Orienting Yourself: The Use of Coordinates Class 9 Solutions Part 1 Chapter 1

Chapter 1 of the new Class 9 Maths book is called Orienting Yourself: The Use of Coordinates. It is the opening chapter of Ganita Manjari Part I, the NCERT book for session 2026-27.

This chapter teaches you how to describe the exact position of a point using two numbers. You learn the Cartesian plane, how to plot and read points, how to find the distance between two points, and how to find the midpoint of a line segment.

On this page you get the topic list, the formulas, solved examples and free step-by-step solutions for both Exercise Sets and the End of Chapter Exercises.

Chapter 1 Video Lesson – Extramarks Shaurya Series

Prefer to watch instead of read? In this Shaurya Series session, Swati Ma’am explains Orienting Yourself: The Use of Coordinates from the very beginning, in simple language.

Ganita Manjari Class 9 Chapter 1 Solutions Orienting Yourself The Use of Coordinates

Ex 1.1 Class 9 Ganita Manjari Solutions (Page 4 – 5)

1: Figure shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.

Referring to Figure, answer the following questions:

(i) If D₁ R₁ represents the door to Reiaan’s room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

Answer: 

From the figure, the door

$D_1R_1$

lies between

$x=8$

and

$x=9$

, and on the x-axis (

$y=0$

)

Therefore:

Distance from the left wall (y-axis): 8 units

Distance from the x-axis: 0 units

The door is 8 units from the left wall and 0 units from the x-axis.

(ii)  What are the coordinates of D₁?

Answer: 

From the figure,

$D_1$

is located at

$x=8$

on the x-axis, so its coordinates are:

D1​=(8,0)

​(iii)  If R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

Answer: 

Given:

$D_1=(8,0)$ $R_1=(11.5,\mathrm{0)}$

The width of the door is the distance between these two points:

$$11.5-8=\boxed{3.5\text{ units}}$$

So, the door is 3.5 units wide.

Yes, this is a comfortable width for a room door. A person using a wheelchair should also be able to enter easily, as the doorway provides sufficient space for wheelchair access.

(iv)  If B₁ (0, 1.5) and B₂ (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Answer: 

Given:$B_2=(0,4)$

B₁ = (0, 1.5)
B₂ = (0, 4)

Bathroom door width:

$$4-1.5=2.5\text{ units}$$

Room door width =

$3.5$

units.

Since

$2.5<3.5$

,

The bathroom door is narrower than the room door.

Ex 1.2 Class 9 Ganita Manjari Solutions (Page 4 – 5)

1. On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (– 7, 0) to (13, 0) on the x-axis and from (0, – 15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.)

1. Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9), and (11, 7).
(i)  Where will the fourth foot of the table be?

Answer:

The given three points form three corners of a rectangle:
A = (8, 9)
B = (11, 9)
C = (11, 7)
To complete the rectangle, the fourth point must have:
– same x-coordinate as A → 8
– same y-coordinate as C → 7
Therefore, the fourth foot is at: (8, 7)

(ii)  Is this a good spot for the table?

Answer:

Yes. The table lies inside Reiaan’s room and does not overlap the bed or wardrobe, so it is a suitable spot.

(iii)  What is the width of the table? The length? Can you make out the height of the table?

Answer:

• Horizontal distance:

   

  $$11-8=3\text{ units}$$ 

• Vertical distance:

   

  $$9-7=2\text{ units}$$ 

Therefore, the table is 3 units long and 2 units wide.

$$\boxed{\text{Length}=3\text{ units},\quad \text{Width}=2\text{ units}}$$

The height cannot be determined from this graph because the graph shows only the table's position and dimensions on the floor (a 2D view), not its vertical height.

2. If the bathroom door has a hinge at B₁ and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Answer:

The bathroom door has its hinge at

$B_1=(0,1.5)$

and extends to

$B_2=(0,4)$

. So its length is:

$$4-1.5=2.5\text{ units}$$

When the door opens into the bedroom, it swings around

$B_1$

. The wardrobe starts at

$x=3$

.

