MCQ Questions for Class 10 Geography: Chapter 7 राष्ट्रीय अर्थव्यववस्था की जीवन रेखाएँ

Chapter 3 Coordinate Geometry

FAQ on Chapter 2 Polynomials

How can I download Chapter 2 Polynomials Class 9 NCERT Solutions PDF

Chapter 2 Polynomials Class 9 Maths NCERT Solutions PDF download is very useful in understanding the basic concepts embedded in the chapter. It is very essential to solve every question before moving further to any other supplementary books.

What do you mean by Algebraic Expressions?

A combination of constants and variables, connected by some or all the basic operations +, –, ×, ÷ is called an algebraic expression.

What is the coefficient of a zero polynomial?

The coefficient of a zero polynomial is 1.

What is Biquadratic Polynomial?

A polynomial of degree 4 is called a biquadratic polynomial.

↧

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

January 15, 2021, 11:53 pm

≫ Next: NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

≪ Previous: NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems| PDF Download

On this page you will get NCERT Solutions for Class 9 Maths Chapter Number Systems that will comfort you as you will get to know various concepts needed for solving various questions. These NCERT Solutions are updated as per the latest pattern of 2020-21 syllabus. Chapter 1 Class 10 Maths NCERT Solutions PDF Download can be really helpful if anyone want to understand the detailed solutions and minimise the errors wherever possible.

For better understanding and application of concepts, one should first need to focus on the Number Systems NCERT Solutions as it will tell you about the difficulty of questions. You can be able to solve more questions related to chapter after practising more questions and thus fetching more marks.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

Page No: 5

Exercise 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q≠ 0?

Answer

Yes. Zero is a rational number as it can be represented as 0/1 or 0/2.

2. Find six rational numbers between 3 and 4.

Answer

There are infinite rational numbers in between 3 and 4.
3 and 4 can be represented as 24/8 and 32/8 respectively.
Therefore, six rational numbers between 3 and 4 are
25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

3. Find five rational numbers between 3/5 and 4/5.

Answer

There are infinite rational numbers in between 3/5 and 4/5
3/5 = 3×6/5×6 = 18/30
4/5 = 4×6/5×6 = 24/30
Therefore, five rational numbers between 3/5 and 4/5 are
19/30, 20/30, 21/30, 22/30, 23/30.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

► True, since the collection of whole numbers contains all natural numbers.

(ii) Every integer is a whole number.
► False, as integers may be negative but whole numbers are always positive.

(iii) Every rational number is a whole number.

► False, as rational numbers may be fractional but whole numbers may not be.

Page No: 8

Exercise 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

► True, since the collection of real numbers is made up of rational and irrational numbers.

(ii) Every point on the number line is of the form√m, where m is a natural number.
► False, since positive number cannot be expressed as square roots.

(iii) Every real number is an irrational number.
► False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Answer

No, the square roots of all positive integers are not irrational. For example √4 = 2.

3. Show how √5 can be represented on the number line.

Answer

Step 1: Let AB be a line of length 2 unit on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.
Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,
AB² + BC² = CA²
⇒ 2² + 1² = CA²
⇒ CA² = 5
⇒ CA = √5
Thus, CA is a line of length √5 unit.
Step 4: Taking CA as a radius and A as a centre draw an arc touching
the number line. The point at which number line get intersected by
arc is at √5 distance from 0 because it is a radius of the circle
whose centre was A.
Thus, √5 is represented on the number line as shown in the figure.

Page No: 14

Exercise 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) 36/100
= 0.36 (Terminating)

(ii) 1/11
0.09090909... = 0.9 (Non terminating repeating)

(iii)

= 33/8 = 4.125 (Terminating)

(iv) 3/13
= 0.230769230769... = 0.230769 (Non terminating repeating)

(v) 2/11
= 0.181818181818... = 0.18 (Non terminating repeating)

(vi) 329/400
= 0.8225 (Terminating)

2. You know that 1/7 = 0.142857.Can you predict what the decimal expansion of 2/7, 3/7, 4/7, 5/7, 6/7 are without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of 1/7 carefully.]

