Solve graphically, the following equations.

x + 2y = 4; 3x - 2y = 4.

Take 2 cm = 1 unit on each axis.

Also, find the area of the triangle formed by the lines and the x-axis.

First equation: x + 2y = 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 0 + 2y = 4 ⇒ y = 2

Let x = 2, then 2 + 2y = 4 ⇒ y = 1

Let x = 4, then 4 + 2y = 4 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 3x - 2y = 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 - 2y = 4 ⇒ y = -2

Let x = 2, then 3 $\times$ 2 - 2y = 4 ⇒ y = 1

Let x = 4, then 3 $\times$ 4 - 2y = 4 ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Both the straight line drawn meet the point A. As it is clear from the graph, co-ordinates of the common point A are (2, 1).

Solution of the given equation x = 2 and y = 1.

The area of the triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x BC x AD

= $\dfrac{1}{2}$ x 2.6 x 1

= 1.3 sq. units

Hence, the area of the triangle formed between the lines and the x-axis = 1.3 sq. units.