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• NCERT Solutions
• NCERT Class 9
• NCERT 9 Maths
• Chapter 1: Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

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In NCERT Solutions for Class 9 Maths Chapter 1 , students are introduced to several important topics that are considered to be very crucial for those who wish to pursue Mathematics as a subject in their higher classes. Based on these NCERT Solutions , students can practise and prepare for their upcoming CBSE exams, as well as equip themselves with the basics of Class 10. These Maths Solutions of NCERT Class 9 are helpful as they are prepared with respect to the latest update on the CBSE syllabus for 2023-24 and its guidelines.

• Chapter 1- Number Systems
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclids Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Inroduction to Probability

NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems

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Exercise 1.1 page: 5.

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

We know that a number is said to be rational if it can be written in the form p/q , where p and q are integers and q ≠ 0.

Taking the case of ‘0’,

Zero can be written in the form 0/1, 0/2, 0/3 … as well as , 0/1, 0/2, 0/3 ..

Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number.

Hence, 0 is a rational number.

2. Find six rational numbers between 3 and 4.

There are infinite rational numbers between 3 and 4.

As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6)

i.e., 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers in between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.

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

There are infinite rational numbers between 3/5 and 4/5.

To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5

with 5+1=6 (or any number greater than 5)

i.e., (3/5) × (6/6) = 18/30

and, (4/5) × (6/6) = 24/30

The numbers in between18/30 and 24/30 will be rational and will fall between 3/5 and 4/5.

Hence,19/30, 20/30, 21/30, 22/30, 23/30 are the 5 rational numbers between 3/5 and 4/5

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

(i) Every natural number is a whole number.

Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals)

i.e., Natural numbers = 1,2,3,4…

Whole numbers – Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers = 0,1,2,3…

Or, we can say that whole numbers have all the elements of natural numbers and zero.

Every natural number is a whole number; however, every whole number is not a natural number.

(ii) Every integer is a whole number.

Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers.

i.e., integers= {…-4,-3,-2,-1,0,1,2,3,4…}

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

Every whole number is an integer; however, every integer is not a whole number.

(iii) Every rational number is a whole number.

Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0.

i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…

All whole numbers are rational, however, all rational numbers are not whole numbers.

Exercise 1.2 Page: 8

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

(i) Every irrational number is a real number.

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Every irrational number is a real number, however, every real number is not an irrational number.

(ii) Every point on the number line is of the form √m where m is a natural number.

The statement is false since as per the rule, a negative number cannot be expressed as square roots.

E.g., √9 =3 is a natural number.

But √2 = 1.414 is not a natural number.

Similarly, we know that there are negative numbers on the number line, but when we take the root of a negative number it becomes a complex number and not a natural number.

E.g., √-7 = 7i, where i = √-1

The statement that every point on the number line is of the form √m, where m is a natural number is false.

(iii) Every real number is an irrational number.

The statement is false. Real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number.

Every irrational number is a real number, however, every real number is not irrational.

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.

No, the square roots of all positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

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

Step 1: Let line AB be of 2 unit on a number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit.

Step 3: Join CA

Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB 2 +BC 2 = CA 2

2 2 +1 2 = CA 2 = 5

⇒ CA = √5 . Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a center 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 center was A.

Thus, √5 is represented on the number line as shown in the figure.

4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP 1 of unit length (see Fig. 1.9). Now draw a line segment P 2 P 3 perpendicular to OP 2 . Then draw a line segment P 3 P 4 perpendicular to OP 3 . Continuing in Fig. 1.9 :

Constructing this manner, you can get the line segment P n-1 Pn by square root spiral drawing a line segment of unit length perpendicular to OP n-1 . In this manner, you will have created the points P 2 , P 3 ,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …

Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral.

Step 2: From O, draw a straight line, OA, of 1cm horizontally.

Step 3: From A, draw a perpendicular line, AB, of 1 cm.

Step 4: Join OB. Here, OB will be of √2

Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.

Step 6: Join OC. Here, OC will be of √3

Step 7: Repeat the steps to draw √4, √5, √6….

Exercise 1.3 Page: 14

1. Write the following in decimal form and say what kind of decimal expansion each has :

= 0.36 (Terminating)

= 4.125 (Terminating)

(vi) 329/400

= 0.8225 (Terminating)

2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions 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.]