Answer

Yes. We can be done this by:

3. Express the following in the form p/q where p and q are integers and q≠ 0.

(i) 0.6

(ii) 0.47

(iii) 0.001

Answer

(i) 0.6 = 0.666...
Let x = 0.666...
10x = 6.666...
10x = 6 + x
9x = 6
x = 2/3

(ii) 0.47 = 0.4777...
= 4/10 + 0.777/10
Let x = 0.777…
10x = 7.777…
10x = 7 + x

x = 7/9

4/10 + 0.777.../10 = 4/10 + 7/90

= 36/90 +7/90 = 43/90

(iii) 0.001 = 0.001001...

Let x = 0.001001...

1000x = 1.001001…
1000x = 1 + x
999x = 1

x = 1/999

4. Express 0.99999…in the form p/q. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Answer

Let x = 0.9999…
10x = 9.9999…
10x = 9 + x
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999 is too much near 1, Therefore, the 1 as answer can be justified.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Answer

1/17 = 0.0588235294117647
There are 16 digits in the repeating block of the decimal expansion of 1/17.
Division Check:

= 0.0588235294117647

6. Look at several examples of rational numbers in the form p/q (q ≠ 0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer

We observe that when q is 2, 4, 5, 8, 10... then the decimal expansion is terminating. For example:
1/2 = 0.5, denominator q = 2¹
7/8 = 0.875, denominator q = 2³
4/5 = 0.8, denominator q = 5¹

We can observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer

Three numbers whose decimal expansions are non-terminating non-recurring are:

0.303003000300003...

0.505005000500005...

0.7207200720007200007200000…

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Answer

5/7 = 0.714285
9/11 = 0.81

Three different irrational numbers are:

0.73073007300073000073…

0.75075007300075000075…

0.76076007600076000076…

9. Classify the following numbers as rational or irrational:

(i) √23

(ii) √225

(iii) 0.3796
(iv) 7.478478

(v) 1.101001000100001…

Answer

(i) √23 = 4.79583152331...

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(ii) √225 = 15 = 15/1
Since the number is rational number as it can represented in p/q form.

(iii) 0.3796

Since the number is terminating therefore, it is an rational number.

(iv) 7.478478 = 7.478

Since the this number is non-terminating recurring, therefore, it is a rational number.

(v) 1.101001000100001…

Since the number is non-terminating non-repeating, therefore, it is an irrational number.

Page No: 18

Exercises 1.4

1. Visualise 3.765 on the number line using successive magnification.

Answer

2. Visualise 4.26 on the number line, up to 4 decimal places.

Answer

4.26 = 4.2626

Page No: 24

Exercise 1.5

1. Classify the following numbers as rational or irrational:

(i) 2 - √5

(ii) (3 + √23) - √23

(iii) 2√7/7√7

(iv) 1/√2

(v) 2π

Answer

(i) 2 - √5 = 2 - 2.2360679… = - 0.2360679…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) - √23 = 3 + √23 - √23 = 3 = 3/1
Since the number is rational number as it can represented in p/q form.

(iii) 2√7/7√7 = 2/7
Since the number is rational number as it can represented in p/q form.

(iv) 1/√2 = √2/2 = 0.7071067811...
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

2. Simplify each of the following expressions:

(i) (3 + √3) (2 + √2)

(ii) (3 + √3) (3 - √3)

(iii) (√5 + √2)²
(iv) (√5 - √2) (√5 + √2)

Answer

(i) (3 + √3) (2 + √2)

⇒ 3 × 2 + 2 + √3 + 3√2+ √3×√2

⇒ 6 + 2√3 +3√2 + √6

(ii) (3 + √3) (3 - √3) [∵ (a + b) (a - b) = a² - b²]

⇒ 3² - (√3)²
⇒ 9 - 3

⇒ 6

(iii) (√5 + √2)²[∵ (a + b)² = a² + b² + 2ab]
⇒ (√5)² + (√2)² + 2 ×√5 × √2
⇒ 5 + 2 + 2 × √5× 2

⇒ 7 +2√10

(iv) (√5 - √2) (√5 + √2) [∵ (a + b) (a - b) = a² - b²]

⇒ (√5)² - (√2)²

⇒ 5 - 2

⇒ 3

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Answer

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realise that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857...

