revisit important concepts and formulas using chapter-specific study materials.

• Self-Assessment: Identify mistakes in plotting coordinates, using negative signs and applying formulas before examinations.

Regular practice with Vedantu's maths class 9 chapter 1 important questions can help students develop a stronger understanding of coordinate geometry and improve their speed and accuracy.

Important Study Materials for Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates

CBSE Class 9 Maths Chapter-wise Important Questions

CBSE Class 9 Maths Chapter-wise Important Questions and Answers cover key concepts from the Ganita Manjari I textbook for the 2026-27 session. 

These chapter-wise resources help students strengthen their mathematical understanding, practise different question types and organise their revision effectively.

Other Important Related Links for CBSE Class 9 Maths