A triangle is formed by the straight lines x + 2y - 3 = 0, 3x - 2y + 7 = 0 and y + 1 = 0. Find graphically :

(i) the co-ordinates of the vertices of the triangle.

(ii) the area of the triangle.

First equation: x + 2y - 3 = 0

Step 1:

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

Let x = -1, then (-1) + 2y - 3 = 0 ⇒ y = 2

Let x = 0, then 0 + 2y - 3 = 0 ⇒ y = 1.5

Let x = 1, then 1 + 2y - 3 = 0 ⇒ y = 1

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 + 7 = 0

Step 1:

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

Let x = -1, then 3 $\times$ (-1) - 2y + 7 = 0 ⇒ y = 2

Let x = 0, then 3 $\times$ 0 - 2y + 7 = 0 ⇒ y = 3.5

Let x = 1, then 3 $\times$ 1 - 2y + 7 = 0 ⇒ y = 5

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.

Third equation: y + 1 = 0

That is, y = - 1, x = 0

Draw a straight line parallel to x-axis with y = -1.

(i) From the graph, the vertices of triangle ABC are A(-1, 2), B(5, -1) and C(-3, -1).

(ii) Area of triangle = $\dfrac{1}{2}$ x base x height

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

= $\dfrac{1}{2}$ x 8 x 3 = 4 x 3 = 12 sq. units.

Hence, area of the triangle = 12 sq. units.