4. Represent √9.3 on the number line.

Answer

Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 unit.
Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.
Step 3: Draw a semi circle with radius OC and centre O.
Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D. Also, Join OD.
Step 5: Now, OBD is a right angled triangle.
Here, OD = 10.3/2 (radius of semi circle), OC = 10.3/2, BC = 1
OB = OC – BC = (10.3/2) – 1 = 8.3/2
Using Pythagoras theorem,
OD² = BD² + OB²
⇒ (10.3/2)² = BD2 + (8.3/2)²
⇒ BD² = (10.3/2)² - (8.3/2)²
⇒ BD² = (10.3/2 – 8.3/2) (10.3/2 + 8.3/2)
⇒ BD² = 9.3
⇒ BD² = √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

5. Rationalise the denominators of the following:

(i) 1/√7

(ii) 1/√7-√6

(iii) 1/√5+√2

(iv) 1/√7-2

Answer

Page No: 26

Exercise 1.6

1. Find:
(i) 64^1/2
(ii) 32^1/5
(iii) 125^1/3

Answer

2. Find:

(i) 9^3/2

(ii) 32^2/5

(iii) 16^3/4

(iv) 125^-1/3

Answer

3. Simplify:

(i) 2^2/3.2^1/5

(ii) (1/3³)⁷

(iii) 11^1/2/11^1/4

(iv) 7^1/2.8^1/2

Answer

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

Number Systems is the first chapter of Class 9 Maths textbook in which we study about natural numbers, whole numbers and integers. This chapter details about rational and irrational numbers. A number ‘r’ is called a rational number, if it can be written in the form p/q, where p and q are integers and q ≠ 0.

• Irrational Numbers: A number ‘s’ is called irrational, if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. The collection of all rational numbers and irrational numbers together make up what we call the collection of real numbers. Every real number is represented by a unique point on the number line.

• Real Numbers and their Decimal Expansions: We will look at the decimal expansions of real numbers and see if we can use the expansions to distinguish between rationals and irrationals and also explain how to visualise the representation of real numbers on the number line using their decimal expansions. The decimal expansion of a rational number is either terminating or nonterminating recurring. The decimal expansion of an irrational number is non-terminating non-recurring

• Representing Real Numbers on the Number Line: The process of visualisation of representation of numbers on the number line, through a magnifying glass is known as the process of successive magnification. Every real number is represented by a unique point on the number line. Further, every point on the number line represents one and only one real number.

• Operations on Real Numbers: We will be dealing with multiple operations like addition, subtraction, multiplication and division of irrational numbers.

• Laws of Exponents for Real Numbers: We will use laws of exponents to solve the questions given in the exercise.

There are total 6 exercises in the chapter which will be help you in revising the chapter properly. We have provided Class 9 Chapter 1 Maths NCERT Solutions step by step on Studyrankers. You only need to click on the exercises given below and start clearing your doubts.

Studyrankers experts have prepared Chapter 1 Class 9 Maths NCERT Solutions in well manner so a student can easily clear their misunderstandings any time.

NCERT Solutions for Class 9 Maths Chapters:

Chapter 2 Polynomials

Chapter 4 Linear Equations in Two Variables

Chapter 4 Linear Equations in Two Variables

FAQ on Chapter 1 Number Systems

Why we should solve NCERT Solutions for Chapter 1 Number Systems Class 9?

Chapter 1 Number Systems Class 9 Maths is very important for every student as it will help you in understanding what is given in the chapter and understand the concepts. It is very essential to solve every question before moving further to any other supplementary books.

Which three integers are equal to their own cube roots?

–1, 0 and 1

Find the sum of (3√3 +7√2) and (√3 - 5√2).

We have (3√3 +7√2) + (√3 - 5√2)

= √3 3+7√2 + √3 - 5√2

= (3√3+√3) + 7√2 - 5√2)

=√3(3+1) + √2(7-2)

=√3(4) + √2(5) = (4√3 + 2√2)

How can I download PDF of Chapter 1 Number Systems Class 9 NCERT Solutions?

