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      Class 9 Maths

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      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        NCERT Solutions – Linear Equations Exercise 4.1 – 4.4

        Exercise 4.1

        Q.1       The cost of a notebook is twice the cost of a pen. Write  a linear equation in two  variables to represent this statement.
        Sol. 

        Let the cost of notebook be  Rs x and that of a pen  be Rs y.
        Since  the cost of a notebook is twice  the cost  of pen. So, the  required linear equation in two
        variables to represent  the above  statement is given by
        x = 2y.
        ⇒ x – 2y = 0


        Q.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−y5−10=0

                        (iii) – 2x + 3y = 6                                    
                     (iv) x = 3y

                        (v) 2x = – 5y     
                     (vi) 3x + 2 = 0

                        (vii) y – 2 = 0                                         
                     (viii)  5 = 2x

        Sol.

        (i)  2x+3y=9 can be written as 2x+3y−9.35¯¯¯=0
        On comparing  it with ax + by + c = 0 , we have
        a = 2, b = 3 and  c=−9.35¯¯¯
        (ii)  On comparing  x−y5−10=0  with  ax  + by + c = 0, we have,
        a=1,b=−15andc=−10
        (iii)  –2x + 3y = 6 can be  written as – 2x + 3y – 6 = 0
        On  comparing  it with  ax + by + c = 0 , we have
        a = – 2, b = 3 and c = – 6
        (iv)  x = 3y can  be written  as x – 3y = 0
        On  comparing  it with  ax + by + c = 0 , we have
        a = 1, b = – 3 and c = 0
        (v)  2x = – 5y can be written  as  2x + 5y = 0
        On  comparing  it with  ax + by + c = 0 , we have
        a = 2, b = 5 and c = 0
        (vi)  On  comparing  3x + 2 = 0 with  ax + by + c = 0 , we  have
        a = 3, b = 0 and c = 2
        (vii) On  comparing  y – 2 = 0 with  ax + by + c = 0,  we  have
        a = 0 , b = 1 and c = – 2.
        (viii)  5 = 2x can be written  as 2x – 5 = 0
        On  comparing  with  ax + by + c = 0 , we have
        a = 2, b = 0  and c = –5.
        or    – 2x + 5 = 0 , we have  a = – 2, b = 0 , c  = 5

        Exercise 4.2

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

        Given equation is y = 3 x + 5
        For y = 0 , 3x + 5 = 0 ⇒x=−53
        Therefore (−53,0) is one of solution.
        For x = 0, y = 0 + 5 = 5
        Therefore, (0, 5) is another solution.
        For x = 1, y = 3 × 1 + 5 = 8
        Therefore (1, 8) is another solution.
        Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with
        this value of y constitutes another solution of the given equation. So, there is no end to different
        solutions of a linear equation in two variables.
        Therefore, A linear equation in two variables has infinitely many solutions.


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

        Sol.

        (i) The given equation can be written as y = 7 – 2x.
        For x = 0 , y = 7 – 2 × 0 = 7 – 0 = 7
        For x = 1, y = 7 – 2 × 1 = 7 – 2 = 5
        For x = 2, y = 7 – 2 × 2 = 7 – 4 = 3
        For x = 3, y = 7 – 2 × 3 = 7 – 6 = 1
        Therefore the four solutions of the given equation are (0, 7), (1, 5) , (2, 3) and (3, 1) .
        (ii) The given equation can be written as  y=9−πx
        For x = 0, y = 9 – 0 = 9
        For x = 1, y=9−π
        For x = 2 , y=9−2π
        For x = 3, y=9−3π
        Therefore the four solutions of the given equation are (0, 9),
        (1,9−π) , (2,9 −2π) and (3, 9 −3π)
        (iii) The given equation can be written as x = 4y
        For x = 0,   y = 0
        For  x = 1 ;  y=14
        For x = 2 ;  y=24=12
        For x = 3 ; y=34=34
        Therefore the four solutions of the given equation are (0,0)(1,14)(2,12)and(3,34)


        Q.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–√,42–√)(v) (1, 1)
        Sol.

