In the previous chapter, you studied how to plot the points on a graph paper. Let us plot the points (0, 3), (2, 2), (4, 1) and (6, 0) on a graph paper. Now join any two of these points and obtain a line. Let us call this as line AB (see Fig. 4.2)

.

Fig. 4.2

Do you see that the other two points also lie on the line AB? Now, pick another point on this line, say (8, –1). Is this a solution? In fact,
8 + 2(–1) = 6. So, (8, –1) is a solution. Pick any other point on this line AB and verify whether its coordinates satisfy the equation or not. Now, take

any point not lying on the line AB, say (2, 0). Do its coordinates satisfy the equation? Check, and see that they do not.

Let us list our observations:

1. Every point whose coordinates satisfy Equation (1) lies on the line AB.

2. Every point (a, b) on the line AB gives a solution x = a, y = b of Equation (1).

3. Any point, which does not lie on the line AB, is not a solution of Equation (1).

So, you can conclude that every point on the line satisfies the equation of the line and every solution of the equation is a point on the line. In fact, a linear equation in two variables is represented geometrically by a line whose points make up the collection of solutions of the equation. This is called the graph of the linear equation. So, to obtain the graph of a linear equation in two variables, it is enough to plot two points corresponding to two solutions and join them by a line. However, it is advisable to plot more than two such points so that you can immediately check the correctness of the graph.

Remark : The reason that a, degree one, polynomial equation ax + by + c = 0 is called a linear equation is that its geometrical representation is a straight line.

Example 5 : Given the point (1, 2), find the equation of a line on which it lies. How many such equations are there?

Solution : Here (1, 2) is a solution of a linear equation you are looking for. So, you are looking for any line passing through the point (1, 2). One example of such a linear equation is x + y = 3. Others are y – x = 1, y = 2x, since they are also satisfied by the coordinates of the point (1, 2). In fact, there are infinitely many linear equations which are satisfied by the coordinates of the point (1, 2). Can you see this pictorially?

Example 6 : Draw the graph of x + y = 7.

Solution : To draw the graph, we need at least two solutions of the equation. You can check that x = 0, y = 7, and x = 7, y = 0 are solutions of the given equation. So, you can use the following table to draw the graph:

Table 2

Draw the graph by plotting the two points from Table 2 and then

by joining the same by a line (see Fig. 4.3).

Example 7 : You know that the force applied on a body is directly proportional to the acceleration produced in the body. Write an equation to express this situation and plot the graph of the equation.

Solution : Here the variables involved are force and acceleration. Let the force applied be y units and the acceleration produced be x units. From ratio and proportion, you can express this fact as y = kx, where k is a constant. (From your study of science, you know that k is actually the mass of the body.)

Now, since we do not know what k is, we cannot draw the precise graph of y = kx. However, if we give a certain value to k, then we can draw the graph. Let us take k = 3, i.e., we draw the line representing y = 3x.

For this we find two of its solutions, say (0, 0) and (2, 6) (see Fig. 4.4).

From the graph, you can see that when the force applied is 3 units, the acceleration produced is 1 unit. Also, note that (0, 0) lies on the graph which means the acceleration produced is 0 units, when the force applied is 0 units.

Remark : The graph of the equation of the form y = kx is a line which always passes through the origin.

Example 8 : For each of the graphs given in Fig. 4.5 select the equation whose graph it is from the choices given below:

(a) For Fig. 4.5 (i),

(i) x + y = 0 (ii) y = 2x (iii) y = x (iv) y = 2x + 1

(b) For Fig. 4.5 (ii),

(i) x + y = 0 (ii) y = 2x (iii) y = 2x + 4 (iv) y = x – 4

(c) For Fig. 4.5 (iii),

(i) x + y = 0 (ii) y = 2x (iii) y = 2x + 1 (iv) y = 2x – 4

(i)

Fig. 4.5

Solution : (a) In Fig. 4.5 (i), the points on the line are (–1, –2), (0, 0), (1, 2). By inspection, y = 2x is the equation corresponding to this graph. You can find that the y-coordinate in each case is double that of the x-coordinate.

(b) In Fig. 4.5 (ii), the points on the line are (–2, 0), (0, 4), (1, 6). You know that the coordinates of the points of the graph (line) satisfy the equation y = 2x + 4. So,
y = 2x + 4 is the equation corresponding to the graph in Fig. 4.5 (ii).

(c) In Fig. 4.5 (iii), the points on the line are (–1, –6), (0, –4), (1, –2), (2, 0). By inspection, you can see that y = 2x – 4 is the equation corresponding to the given graph (line).

4.5 Equations of Lines Parallel to the x-axis and y-axis

You have studied how to write the coordinates of a given point in the Cartesian plane. Do you know where the points (2, 0), (–3, 0), (4, 0) and (n, 0), for any real number
n, lie in the Cartesian plane? Yes, they all lie on the x-axis. But do you know why? Because on the x-axis, the y-coordinate of each point is 0. In fact, every point on the
x-axis is of the form (x, 0). Can you now guess the equation of the x-axis? It is given by y = 0. Note that y = 0 can be expressed as 0.x + 1.y = 0. Similarly, observe that the equation of the y-axis is given by x = 0.

Now, consider the equation x – 2 = 0. If this is treated as an equation in one variable x only, then it has the unique solution x = 2, which is a point on the number line. However, when treated as an equation in two variables, it can be expressed as

x + 0.y – 2 = 0. This has infinitely many solutions. In fact, they are all of the form
(2, r), where r is any real number. Also, you can check that every point of the form (2, r) is a solution of this equation. So as, an equation in two variables, x – 2 = 0 is represented by the line AB in the graph in Fig. 4.8.

Fig. 4.8

Example 9 : Solve the equation
2x + 1 = x – 3, and represent the solution(s) on (i) the number line,

(ii) the Cartesian plane.

Solution : We solve 2x + 1 = x – 3, to get

2x – x = –3 – 1

i.e., x = –4

(i) The representation of the solution on the number line is shown in Fig. 4.9, where
x = – 4 is treated as an equation in one variable.

Fig. 4.9

ii) We know that x = – 4 can be written as
x + 0.y = – 4

which is a linear equation in the variables
x and y. This is represented by a line. Now all the values of y are permissible because 0.y is always 0. However, x must satisfy the equation x = – 4. Hence, two solutions of the given equation are x = – 4, y = 0 and x = – 4, y = 2.

Note that the graph AB is a line parallel to the y-axis and at a distance of 4 units to the left of it (see Fig. 4.10).

Fig. 4.10

Similarly, you can obtain a line parallel to the x-axis corresponding to equations of the type

y = 3 or 0.x + 1.y = 3

4.6 Summary

In this chapter, you have studied the following points:

1. An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.

2. A linear equation in two variables has infinitely many solutions.

3. The graph of every linear equation in two variables is a straight line.

4. x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.

5. The graph of x = a is a straight line parallel to the y-axis.

6. The graph of y = a is a straight line parallel to the x-axis.

7. An equation of the type y = mx represents a line passing through the origin.

8. 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.