You can easily download PDF of Class 9 NCERT Solutions Chapter 1 here through which you can easily get what you're looking for.

↧

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

January 15, 2021, 11:55 pm

≫ Next: NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

≪ Previous: NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry| PDF Download

On this page, you will get Chapter 3 Coordinate Geometry Class 9 Maths NCERT Solutions which are accurate and detailed and as per the latest syllabus released by CBSE. You can also Download PDF of Class 9 Maths Chapter 3 Coordinate Geometry NCERT Solutions which will develop you understanding of the chapter and help in gaining good marks in the examinations.

You can also complete your homework on time through the help of these NCERT Solutions and able to solve the difficult problems in a given in a exercise. A student will learn how to solve a questions efficiently and obtain maximum marks in the exams. These NCERT Solutions for Class 9 will prove useful guide in making a student confident.

NCERT Solutions for Class 9 Maths Chapter 1 3 Coordinate Geometry

Page No: 53

Exercise 3.1

1. How will you describe the position of a table lamp on your study table to another person?

Answer

To describe the position of a table lamp on the study table, we have two take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates.
Let the distance of point from X axis is x and from Y axis is y then the the position of the table lamp in terms of coordinates is (x,y).

2. (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5 th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross - streets can be referred to as (4, 3).
(ii) how many cross - streets can be referred to as (3, 4)

Answer

(i) Only one street can be referred to as (4, 3) as we see from the figure.

(ii) Only one street can be referred to as (3, 4) as we see from the figure.

Page No: 60

Exercise 3.2

1. Write the answer of each of the following questions:

(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.

Answer

(i) The name of horizontal lines and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively.

(ii) The name of each part of the plane formed by these two lines x-axis and y-axis is quadrants.

(iii) The point where these two lines intersect is called origin.

2. See Fig.3.14, and write the following:

(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii)The coordinates of the point L.
(viii) The coordinates of the point M.

Answer

(i) The coordinates of B is (-5, 2).

(ii) The coordinates of C is (5, -5).

(iii) The point identified by the coordinates (-3, -5) is E.

(iv) The point identified by the coordinates (2, -4) is G.

(v) Abscissa means x coordinate of point D. So, abscissa of the point D is 6.

(vi) Ordinate means y coordinate of point H. So, ordinate of point H is -3.

(vii) The coordinates of the point L is (0, 5).

(viii) The coordinates of the point M is (- 3, 0).

Page No: 65

Exercise 3.3

1. In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.

Answer

(-2, 4) → Second quadrant
(3, -1) → Fourth quadrant
(-1, 0) → Second quadrant
(1, 2) → First quadrant
(-3, -5) → Third quadrant

2. Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

x	-2	-1	0	1	3
y	8	7	-1.25	3	-1

Answer

Points (x,y) on the plane. 1unit = 1 cm

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

Chapter 3 Coordinate Geometry Class 9 Maths NCERT Solutions deals with some basic concepts of coordinate geometry. The system used for describing the position of a point in a plane is also known as the Cartesian system. The mutually perpendicular lines are called axes. The horizontal line is called the x-axis and the vertical line is called the y-axis. The x-coordinate of a point is called the abscissa. The y-coordinate of a point is called the ordinate.

• Cartesian System: The coordinates of the origin O? It has zero distance from both the axes so that its abscissa and ordinate are both zero. Therefore, the coordinates of the origin are (0, 0).You will learn to find a coordinates given a Cartesian plane.

• Plotting a Point in the Plane if its Coordinates are Given: If x ≠ y, then the position of (x, y) in the Cartesian plane is different from the position of (y, x). So, if we interchange the coordinates x and y, the position of (y, x) will differ from the position of (x, y). This means that the order of x and y is important in (x, y). Therefore, (x, y) is called an ordered pair. The ordered pair (x, y) ≠ ordered pair (y, x), if x ≠ y. Also (x, y) = (y, x), if x = y. You will learn to plot the points on a cartesian plane.