        (i) Putting x = 0 , y = 2 in L.H.S. of x – 2y = 4, we have
        L.H.S. = 0 – 2 × 2 = – 4 ≠ R.H.S.
        Therefore x = 0 , y = 2 is not its solution.
        (ii) Putting x = 2, y = 0 in L.H.S. of x – 2y = 4, we have
        L.H.S. = 2 – 2 × 0 = 2 – 0 = 2 ≠ R.H.S.
        Therefore x = 2, y = 0 is not its solution.
        (iii) Putting x = 4, y = 0 in the L.H.S. of x – 2y = 4, we have
        L.H.S. = 4 – 0 = 4 = R.H.S.
        Therefore x = 4, y = 0 is its solution.
        (iv) Putting x=2–√,y=42–√ in the L.H.S. of x – 2y = 4, we have
        L.H.S. x=2–√−2×42–√=2–√−82–√
        =−72–√≠ R.H.S.
        Therefore (2–√,42–√) is not its solution.
        (v) Putting x = 1, y = 1 in the L.H.S. of x – 2y = 4, we have
        L.H.S. = 1 – 2 × 1 = 1 – 2 = – 1 ≠ R.H.S.
        Therefore x = 1, y = 1 is not its solution.


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

        If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation.
        Therefore 2 × 2 + 3 × 1 = k ⇒ k = 4 + 3 = 7.

        Exercise 4.3

        Q.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
        Sol.

        (i) We have x + y = 4 ⇒ y = 4 – x
        When x = 0 , we have : y = 4 – 0 = 4
        When x = 2, we have : y = 4 – 2 = 2
        When x = 4, we have : y = 4 – 4 = 0
        Thus, we have the following table :
        1
        Plotting the points (0, 4) (2, 2) and (4, 0) on the graph paper and drawing a line joining them, we obtain
        the graph of the line represented by the given equation as shown.
        2

        (ii) We have, x – y = 2 ⇒ y = x – 2
        When x = 0 , we have y = 0 – 2 = – 2
        When x = 2, we have y = 2 – 2 = 0
        When x = 4, we have y = 4 – 2 = 2
        Thus, we have the following table :
        3
        Plotting the points (0, –2), (2, 0) and (4, 2) on the graph paper and drawing a line joining them, we obtain
        the graph of the line represented by the given equation as shown.
        4
        (iii)We have y = 3x
        When x = 0 , we have : y = 3(0) = 0
        When x = 1, we have : y = 3 (1) = 3
        When x = –1, we have : y = 3(–1) = – 3
        Thus , we have the following table :
        5

        Plotting the points (0, 0), (1, 3) and (–1, –3) on the graph paper and drawing a line joining them, we
        obtain the graph of the line represented by the given equation as shown.
        6(iv) We have , 3 = 2x + y ⇒ y = 3 – 2x
        When x = 0, we have :
        y = 3 – 2(0) = 3 – 0 = 3
        When x = 3, we have :
        y = 3 – 2(3) = 3 – 6 = – 3
        When x = – 1, we have :
        y = 3 – 2 (–1) = 3 + 2 = 5
        Thus , we have the following table :
        7
        Plotting the points (0, 3), (3, –3) and (–1, 5) on the graph paper and drawing a line joining them, we
        obtain the graph of the line represented by the given equation as shown
        8


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

        Here (2, 14) is a solution of a linear equation we are looking for. One example of such a linear equation is 7x – y = 0.  Note that  x + y = 16, 2x + y = 18 and 7x + y = 28 are also satisfied by the co-ordinates of the point (2, 14). So any line passing through the point (2, 14) is an example of a linear equation for which (2, 14) is a solution. Thus there are infinite number of lines through (2, 14).


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

        Since (3, 4) lies on the graph corresponding to 3y = ax + 7. Therefore, (3, 4) satisfies the given equation.
        That is , 3(4) = a(3) + 7
        ⇒ 12 – 7 = 3a
        ⇒ 3a = 5
        ⇒ a=53
        Hence for a=53, (3, 4) lie on the graph of equation 3y = ax + 7.


        Q.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.
        Sol.

        Taxi fare for first km = Rs 8
        Taxi fare for the subsequent km = Rs 5
        Total fare = Rs y
        Total distance = x km
        The linear equation for the above information is given by
        y = 8 × 1 + 5(x – 1)
        ⇒ y = 8 + 5x – 5
        ⇒ y = 5x + 3
        ⇒ 5x – y + 3 = 0
        When x = 0, y = 5 × 0 + 3 = 0 + 3 = 3
        When x = – 1, y = 5 ×( – 1) + 3 = – 5 + 3 = – 2
        When x = – 2, y = 5 × (– 2) + 3 = – 10 + 3 = – 7
        Thus, we have the following table :
        9
        Plotting the points (0, 3), (–1, –2) and (–2, –7) on the graph paper and drawing a line joining them, we,obtain the required graph.
        10


        Q.5     From the choices given below, choose the equation whose graphs are given in Fig(a) and Fig(b)
                   
        For figure (a)                 For figure (b)
                      (i) y = x b                        (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

        11                  12

        Sol.

