NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Maths Orienting Yourself: The Use of Coordinates

Exercise Set 1.1

Question 1:
Fig. 1.3 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.

Thank you for reading this post, don't forget to subscribe!

Referring to Fig. 1.3, answer the following questions:
(i) If D₁R₁ 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?
(ii) What are the coordinates of D₁?
(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?
(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?
Solution
(i) From the figure, D₁ is on the x-axis and 8 units to the right of O, its coordinates are (8, 0).
Since D₁ = (8, 0) and R₁ =(11.5, 0), the door is 8 units away from the left wall (y-axis).
The door lies on the x-axis, so its distance from the x-axis = 0 units.

(ii) Since D₁ is on the x-axis and 8 units to the right of O, its coordinates are (8, 0).

(iii) D₁ R₁, represents the door, coordinates of D₁ = (8, 0) and R₁ = (11.5, 0).
So the width of the door = distance between x-coordinates of D₁ and R₁ points
= 11.5 – 8 = 3.5 units

(iv) Coordinates of B₁ = (0, 1.5) and B₂ = (0, 4).
So the width of the bathroom door = distance between B₁ and B₂ = 4 – 1.5 = 2.5 units And width of room door is 3.5 units.
Since 2.5 < 3.5, the bathroom door is narrower than the room door.

NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Maths Orienting Yourself: The Use of Coordinates

Exercise Set 1.2

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.) Using Fig. 1.5, answer the given questions.

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?
(ii) Is this a good spot for the table?
(iii) What is the width of the table? The length? Can you make out the height of the table?
Solution

(i) Given three feet points of the table: (8, 9), (11, 9), and (11, 7)
Since the table is rectangular in shape,
Fourth feet point = (8, 7)

(ii) Yes, this is a good spot because it is placed near the wall and does not obstruct movement or furniture.

(iii) Width = 9 −7 = 2 units
Length = 11 − 8 = 3 units
Height cannot be determined from the top view.

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?
Solution
B₁ coordinates = (0, 1.5)
B₂ coordinates = (0, 4)
Door width = 4 – 1.5 = 2.5 units
The swinging arc has radius 2.5, the wardrobe sits at x = 3, so the door clears by 0.5 units
Since 2.5 < 3, the door will not hit the wardrobe.
If the door is made wider, it is better to open it outwards.

3. Look at Reiaan’s bathroom.
(i) What are the coordinates of the four corners O, F, R, and P of the bathroom?
(ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.
(iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.
Solution
(i) Corners of the bathroom:
O = (0, 0)
F = (0, 9)
R = (−6, 9)
P = (−6, 0)

(ii) Shape of showering area SHWR is a trapezium.
Coordinates:
S = (−6, 6)
H = (−3, 6)
W = (−2, 9)
R = (−6, 9)

(iii) Washbasin (3 ft × 2 ft):
Example placement: (−6, 0), (−4, 0), (−4, 3), (−6, 3)
Toilet (2 ft × 3 ft):
Example placement: (−4, 3), (−2, 3), (−2, 6), (−4, 6)

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 length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.
(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.
Solution
(i)

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

(ii) The room runs from x = −6 to x = 12, and y = 0 to y = −15.
Centre of the room = $(\frac{- 6 + 12}{2}, \frac{0 + (- 15)}{2})$ = (3, 7.5)
The table is 5 ft × 3 ft, so:
Half length = 2.5
Half width = 1.5
The table is 5 ft × 3 ft, extending 2.5 ft either side in x and 1.5 ft either side in y.
The coordinates of the four feet:
(3 − 2.5, −7.5 + 1.5) = (0.5, −6)
(3 + 2.5, −7.5 + 1.5) = (5.5, −6)
(3 + 2.5, −7.5 − 1.5) = (5.5, −9)
(3 − 2.5, −7.5 − 1.5) = (0.5, −9)
Therefore, the coordinates of the feet of the table are: (0.5, −6), (5.5, −6), (5.5, −9), (0.5, −9).

NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Maths Orienting Yourself: The Use of Coordinates

End of Chapter Exercise

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.
x-coordinate = 0; y-coordinate = 0
So, the coordinates of the point of intersection are (0, 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?
Solution
Since point H lies on the line through W parallel to the y-axis, its x-coordinate remains the same as that of W.
So, coordinates of H = (−5, y), where y can be any real number.
If y > 0, H lies in Quadrant II.
If y < 0, H lies in Quadrant III.