There are only three exercises in Coordinate Geometry Class 9 Maths Textbook. If you're facing any type of difficulty while finding a given question then you can look at the exerciseiwse NCERT Solutions given below.

Studyrankers experts have taken every while preparing these Chapter 3 Coordinate Geometry NCERT Solutions and detailed every step so if you find any doubt while solving a question then you can always look from here and move further.

NCERT Solutions for Class 9 Maths Chapters:

Chapter 1 Number Systems

Chapter 2 Polynomials

Chapter 3 Coordinate Geometry

FAQ on Chapter 3 Coordinate Geometry

What are the benefits of NCERT Solutions for Chapter 3 Coordinate Geometry Class 9 NCERT Solutions?

Chapter 3 Coordinate Geometry Class 9 Maths NCERT Solutions will be very useful in revising the chapter thoroughly and understand all the basic concepts. Through the help of these, you will get to know about various terms such as Cartesian plane, quadrants, origin and many others.

How to plot a point in the plane when coordinates of the point are given?

To plot a point in the coordinate plane we draw the coordinate axes and choose our units such that we can mark equal distances on the x-axis or on the y-axis or on both the axes. We take origin as zero for x-axis as well as on y-axis. The distances marked along OX and OY are taken as positive and those along OX' and OY' are taken as negative.

What are the coordinates of origin?

(0, 0) are the coordinates of origin.

What is the sign of the y-coordinate of a point lying in the III-quadrant?

The y-coordinate of a point in III-quadrant is always negative.

↧

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

January 16, 2021, 1:15 am

≫ Next: MCQ Questions for Class 9 History: Chapter 5 आधुनिक विश्व में चरवाहे

≪ Previous: NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables| PDF Download

Here you will find Chapter 4 Linear Equations In Two Variables Class 9 Maths NCERT Solutions which are very much essential in completing homework and prepare for the exams with ease. Here you can download PDF of NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables that will serve as beneficial tool that can be used to recall various questions any time.

These NCERT Solutions for Class 9 which are accurate and detailed are prepared by Studyrankers experts which will make learning exciting and cover the syllabus in less time. NCERT Class 9 Maths Chapter 4 will help you in rectifying all those mistakes that can happen in the exams.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Page No: 68

Exercise 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be x and that of a pen to be y).

Answer

Let the cost of pen be y and the cost of notebook be x.
A/q,
Cost of a notebook = twice the pen = 2y.
∴2y = x
⇒ x - 2y = 0
This is a linear equation in two variables to represent this statement.

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35          (ii) x - y/5 - 10 = 0    (iii) -2x + 3y = 6                (iv) x = 3y
(v) 2x = -5y                        (vi) 3x + 2 = 0                        (vii) y - 2 = 0                     (viii) 5 = 2x

Answer

(i) 2x + 3y = 9.35
⇒ 2x + 3y - 9.35 = 0
On comparing this equation with ax + by + c = 0, we get
a = 2x, b = 3 and c = -9.35

(ii) x - y/5 - 10 = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -1/5 and c = -10

(iii) -2x + 3y = 6
⇒ -2x + 3y - 6 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 3 and c = -6

(iv) x = 3y
⇒ x - 3y = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -3 and c = 0

(v) 2x = -5y
⇒ 2x + 5y = 0
On comparing this equation with ax + by + c = 0, we get
a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 3, b = 0 and c = 2

(vii) y - 2 = 0
⇒ 0x + y - 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 0, b = 1 and c = -2

(viii) 5 = 2x
⇒ -2x + 0y + 5 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 0 and c = 5

Page No: 70

Exercise 4.2

1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,               (ii) only two solutions, (iii) infinitely many solutions

Answer

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

2. Write four solutions for each of the following equations:
    (i) 2x + y = 7       (ii) πx + y = 9            (iii) x = 4y