        For Fig (a), the correct equation from the choices given is clearly x + y = 0 as it is satisified by the points (–1,1) and (1, –1) given on the graph.
        For Fig (b), the correct equation from the choices given is clearly y = – x + 2 as it is satisfied by the points (–1, 3) (0, 2) and (2, 0) given on the graph.


        Q.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.

        Sol.

        Let x be the distance and y be the work done. Therefore, according to the problem the equation will be y = 5x.
        To draw its graph :
        When x = 0 , we have , y = 5(0) = 0
        When x = 1, we have y = 5 (1) = 5
        When x = – 1, we have y = 5 (–1) = – 5
        Therefore the table is as under :
        13
        Plotting (0, 0) , (1,5) and (–1, –5) on the graph paper and drawing a line joining them we obtain the graph of
        the line y = 5x as shown.
        14From the graph , clearly,
        (i) when distance travelled is 2 units. i.e. x = 2, then y = 10.
        Therefore Work done = 10
        (ii) When distance travelled is 0 units i.e., x = 0 , then y = 0
        Therefore Work done = 0.


        Q.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.
        Sol.

        Let Yamini and Fatima contributed Rs x and Rs y towards the P.M.’s Relief Fund totally Rs 100.
        Therefore the linear equation using the above data is x + y = 100, i.e. y = 100 – x.
        To draw its graph :
        When x = 0, we have y = 100 – 0 = 100
        When x = 100 we have y = 100 – 100 = 0
        When x = 50 , we have y = 100 – 50 = 50
        The table for these values is as under
        15
        Plotting the points (0, 100), (100, 0) and (50, 50) on the graph paper and drawing a line joining them, we obtain the graph of the line x + y = 100 as shown.
        16


        Q.8     In countries like the 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=(95)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 30ºC, what is the temperature in Fahrenheit ?
        (iii) If the temperature is 95º F, what is the temperature in Celsius ?
        (iv) If the temperature is 0º C, what is the temperature in Fahrenheit and if the temperature is 0ºF, 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.
        Sol.

        (i) We have F=95C+32⇒C=59(F−32)
        We calculate the vales of F for different values of C.
        When C=−40,F=95×(−40)+32=−72+32=−40
        When C=10,F=95×10+32=18+32=50
        The table of values is as under :
        17
        We choose a suitable scale on the x-axis (for Celsius) and on the same scale on the y-axis (for Fahrenheit).
        18Plotting the points (– 40, – 40) and (10, 50) on the graph point. On joining these points by line segment, we obtain the graph of  F=95C+32.
        (ii) From the graph, we see that when C = 30º shown C1 on the x-axis in the positive direction, then F =86º shown by the point F1 on the y-axis in the positive direction.
        Hence 30ºC = 86ºF
        (iii) From the graph, we see that when F = 95º shown F2 on the y-axis in the positive direction, then C = 35º shown by the point C2 on the x-axis in the positive direction.
        Hence 95º F = 35ºC
        (iv) From the graph, clearly 0ºC = 32ºF and 0ºF = – 17.8ºF.
        (v) Clearly from the graph the temperature which is numerically the same in both Fahrenheit and Celsius is – 40º i.e. – 40ºC = – 40º F.

        Exercise 4.4

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

        (i) The representation of the solution on the  number when y = 3 is treated as an equation in one variable is as under
        19
        (ii)  We know that y = 3 can be written as 0. x + y = 3 as a linear equation in variables x and y. Now all the values of x are permissible as 0. x is always 0. However, y must satisfy the  relation y = 3.
        Hence , three  solution of the given equation are
        x = 0, y = 3 ;      x = 2, y = 3   ;    x = – 2 , y = 3
        22
        Plotting the  points (0, 3) (2, 3) and  (–2, 3) and on joining them we get the graph AB as a line  parallel to x-axis at a distance of 3 units above it.


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

        (i) The  representation of the solution on the number line 2x + 9 = 0 i.e., x=−92 is treated as an  equation in one variable is as under. 20(ii) We know  that 2x + 9 = 0 can be written as 2x + 0y + 9  = 0 as a linear equation in variables x and y. Now all the  values of x are  permissible  as 0. y is always 0. However, x must  satisfy the relation 2x + 9 = 0
        i.e., x=−92.
        Hence three solution of the  given  equation are  x=−92,y = 0 ; x=−92,y = 2 and x=−92,y =– 2.
        Therefore  Plotting  the point (−92,0),(−92,2)and(−92,−2)and on  joining them we get the graph AB as a line  parallel to y-axis at a distance of 92 on the left of y-axis.
        21

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