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

Assume that   x  = 0.666…

Then,10 x  = 6.666…

10 x  = 6 +  x

(ii) \(\begin{array}{l}0.4\overline{7}\end{array} \)

= (4/10)+(0.777/10)

Assume that  x  = 0.777…

Then, 10 x  = 7.777…

10 x  = 7 +  x

(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )

= (36/90)+(7/90) = 43/90

Assume that   x  = 0.001001…

Then, 1000 x  = 1.001001…

1000 x  = 1 +  x

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.

Assume that x  = 0.9999…..Eq (a)

Multiplying both sides by 10,

10 x  = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

10 x  = 9.9999

– x  = -0.9999…

_____________

The difference between 1 and 0.999999 is 0.000001 which is negligible.

Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the 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.

Dividing 1 by 17:

There are 16 digits in the repeating block of the decimal expansion of 1/17.

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?

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 = 2 1

7/8 = 0. 875, denominator q =2 3

4/5 = 0. 8, denominator q = 5 1

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

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

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

• √3 = 1.732050807568
• √26 =5.099019513592
• √101 = 10.04987562112

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

Three different irrational numbers are:

• 0.73073007300073000073…
• 0.75075007300075000075…
• 0.76076007600076000076…

9.  Classify the following numbers as rational or irrational according to their type:

√23 = 4.79583152331…

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

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

Exercise 1.4 Page: 18

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

Exercise 1.5 Page: 24

1. Classify the following numbers as rational or irrational:

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23

(3 + √ 23) –√23 = 3+ √ 23–√23

Since the number 3/1 is in p/q form, ( 3 +√23)- √23 is rational.

(iii) 2√7/7√7

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

We know that, the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

2. Simplify each of the following expressions:

(i) (3+√3)(2+√2)

(3+√3)(2+√2 )

Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )

(3+√3)(3-√3 ) = 3 2 -(√3) 2 = 9-3

(iii) (√5+√2) 2

(√5+√2) 2 = √5 2 +(2×√5×√2)+ √2 2

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

(√5-√2)(√5+√2) = (√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?

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 realize whether 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.

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, 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,

OD 2 =BD 2 +OB 2

⟹ (10.3/2) 2 = BD 2 +(8.3/2) 2

⟹ BD 2 = (10.3/2) 2 -(8.3/2) 2

⟹ (BD) 2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD 2  = 9.3

⟹ BD =  √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. Rationalize the denominators of the following:

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)

Multiply and divide 1/(√7-√6) by (√7+√6)

= (√7+√6)/√7 2 -√6 2 [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]

= (√7+√6)/(7-6)

= (√7+√6)/1

(iii) 1/(√5+√2)

Multiply and divide 1/(√5+√2) by (√5-√2)

= (√5-√2)/(√5 2 -√2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√7 2 -2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]

= (√7+2)/(7-4)

Exercise 1.6 Page: 26

64 1/2 = (8×8) 1/2

= 8 1 [⸪2×1/2 = 2/2 =1]

32 1/5 = (2 5 ) 1/5

= 2 1 [⸪5×1/5 = 1]

(iii)125 1/3

(125) 1/3 = (5×5×5) 1/3

= 5 1 (3×1/3 = 3/3 = 1)

9 3/2 = (3×3) 3/2

= (3 2 ) 3/2

= 3 3 [⸪2×3/2 = 3]

(ii) 32 2/5

32 2/5 = (2×2×2×2×2) 2/5

= (2 5 ) 2⁄5

= 2 2 [⸪5×2/5= 2]

(iii)16 3/4

16 3/4 = (2×2×2×2) 3/4

= (2 4 ) 3⁄4

= 2 3 [⸪4×3/4 = 3]

(iv) 125 -1/3

125 -1/3 = (5×5×5) -1/3

= (5 3 ) -1⁄3

= 5 -1 [⸪3×-1/3 = -1]

3. Simplify :

(i) 2 2/3 ×2 1/5

2 2/3 ×2 1/5 = 2 (2/3)+(1/5) [⸪Since, a m ×a n =a m+n ____ Laws of exponents]

= 2 13/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/3 3 ) 7

(1/3 3 ) 7 = (3 -3 ) 7 [⸪Since,(a m ) n = a m x n ____ Laws of exponents]

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

11 1/2 /11 1/4 = 11 (1/2)-(1/4)

= 11 1/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 7 1/2 ×8 1/2

7 1/2 ×8 1/2 = (7×8) 1/2 [⸪Since, (a m ×b m = (a×b) m ____ Laws of exponents]

As the Number System is one of the important topics in Maths, it has a weightage of 8 marks in Class 9 Maths CBSE exams. On an average three questions are asked from this unit.