Answer

(i) 2x + y = 7
⇒ y = 7 - 2x
→ Put x = 0,
y = 7 - 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 - 2 × 1 ⇒ y = 5
(1, 5) is the solution.
→ Now, put x = 2
y = 7 - 2 × 2 ⇒ y = 3
(2, 3) is the solution.
→ Now, put x = -1
y = 7 - 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9
⇒ y = 9 - πx
→ Put x = 0,
y = 9 - π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 - π×1 ⇒ y = 9-π
(1, 9-π) is the solution.
→ Now, put x = 2
y = 9 - π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution.
→ Now, put x = -1
y = 9 - π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution.
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution.
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:
    (i) (0, 2)          (ii) (2, 0)             (iii) (4, 0)      (iv) (√2, 4√2)    (v) (1, 1)

Answer

(i) Put x = 0 and y = 2 in the equation x - 2y = 4.
0 - 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x - 2y = 4.
2 - 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x - 2y = 4.
4 - 2×0 = 4
⇒ 4 = 4
∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x - 2y = 4.
√2 - 2×4√2 = 4 ⇒ √2 - 8√2 = 4 ⇒ √2(1 - 8) = 4
⇒ -7√2 ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x - 2y = 4.
1 - 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer

Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7

Page No: 74

Exercise 4.3

1. Draw the graph of each of the following linear equations in two variables:
   (i) x + y = 4                  (ii) x - y = 2               (iii) y = 3x    (iv) 3 = 2x + y

Answer

(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0

x	0	4
y	4	0

(ii) x - y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0

x	0	2
y	-2	0

(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3

x	0	1
y	0	3

(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1

x	0	1
y	3	1

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer

Here, x = 2 and y =14.
Thus, x + y = 1
also, y = 7x ⇒ y - 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y - 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.

3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Answer

The point (3, 4) lies on the graph of the equation.
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 - 7
⇒ a = 5/3

4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer

Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x - 5
⇒ y = 5x + 3

x	0	-3/5
y	3	0

5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
   For Fig. 4. 6                           For Fig. 4.7
   (i) y = x                                  (i) y = x + 2
   (ii) x + y = 0                           (ii) y = x – 2
   (iii) y = 2x                              (iii) y = –x + 2
   (iv) 2 + 3y = 7x (iv) x + 2y = 6

Answer

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.

Page No: 75

6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units (ii) 0 unit

Answer

Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit

x	2	0
y	10	0

7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Answer

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0

x	0	50	100
y	100	50	0

8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

F = (9/5)C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer

(i) F = (9/5)C + 32

When C = 0 then F = 32

also, when C = -10 then F = 14

C	0	-10
F	32	14

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30 + 32
⇒ F = 54 + 32
⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C + 32
⇒ (9/5)C = 95 - 32
⇒ C = 63 × 5/9
⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0 + 32
⇒ F = 32

(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F - 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.

Page No: 77

Exercise 4.4

1. Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
y = 3

(ii) in two variables, it is represented as a line parallel to X-axis.

0x + y = 3

2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
x = -9/2

(ii) in two variables, it is represented as a line parallel to Y-axis.

2x + 0y + 9 = 0

Go Back to NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

NCERT Solutions of Chapter 4 Linear Equations in Two Variables will help in developing basic problem solving skills and scoring more marks in the exams. Any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables.

• Linear Equations: The solution of a linear equation is not affected when: (i) the same number is added to (or subtracted from) both the sides of the equation. (ii) you multiply or divide both the sides of the equation by the same non-zero number.

• Solution of a Linear Equation: A linear equation in two variables has infinitely many solutions. You need to find solutions of different linear equations in two variables.

• Graph of a Linear Equation in Two Variables: Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation. In this section we will look geometrical representation of the solutions of linear equations in two variables.

• Equations of Lines Parallel to the x-axis and y-axis: We need to give the geometric representations of given equations in one variable and in two variables.

There are total four exercises in the Chapter 4 Class 9 Maths textbook. You will also find exercisewise NCERT Solutions by clicking on the exercise number.

NCERT Solutions are very important to keep revising the syllabus and clearing doubts wherever occur. Through the help of the concepts, you can easily tackle the questions given the supplementary books.

NCERT Solutions for Class 9 Maths Chapters:

Chapter 1 Number Systems

Chapter 2 Polynomials