3. Consider the points R (3, 0), A (0, – 2), M (– 5, – 2) and P (– 5, 2). If they are joined in the same order, predict:
(i) Two sides of RAMP that are perpendicular to each other.
(ii) One side of RAMP that is parallel to one of the axes.
(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.
Solution

Given points: R(3, 0), A(0, −2), M(−5, −2), P(−5, 2)
(i) Points A and M both have the same y = −2.
So, AM is the horizontal line
Points M and P both have the same x = −5.
So, MP is the vertical line.
Horizontal ⟂ Vertical
AM⊥MP

(ii) AM: y = −2 → parallel to x-axis
MP: x = −5 → parallel to y-axis
Therefore, AM is parallel to the x-axis (or MP to y-axis).

(iii) M(−5, −2) and P(−5, 2)
They have the same x-coordinate and opposite y-coordinates.
Therefore, points M and P are mirror images on the x-axis.

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.)
Solution

The coordinates of the triangle IZN are:
I = (5, 0), Z = (5, -6), N = (0, -6)
The lengths of the sides are:
IZ: Distance between (5,0) and (5, -6) = 0 – (-6) = 6 units
ZN: Distance between (5, -6) and (0, -6) = 5-0 = 5 units
IN: Using distance formula:
IN = $\sqrt{(5 – 0)^2 + (0 – (-6)}^2)$
$\quad$ = $\sqrt{(5)^2 + (6)^2}$
$\quad$ = $\sqrt{25 + 36}$
$\quad$ = $\sqrt{61}$ 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?
Solution
If negative numbers were not used, the coordinate system would include only the first quadrant, where both x- and y-coordinates are positive.
Such a system would not allow us to locate all points on a 2-D plane, because points in the second, third, and fourth quadrants (which require negative coordinates) cannot be represented.

*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.
Solution
Given points: M(−3, −4), A(0, 0), G(6, 8)
Using the distance formula: $M A = \sqrt{(0 - (- 3))^{2} + (0 - (- 4))^{2}} = \sqrt{3^{2} + 4^{2}} = 5$ $A G = \sqrt{(6 - 0)^{2} + (8 - 0)^{2}} = \sqrt{36 + 64} = 10$ $M G = \sqrt{(6 - (- 3))^{2} + (8 - (- 4))^{2}} = \sqrt{9^{2} + 12^{2}} = 15$
Since MA + AG = 5 + 10 = 15 = MG
Therefore, the three points lie on a 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.
Solution
Given points: R(−5, −1), B(−2, −5), C(4, −12)
Using the distance formula:

$R B = \sqrt{(- 2 - (- 5))^{2} + (- 5 - (- 1))^{2}} = \sqrt{3^{2} + (- 4)^{2}} = \sqrt{9 + 16} = 5$ $B C = \sqrt{(4 - (- 2))^{2} + (- 12 - (- 5))^{2}} = \sqrt{6^{2} + (- 7)^{2}} = \sqrt{36 + 49} = \sqrt{85}$ $R C = \sqrt{(4 - (- 5))^{2} + (- 12 - (- 1))^{2}} = \sqrt{9^{2} + (- 11)^{2}} = \sqrt{81 + 121} = \sqrt{202}$
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.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
Solutiion
(i) Take one vertex at the origin O(0, 0).
Choose the other two points at equal distances along the x and y axes: A(4, 0), B(0, 4).
Here,
OA = OB = 4 units
∠AOB = 90
So, △OAB is a right-angled isosceles triangle.

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 6

(ii) Take origin O(0, 0).
Choose:
C(−3,−4) (Quadrant III),
D(3,−4) (Quadrant IV)
Here,
OC = OD = 5 units
So, △OCD is an isosceles triangle.