• One out of three questions in part A (1 marks).
• One out of three questions in part B (2 marks).
• One out of three questions in part C (3 marks).

This chapter talks about:

• Introduction of Number Systems
• Irrational Numbers
• Real Numbers and their Decimal Expansions
• Representing Real Numbers on the Number Line.
• Operations on Real Numbers
• Laws of Exponents for Real Numbers

List of Exercises in NCERT Solutions for Class 9 Maths Chapter 1:

Exercise 1.1 Solutions 4 Questions ( 2 long, 2 short)

Exercise 1.2 Solutions 4 Questions ( 3 long, 1 short)

Exercise 1.3 Solutions 9 Questions ( 9 long)

Exercise 1.4 Solutions 2 Questions ( 2 long)

Exercise 1.5 Solutions 5 Questions ( 4 long 1 short)

Exercise 1.6 Solutions 3 Questions ( 3 long)

NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number System is the first chapter of Class 9 Maths. The Number System is discussed in detail in this chapter. The chapter discusses the Number Systems and their applications. The introduction of the chapter includes whole numbers, integers and rational numbers.

The chapter starts with the introduction of Number Systems in section 1.1, followed by two very important topics in sections 1.2 and 1.3

• Irrational Numbers – The numbers which can’t be written in the form of p/q.
• Real Numbers and their Decimal Expansions – Here, you study the decimal expansions of real numbers and see whether it can help in distinguishing between rational and irrational.

Next, it discusses the following topics.

• Representing Real Numbers on the Number Line – In this, the solutions for 2 problems in Exercise 1.4.
• Operations on Real Numbers – Here, you explore some of the operations like addition, subtraction, multiplication and division on irrational numbers.
• Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.

Explore more about Number Systems and learn how to solve various kinds of problems only on  NCERT Solutions For Class 9 Maths . It is also one of the best academic resources to revise for your CBSE exams.

Key Advantages of NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems

• These NCERT Solutions for Class 9 Maths help you solve and revise the whole CBSE syllabus of Class 9.
• After going through the step-wise solutions given by our subject expert teachers, you will be able to score more marks in the board exams.
• It follows NCERT guidelines.
• It contains all the important questions from the examination point of view.

The faculty have curated the solutions in a lucid manner to improve the problem-solving abilities of the students. For a more clear idea about Number Systems, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 9 Maths Number Systems

Disclaimer: 

Dropped Topics – 1.4 Representing real numbers on the number line.

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Assignments for Class 9, 10, 11 & 12

Assignments for all Classes and all chapters in PDF format will be available very soon, Revision questions, key points of each chapters and other practice material updated for new academic session 2024-25. Download NCERT Books and offline apps for all class updated for new academic year 2024-25. Assignments for the academic year 2024-25 based on latest CBSE Curriculum will be uploaded very soon.

Chapter wise assignments for class 9 Maths are given below updated for new academic session 2024-25. These assignments will be available in updated form along with new assignments and chapter wise tests with solutions. After the completion of chapters from NCERT Books and Exemplar Books, students should go for Assignments. Visit to Discussion Forum to ask your doubts related to NIOS and CBSE Board. Download NCERT Books and Offline Apps based on latest CBSE Syllabus.

Assignments Chapter 4:  1   2   3   4

Assignments Chapter 8:  1   2   3   4

Assignments Chapter 10:  1   2   3

Assignments Chapter 13:  1   2   3   4   5

Assignments Chapter 14:  1   2   3

Assignments Chapter 15:  1   2

Revision notes are very help full for practice. In the assignments of  Chapter 1  and  Chapter 2 , a lots of extra questions with answers are given for practice.