Since the door is only 2.5 units long, its outer edge reaches at most

$2.5$

units from

$B_1$

, while the wardrobe begins 3 units away.

Answer:

No, the door will not hit the wardrobe. There is a small gap between the swinging door and the wardrobe.

If the door is made wider, its swinging radius will increase. If it becomes wide enough to reach the wardrobe, it could hit it. Therefore, a suitable change would be to move the wardrobe farther to the right (or redesign the door to open outward) if a wider door is required.

4. Other rooms in the house:
(i)  Reiaan’s room door leads from the dining room which has the length 18 ft and width 15 ft. The dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

Answer:

Coordinates of corners:
P = (-6, 0)
A = (12, 0)
B = (12, -15)
C = (-6, -15)

(ii)  Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Answer:

To place a rectangular

$5\text{ ft} \times 3\text{ ft}$

dining table exactly at the centre of the dining room:

1) Coordinates of the dining room

The dining room corners are:

$$(-6,0),\ (12,0),\ (12,-15),\ (-6,-15)$$

2) Centre of the dining room

The centre is the midpoint of the room:

$$\left(\frac{-6+12}{2},\frac{0+(-15)}{2}\right)=\left(3,-7.5\right)$$

3) Place the table at the centre

Assume the 5 ft side is parallel to the x-axis and the 3 ft side is parallel to the y-axis.

So from the centre

$(3,-7.5)$

:

• Half of 5 ft =
  $2.5$ 
• Half of 3 ft =
  $1.5$ 

Hence the four feet (corners) of the table are:

$$\boxed{(0.5,-6),\ (5.5,-6),\ (5.5,-9),\ (0.5,-9)}$$

Answer:

$$\boxed{(0.5,-6),\ (5.5,-6),\ (5.5,-9),\ (0.5,-9)}$$

If you want, I can also draw the table placement on the coordinate diagram.

End of Chapter Exercise (Page 12 – 14)

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

Answer:

The point where the x-axis and y-axis intersect is called the origin.

Its coordinates are:

$$\boxed{(0,0)}$$

So, the x-coordinate = 0 and the y-coordinate = 0

2. Point W has x-coordinate equal to – 5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Answer:

$H$

lies in Quadrant II.

• If
  $y<0$ 

  ,

$H$

lies in Quadrant III.

• If
  $y=0$ 

  ,

$H$

lies on the negative x-axis.

Therefore,

$H$

can lie in:

$$\boxed{\text{Quadrant II or Quadrant III}}$$

$$\boxed{AM\text{ is parallel to the x-axis}}$$

(Also,

$MP$

is parallel to the y-axis.)

(iii)  Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

4. Plot point Z (5, – 6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
(Comment: Answers may differ from person to person.)

Answer:

One possible construction is to choose

$$Z(5,-6),\qquad I(5,0),\qquad N(0,-6).$$

Here,

$IZ$

is vertical and

$ZN$

is horizontal, so they are perpendicular. Therefore,

$\triangle IZN$

is right-angled at

$Z$

.

The side lengths are:

$$IZ=|0-(-6)|=\boxed{6\text{ units}}$$

$$ZN=|5-0|=\boxed{5\text{ units}}$$

Using the Pythagorean theorem,

$$IN=\sqrt{IZ^2+ZN^2} =\sqrt{6^2+5^2} =\sqrt{61}$$

$$\boxed{IN=\sqrt{61}\text{ units}\approx7.81\text{ units}}$$

IZ = 6 units, NZ = 5 units, IN =

$\sqrt{61}$

​units​​​

5. What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Answer:

If we did not have negative numbers, the coordinates of points could only be zero or positive.

So we could locate points only where

$$x \ge 0 \quad \text{and} \quad y \ge 0$$

This would cover only the first quadrant and the positive parts of the two axes.

Therefore, such a coordinate system would not allow us to locate all the points on a 2-D plane, because points lying to the left of the y-axis or below the x-axis require negative coordinates.

Negative numbers are necessary to represent all points on a 2-D plane.

6. Are the points M (– 3, – 4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.

Answer:

Yes, the points

$M(-3,-4)$

,

$A(0,0)$

, and

$G(6,8)$

lie on the same straight line.