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 12

*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?
Soution
Row 1:
S(−3, 0), M(0, 0), T(3, 0)
SM = $\sqrt{(0 + 3)^{2} + (0 - 0)^{2}}$ = $\sqrt{(3)^{2}}$ = 3
MT = $\sqrt{(3 - 0)^{2} + (0 - 0)^{2}}$ = $\sqrt{(3)^{2}}$ = 3
Since SM = MT
Yes, M is the midpoint.
Row 2:
S(2, 3), M(3, 4), T(4, 5)
SM = $\sqrt{(3 - 2)^{2} + (4 - 3)^{2}}$ = $\sqrt{(1)^{2} + (1)^{2}}$ = $\sqrt{2}$
MT = $\sqrt{(4 - 3)^{2} + (5 - 4)^{2}}$ = $\sqrt{(1)^{2} + (1)^{2}}$ = $\sqrt{2}$
Since SM = MT
Yes, M is the midpoint.
Row 3:
S(0, 0), M(0, 5), T(0, −10)
SM = $\sqrt{(0 - 0)^{2} + (5 - 0)^{2}}$ = $\sqrt{(5)^{2}}$ = 5
MT = $\sqrt{(0 - 0)^{2} + (- 10 - 5)^{2}}$ = $\sqrt{(- 15)^{2}}$ = 15
Since SM ≠ MT
No, M is the midpoint.
Row 4:
S(−8, 7), M(0, −2), T(6, −3)
SM = $\sqrt{(0 + 8)^{2} + (- 2 - 7)^{2}}$ = $\sqrt{(8)^{2} + (- 9)^{2}}$ = $\sqrt{64 + 81}$ = $\sqrt{145}$
MT = $\sqrt{(6 - 0)^{2} + (- 3 + 2)^{2}}$ = $\sqrt{(6)^{2} + (- 1)^{2}}$ = $\sqrt{36 + 1}$ = $\sqrt{37}$
Since SM ≠ MT
No, M is the midpoint.
Conclusion: When M is the midpoint of ST, then SM = MT.

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 9

*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).
Solution

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 36

Given that M(−7, 1) is the midpoint of A(3, −4) and B(x, y).
Using the midpoint relation:
$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
(-7, 1) = $(\frac{3 + x}{2}, \frac{- 4 + y}{2})$
-7 = $\frac{3 + x}{2}$ and 1 = $\frac{- 4 + y}{2}$
-14 = 3 + x and 2 = -4 + y
x = -14 – 3 and y = 2 + 4
x = -17 and y = 6
Therefore, 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).
Solution

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 35

Let A (4, 7) and B (16, –2).
Let P(x₁, y₁) and Q(x₂, y₂) divide AB into three equal parts.
Since P is the midpoint of AQ, $x_{1} = \frac{4 + x_{2}}{2} (1)$ $y_{1} = \frac{7 + y_{2}}{2} (2)$ Since Q is midpoint of PB, $x_{2} = \frac{x_{1} + 16}{2} (3)$ $y_{2} = \frac{y_{1} - 2}{2} (4)$
Substitute (1) into (3): $x_{2} = \frac{\frac{4 + x_{2}}{2} + 16}{2}$ $x_{2} = \frac{4 + x_{2} + 32}{4} = \frac{x_{2} + 36}{4}$ Multiply both sides by 4: $4 x_{2} = x_{2} + 36$ $3 x_{2} = 36 \Rightarrow x_{2} = 12$ Now from (1): $x_{1} = \frac{4 + 12}{2} = 8$ Substitute (2) into (4): $y_{2} = \frac{\frac{7 + y_{2}}{2} - 2}{2}$ $y_{2} = \frac{7 + y_{2} - 4}{4} = \frac{y_{2} + 3}{4}$ Multiply both sides by 4: $4 y_{2} = y_{2} + 3$ $3 y_{2} = 3 \Rightarrow y_{2} = 1$ Now from (2): $y_{1} = \frac{7 + 1}{2} = 4$ Hence,
P = (8, 4), Q = (12, 1)

*12. (i) Given the points A (1, – 8), B (– 4, 7) and C (–7, – 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?
(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.
Solution
(i) OA = $\sqrt{(1)^{2} + (- 8)^{2}}$ = $\sqrt{1 + 64}$ = $\sqrt{65}$
OB = $\sqrt{((- 4))^{2} + (7)^{2}}$ = $\sqrt{16 + 49}$ = $\sqrt{65}$
OC = $\sqrt{((- 7))^{2} + (- 4)^{2}}$ = $\sqrt{49 + 16}$ = $\sqrt{65}$
Since,
OA = OB = OC = $\sqrt{65}$
All three points are at the same distance from the origin.
Hence, they lie on a circle with centre O(0, 0).
Radius of circle $K = \sqrt{65}$

(ii) For D (-5, 6):
OD = $\sqrt{(- 5)^{2} + (6)^{2}}$ = $\sqrt{25 + 36}$ = $\sqrt{61}$
Since $\sqrt{61}$ < $\sqrt{65}$
∴ D lies inside the circle.
For E (0, 9),
OE = $\sqrt{(0)^{2} + (9)^{2}}$ = $\sqrt{81}$ = 9
Since 9 > $\sqrt{65}$
∴ E lies 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.
Solution

