Graph of a Linear Equation in 2 Variables - Linear Equations in 2 Variables, Class 9, Mathematics | Extra Documents & Tests for Class 9 PDF Download

GRAPH OF A LINEAR EQUATION IN ONE VARIABLES

In order to draw the graph of a linear equation in one variable we may follow the following algorithm.

STEP-I : Obtain the linear equation.
STEP-II : If the equation is of the form ax = b, a ≠ 0 then plot the point ba , and one more point   where α is any real number on the graph paper. If the equation is of the form ay = b, a ≠ 0, then plot the point , where β is any real number on the graph paper.

STEP-III : Join the points plotted in step-II to obtain the required line.

Ex. Draw the graph of each of the following linear equations :
(1) x – 3 = 0  

(2) x + 2 = 0  

(3) 2x – 5 = 0  

(4) 2x + 1 = 3x + 2

Sol. (1) The given equation x – 3 = 0 ⇒ x = 3.

For each value of y, x = 3. Thus, we have the following table :

Plot the points (3, –1), (3, 0), (3, 1) and (3, 2) on a graph and draw a line AB passing through these points.
Line AB is the required graph of the equation x – 3 = 0

(2 ) The given equation is x + 2 = 0 ⇒ x – 2

For each value of y, x = – 2 Thus, we have the following table :

Plot the points (–2, 2), (–2, 3) and (–2, –1) on the graph and draw a line AB passing through these points.
Line AB is the required graph of the equation x = –2.

(3) The given equation is 2x – 5 = 0

 on a graph draw a line AB passing through these points. Line AB is the required graph of the equation 2x – 5 = 0

(4) The given equation is 2x + 1 = 3x + 2 ⇒ 1 – 2 = 3x – 2x ⇒ –1 = x ⇒ x = –1.

Plot the points on a graph draw a line AB passing through these points. Line AB is the required graph of the equation 2x – 5 = 0

Plot the points (–1, 1), (–1, 2) and (–1, –1) on the graph and draw a line AB passing through these points. Line AB is the required graph of the equation x = – 1.

Ex. Draw the graph of each of the following linear equation :

(1) y – 2 = 0

(2) 2y + 3 = y + 6

Sol. (1) The given equation is y – 2 = 0 ⇒ y = 2.

For each value of x, y = 2.
Thus, we have the following table :

Plot the points (1, 2), (–2, 2) and (–1, 2) on the graph and draw a line AB passing through these points. Line AB is the required graph of the equation y – 2 = 0.

(2) The given equation is 2y + 3 = y + 6.

⇒ 2y – y = 6 – 3 ⇒ y = 3
For each value of x, y = 3.
Thus, we have the following table :

Plot the points (–2, 3), (1, 3) and (2, 3) on the graph and draw a line AB passing through these points. Line AB is the required graph of the equation y = 3.

GRAPH OF A LINEAR EQUATION IN TWO VARIABLES

In order to draw the graph of a linear equation ax + by + c = 0, a ≠ 0, b ≠ 0, we may follow the following algorithm.

STEP-I : Obtain the linear equation let the equation be ax + by + c = 0.
STEP-II : Express y in terms of x to obtain

STEP-III :Put any two or three values of x and calculate the corresponding values of y from the expression in step-II to obtain two solution say (α₁, β₁) and (α₂, β₂) if possible take values of x as integers in such a manner that the corresponding values of y are also integers.

STEP-IV : Plot points (α₁, β₁) and (α₂, β₂).

STEP-V :Join the points marked in step IV to obtain a line.

The line obtained is the graph of the equation ax + by + c = 0.

Ex. Draw the graph of the equation 3x – 2y = 7.

Thus, we have the following table exhibiting the abscissa and ordinates of the points on the line represented by the given equation.

Plotting the points A(3, 1), B(5, 4) and C(7, 7) on the graph paper and Joining the points A, B and C, we get a straight line.

Ex. Draw the graph of the equation 2x + 3y = 5. Check whether the points (–3, 4) and (7, –3) are solutions of the given equation.

∴ Table of values of x and y for the equation is :

Now plot the point A(–2, 3), B(1, 1) and C(4, –1) in the plane. Joining these points, we get line AC, the graph of the given equation.

As the point P(–3, 4) does not lie on the graph of the equation 2x + 3y = 5, so it is not a solution.

As the point Q(7, –3) lies on the graph of the equation 2x + 3y = 5, so it is a solution.

Alternatively, we can check whether a given point is a solution by substituting the coordinates of the point in the given equation – if it satisfies, we get a solution otherwise it is not a solution.

Now substituting x = – 3, y = 4 in the equation 2x + 3y = 5 ; we get

Ex. Draw the graph of the line x – 3y = 4. From the graph, find the ordered pair of the points when 

(i) y = – 1, 

(ii) x = –2.

When y = 0, x = 4
When y = 1, x = 7
and when y = 2, x = 10
∴ Table of points is

Plot the points A(4, 0), B(7, 1) and C(10, 2). Join these points. We get line AC, the graph of the equation x – 3y = 4.

(i) When y = –1, draw LM || X'OX, meeting the line AC at M. From M draw MN ⊥ X'OX.

We get ON = 1          ∴            When y = –1, x = 1

(ii) When x = –2, draw PQ || YOY', meeting the line AC at Q. From Q draw QR ⊥ Y'OY

We get OR = 2 and R is on OY'

∴ x = –2 gives y = –2

Thus when y = –1, x = 1 i.e., the point is (1, –1)

and when x = –2, y = –2, i.e., the point is (–2, –2)