A method to check this without plotting is to compare the slopes.

Slope of

$MA$

:

$$\frac{0-(-4)}{0-(-3)}=\frac{4}{3}$$

Slope of

$AG$

:

$$\frac{8-0}{6-0}=\frac{8}{6}=\frac{4}{3}$$

Since the two slopes are equal,

$$\boxed{\text{M, A and G are collinear.}}$$

So, comparing the slopes of the line segments is one way to check whether three points lie on the same straight line.

7. Use your method (from Problem 6) to check if the points R (– 5, – 1), B (– 2, – 5), and C (4, – 12) are on the same straight line. Now plot both sets of points and check your answers.

Answer:

Given points: R(−5, −1), B(−2, −5), C(4, −12)
Using the distance formula:

$$RB=\sqrt{(-2-(-5){)}^2+(-5-(-1){)}^2}=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=5$$ $$BC=\sqrt{(4-(-2){)}^2+(-12-(-5){)}^2}=\sqrt{6^2+(-7{)}^2}=\sqrt{36+49}=\sqrt{85}$$ $$RC=\sqrt{(4-(-5){)}^2+(-12-(-1){)}^2}=\sqrt{9^2+(-11{)}^2}=\sqrt{81+121}=\sqrt{202}$$

RB + BC = 5 +

$\sqrt{85}$

​ ≠

$\sqrt{202}$

​ = RC
Therefore, the points R, B, and C are not collinear (do not lie on the same straight line).
Verification: On plotting, the three points will not lie on a single straight line

8. Using the origin as one vertex, plot the vertices of:
(i)  A right-angled isosceles triangle.

Answer:

A right-angled isosceles triangle

One possible set of vertices is:

$$\boxed{O(0,0),\ A(4,0),\ B(0,4)}$$

Here,

$$OA=OB=4$$

and

$OA\perp OB$

. Therefore,

$\triangle OAB$

is a right-angled isosceles triangle.

(ii)  An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Answer:

One possible set of vertices is:

$$\boxed{O(0,0),\ P(-3,-4),\ Q(3,-4)}$$

Here,

$P(-3,-4)$

lies in Quadrant III and

$Q(3,-4)$

lies in Quadrant IV.

Also,

$$OP=OQ=5$$

so

$\triangle OPQ$

is an isosceles triangle.

9. The following table shows the coordinates of points S, M, and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Answer:

Conclusion

When

$M$

is the midpoint of

$ST$

, it lies on

$ST$

and divides it into two equal parts:

$$\boxed{SM=MT}$$

If

$S(x_1,y_1)$

and

$T(x_2,y_2)$

, then the coordinates of the midpoint

$M$

are:

M(2x1​+x2​​,2y1​+y2​​)​

10. Use the connection you found to find the coordinates of B given that M (–7, 1) is the midpoint of A (3, – 4) and B (x, y).

Answer:

Using the midpoint formula,

 

$$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$ 

Given:

 

$$M(-7,1),\quad A(3,-4),\quad B(x,y)$$ 

For the x-coordinate:

 

$$-7=\frac{3+x}{2}$$ 

$$-14=3+x$$ 

$$x=-17$$ 

For the y-coordinate:

 

$$1=\frac{-4+y}{2}$$ 

$$2=-4+y$$ 

$$y=6$$ 

Therefore,

 

$$\boxed{B(-17,6)}$$ 

11. Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, –2).

Answer:

$$OA=\sqrt{(1-0)^2+(-8-0)^2} =\sqrt{1+64} =\sqrt{65}$$

For

$B(-4,7)$

:

$$OB=\sqrt{(-4)^2+7^2} =\sqrt{16+49} =\sqrt{65}$$

For

$C(-7,-4)$

:

$$OC=\sqrt{(-7)^2+(-4)^2} =\sqrt{49+16} =\sqrt{65}$$

Thus,

$$OA=OB=OC=\sqrt{65}$$

Therefore,

$A$

,

$B$

, and

$C$

lie on the same circle with centre

$O(0,0)$

.

$$\boxed{\text{Radius of circle }K=\sqrt{65}\text{ units}}$$

(ii) Given the points D (– 5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.