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 37

Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)
Given midpoints: D(5, 1), E(6, 5), F(0, 3)
Using the midpoint formula:
D = $(\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$
⇒ (5, 1) = $(\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$
5 = $\frac{x_{2} + x_{3}}{2}$ and 1 = $\frac{y_{2} + y_{3}}{2}$
x₂ + x₃ = 10, y₂ + y₃ = 2…….(1)
E = $(\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$
⇒ (6, 5) = $(\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$
6 = $\frac{x_{1} + x_{3}}{2}$ and 5 = $\frac{y_{1} + y_{3}}{2}$
x₁ + x₃ = 12, y₁ + y₃ = 10…….(2)
F = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
⇒ (0, 3) = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
0 = $\frac{x_{1} + x_{2}}{2}$ and 3 = $\frac{y_{1} + y_{2}}{2}$
x₁ + x₂ = 0, y₁ + y₂ = 6………..(3)

For x coordinates:
Add (2) and (3):
(x₁ + x₃) + (x₁ + x₂) = 12 + 0
⇒2x₁ + x₂ + x₃ = 12
Using (1): x₂ + x₃ = 10
2x₁ + 10 = 12
2x₁ = 12 – 10
⇒ x₁ = 1
From (3): x₁ + x₂ = 0
⇒1 + x₂ = 0
⇒x₂ = −1
From (1): −1 + x₃ = 10
⇒ x₃ = 11

Now for y-coordinates:
Add (2) and (3):
(y₁ + y₃) + (y₁ + y₂) = 10 + 6
⇒2y₁ + y₂ + y₃ = 16
Using (1): y₂ + y₃ = 2
2y₁ + 2 = 16
⇒ y₁ = 7
From (3): y₁ + y₂ = 6
⇒ 7 + y₂ = 6
⇒ y₂ = −1
From (1): −1 + y₃ = 2
⇒ y₃ = 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.
(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).
Solution
(i)

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 23

(ii)
(a) There is only one intersection where 4th N–S street meets 3rd E–W street, referred as (4, 3).
(b) Similarly, there is only one intersection where 3rd N–S street meets 4th E–W street, referred as (3, 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.
(ii) whether the two circles intersect each other.
Solution

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 24

(i) Both circles lie completely inside the screen.
(ii) Yes, the circles intersect each other.

16. Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?
Solution
Plot the points A(2, 1), B(−1, 2), C(−2, −1), D(1, −2) and join them in order.

orienting yourself the use of coordinates class 9 Maths Ganita manjari part 1 chapter 1 NCERT solutions image 25

Length of all sides:
AB = $\sqrt{(- 1 - 2)^{2} + (4 - 3)^{2}}$ = $\sqrt{(- 3)^{2} + (1)^{2}}$ = $\sqrt{9 + 1}$ = $\sqrt{10}$
BC = $\sqrt{(- 2 + 1)^{2} + (- 1 - 2)^{2}}$ = $\sqrt{(- 1)^{2} + (- 3)^{2}}$ = $\sqrt{1 + 9}$ = $\sqrt{10}$
CD = $\sqrt{(1 + 2)^{2} + (- 2 + 1)^{2}}$ = $\sqrt{(3)^{2} + (- 1)^{2}}$ = $\sqrt{9 + 1}$ = $\sqrt{10}$
DA = $\sqrt{(2 - 1)^{2} + (1 + 2)^{2}}$ = $\sqrt{(1)^{2} + (3)^{2}}$ = $\sqrt{1 + 9}$ = $\sqrt{10}$
Length of diagonals:
AC = $\sqrt{(- 2 - 2)^{2} + (- 1 - 1)^{2}}$ = $\sqrt{(- 4)^{2} + (- 2)^{2}}$ = $\sqrt{16 + 4}$ = $\sqrt{20}$
BD = $\sqrt{(1 + 1)^{2} + (- 2 - 2)^{2}}$ = $\sqrt{(2)^{2} + (- 4)^{2}}$ = $\sqrt{4 + 16}$ = $\sqrt{20}$
Since all sides are equal and diagonals are equal,
ABCD is a square.
Side $= \sqrt{10}$
Area = $\sqrt{(10)^{2}}$ = 10 square units.

NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Maths Orienting Yourself: The Use of Coordinates

BrightWay Coaching Academy

NCERT Solutions for Class 9 Ganita Manjari Chapter 1