• Revision Notes for Chapter 1
• Revision Notes for Chapter 2
• Revision Notes for Chapter 3
• Revision Notes for Chapter 5
• Revision Notes for Chapter 6
• Revision Notes for Chapter 7
• Revision Notes for Chapter 12

We are placing all NCERT solutions in Hindi Medium as well as English Medium for CBSE Board, UP Board, Uttarakhand Board, Bihar Board, Gujrat Board and all other boards who are following NCERT Books 2024-25 for their exams. You can directly go for HINDI MEDIUM or ENGLISH MEDIUM solutions by just clicking on Exercise or Prashnavali as per your requirement. For example, Exercise 1.1 shows English Medium solutions for ex. 1.1 and Prashnavali 1.1 shows Hindi Medium solutions for Exercise 1.1. If you find still some difficulty, please inform us, we will try to rectify as soon as possible. Our only aim is to help the students. You may call us in Hindi or in English or leave message on Whats App, we will call you. You can post your questions through Discussion Forum to get proper answers.

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Basic concepts, definitions and formulas of mathematics, mathematics assignments for 9th standard to 10+2 standard, maths study material for 8th, 9th, 10th, 11th, 12th classes, Mathematics lesson plan for classes 8th,10th and 12th standard, Interesting maths riddles and maths magic, Class-wise mathematics study material for students from 8th to 12. A complete resource centre of mathematics for students and teachers

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Maths assignment class viii | quadrilateral, math assignment class ix ch-3 | coordinate geometry, mathematics assignment  class ix  chapter 3  coordinate geometry.

What   is Cartesian coordinate system

Answer: A system which involves the x-axis and y-axis on the two dimensional surface is called a cartesian coordinate system.

What we call the horizontal line in Cartesian coordinate system.

Answer: Horizontal line is called the x - axis.

What we call the vertical   line in Cartesian coordinate system.

Answer: Vertical Line is called the y - axis.

Which point is called the origin and what are its coordinates.

Answer: The point of intersection of x - axis and y - axis is called the Origin. Coordinates of origin are (0,0)

What is the point on x – axis.

Answer: Any point on the x - axis is (a, 0)

What is the point on y-axis.

Answer:  Any point on the y - axis is (0, b)

What do you mean by ordinates.

Answer: Point on y - axis is called ordinate.

What do you mean by abscissa.

Answer: Point on the x-axis is called abscissa.

Write the coordinates of the points given in the following graph. Also write In which quadrant or on which axis the following points lie.

Answer: 

A(3, 4) lie in I quadrant

B(-4, 3) lie in II quadrant

C(-2. -3) lie in III quadrant

D(3, -2) lie in IV quadrant

E(2, 0) lie on the x-axis

F(-2, 0) lie on the x-axis

G(0, 2) lie on y-axis

H(0, -2) lie on y-axis

Write the abscissa and ordinate of the following points

(-3, 5),     (0, 9),    (-5, 0),     (3,3),    (7, -9),   (12, 13)

Abscissa:  -3, 0, -5, 3, 7, 12

Ordinate:  5, 9, 0, 3, -9, 13

Write the coordinates of the point :

a) Whose ordinate is -5 and which lies on y- axis.

b) Which lies on x-axis and y – axes both.

c) Whose abscissa is -3 and which lies on x – axis.

b) Point on the x-axis is (a, 0) and Point on the y-axis is (0, b)

  Question 13:

What is the perpendicular distance of the point (3, 4) from

           a)   x – axis           b) Y – axis

a) 4 unit   b) 3 unit

What is the perpendicular distance of the point (7, 26) from

a) 26 unit   b) 7 unit

Name the quadrant in which abscissa and ordinate have different sign.

Ans: In II and IV quadrant abscissa and ordinate have different sign.

Plot the points P(1,0), Q(4,0), and S(1, 3). Find the coordinates of point R such that PQRS is a square. Also find the area of the square.     

Coordinates of R(4, 3),  Area of square PQRS = 9 square units)

Mark the points A(2, 2), B(2, -2), C(-2, -2) and D(-2, 2) on the graph paper and join these points in order. Identify the figure so obtained. Also find the area of the figure.  

Square, Area = 16cm 2 ]

Plot the points P(-1, 0), Q(0, 1) and R(2, 3) on the graph paper and check whether they are collinear or not.

Answer: If P, Q, R all points lie on a straight line then points are collinear.

If the points A(0, -6), B(0, 2 and C (a, 3) lie on the y-axis, then find the value of a.

Answer: For points on the y-axis the abscissa is "0" So the value of a = 0

From the given graph write :

a) The coordinates of the points B and F

b) The abscissas of points D and H

c) The ordinates of the points A and C

d) The perpendicular distance of the point G from x-axis 

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