Answer:

For

$D(-5,6)$:

 

$$OD=\sqrt{(-5)^2+6^2} =\sqrt{25+36} =\sqrt{61}$$ 

Since

 

$$\sqrt{61}<\sqrt{65},$$ 

$$\boxed{D\text{ lies inside circle }K.}$$ 

For

$E(0,9)$:

 

$$OE=\sqrt{0^2+9^2}=9$$ 

Since

 

$$9>\sqrt{65},$$ 

$$\boxed{E\text{ lies outside circle }K.}$$ 

Hence,

D is inside the circle and E is outside the circle.

​

13. The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B, and C.

Let

$$A(x_1,y_1),\quad B(x_2,y_2),\quad C(x_3,y_3)$$

Given that

$D(5,1)$

,

$E(6,5)$

, and

$F(0,3)$

are the midpoints of

$BC$

,

$CA$

, and

$AB$

, respectively.

Using the midpoint formula:

$$D=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$$

Therefore,

$$x_2+x_3=10,\qquad y_2+y_3=2 \tag{1}$$

Similarly, from

$E(6,5)$

,

$$x_1+x_3=12,\qquad y_1+y_3=10 \tag{2}$$

and from

$F(0,3)$

,

$$x_1+x_2=0,\qquad y_1+y_2=6 \tag{3}$$

For the x-coordinates, adding (2) and (3):

$$2x_1+x_2+x_3=12$$

Using

$x_2+x_3=10$

,

$$2x_1+10=12$$

$$x_1=1$$

Then,

$$x_2=-1,\qquad x_3=11$$

For the y-coordinates, adding (2) and (3):

$$2y_1+y_2+y_3=16$$

Using

$y_2+y_3=2$

,

$$2y_1+2=16$$

$$y_1=7$$

Then,

$$y_2=-1,\qquad y_3=3$$

Therefore,

A(1,7),B(−1,−1),C(11,3)

14. A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction.
(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.

Answer:

Take the point where the two main roads cross as the centre

$O$

.

Using the scale

$$\boxed{1\text{ cm}=200\text{ m}}$$

draw one horizontal line for the E–W main road and one vertical line for the N–S main road, intersecting at

$O$

.

Then draw:

• 5 parallel streets on each side of the N–S road, each
  $1$ 

  cm apart.

• 5 parallel streets on each side of the E–W road, each
  $1$ 

  cm apart.

Thus, there are 10 streets in each direction, all

$200$

m apart.

(ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find:
(a) how many street intersections can be referred to as (4, 3).
(b) how many street intersections can be referred to as (3, 4).

Answer:

Because the streets are numbered according to their distance from the main roads, there is a street with a given number on both sides of each main road.

(a) Intersections referred to as

$(4,3)$

There are two 4th N–S streets and two 3rd E–W streets.

Therefore,

$$2\times2=4$$

4 intersections​

(b) Intersections referred to as

$(3,4)$

Similarly, there are two 3rd N–S streets and two 4th E–W streets.

$$2\times2=4$$

$$\boxed{4\text{ intersections}}$$

Final Answer:

(a) 4(b) 4

15. A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:
(i)  whether any part of either circle lies outside the screen.

Answer:

The screen extends from

$$0\le x\le 800,\qquad 0\le y\le 600$$

For the circle centred at

$A(100,150)$

with radius

$80$

:

$$x:100-80=20\text{ to }100+80=180$$

$$y:150-80=70\text{ to }150+80=230$$

All these values lie within the screen. Hence, the first circle is completely inside the screen.

For the circle centred at

$B(250,230)$

with radius

$100$

:

$$x:250-100=150\text{ to }250+100=350$$

$$y:230-100=130\text{ to }230+100=330$$

This circle is also completely inside the screen.

No part of either circle lies outside the screen.
(ii)  whether the two circles intersect each other.

Answer:

Class 9 Maths Ganita Manjari Chapter 1 Exercises:

Every question is solved step by step, with the reason for each step. Click the exercise you need.