Page 45 - Mathematics Class - XI
P. 45

ACTIVITY 5.2






        OBJECTIVE
        To verify that the graph of a given inequality, say 5x + 4y – 40 < 0, of the form ax + by + c < 0, a, b > 0 and
        c < 0 represents only one of the two half planes.


        MATERIAL REQUIRED
            Cardboard                           Graph paper                        Ruler
            White chart paper                   Sketch pen                         Glue


        PRE-REQUISITE KNOWLEDGE
            1.  Knowledge of linear equations in one variable and two variables

            2.  Knowledge of inequation and sign of inequality
            3.  Knowledge of making a graph of linear equation in two variables


        PROCEDURE
            1.  Take a cardboard of a convenient size and paste a white chart paper on it.
            2.  Paste a graph paper on the white chart paper.
            3.  On the  graph  paper, draw two  perpendicular  lines  X′OX  and  YOY′  to  represent  x-axis  and  y-axis,
               respectively.

            4.  Find the ordered pairs of the form (x,  y) which
               satisfy  the  given  equation  5x + 4y  –  40  =  0,
               corresponding to the given inequality.

                                A           B          C


                     x          0           4          8

                     y          10          5          0



            5.  Plot the points A (0, 10), B (4, 5) and C (8, 0) on
               the graph as shown in Fig. (a).
            6.  On the graph, mark the two half planes as I and II
               as shown in Fig. (a).

                                                                                          Fig. (a)
        DEMONSTRATION
            1.  Mark some points O (0, 0), D (1, 2), E (3, 2), F (4, 3), G (–2, –2) in half plane I and points H (5, 7), I (7, 5),
               J (8, 4), K (9, 5) in half plane II, as shown in Fig. (b).
            2.  Put these coordinates in the left hand side of the inequality 5x + 4y – 40 < 0 to verify.
                 (i) For point O (0, 0),
                    Value of LHS = 5(0) + 4(0) – 40 = – 40 < 0

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