Class 9 LINEAR EQUATIONS IN TWO VARIABLES FOR CBSE-NCERT

### Graph of a Linear Equation in Two Variables

So far, you have obtained the solutions of a linear equation in two variables algebraically. Now, let us look at their geometric representation. You know that each such equation has infinitely many solutions.

How can we show them in the coordinate plane? You may have got some indication in which we write the solution as pairs of values. The solutions of the linear equation in Example 3, namely,

x + 2y = 6

can be expressed in the form of a table as follows by writing the values of y below the corresponding values of x :

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).

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.
Q 3200267118

Given the point (1, 2), find the equation of a line on which it lies. How many such equations are there?
Class 9 Chapter 4 Example 5
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?
Q 3210267119

Draw the graph of x + y = 7.
Class 9 Chapter 4 Example 6
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:

Draw the graph by plotting the

two points from Table 2 and then

by joining the same by a line
Q 3210367210

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.
Class 9 Chapter 4 Example 7
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.
Q 3220367211

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
Class 9 Chapter 4 Example 8
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).