The sides of a triangle are given by the equations y - 2 = 0; y + 1 = 3 (x - 2) and x + 2y = 0.

Find, graphically :

(i) the area of triangle;

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

(i)

First equation: y - 2 = 0

y = 2

Second equation: y + 1 = 3 (x - 2)

Step 1:

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

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

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

Let x = 4, then y + 1 = 3 (4 - 2) ⇒ 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: x + 2y = 0

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 = 0 ⇒ y = 0

Let x = 2, then 0 + 2y = 0 ⇒ y = -1

Let x = 4, then 4 + 2y = 0 ⇒ y = -2

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.

(i) The area of the triangle formed by the lines = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x AB x CD

= $\dfrac{1}{2}$ x 7 x 3

= $\dfrac{21}{2}$

= 10.5 sq. units

Hence, area of triangle = 10.5 sq. units.

(ii) The co-ordinates of A = (-4, 2)

The co-ordinates of B = (3, 2)

The co-ordinates of C = (2, -1)

Hence, the co-ordinates of the vertices of the triangle are (-4, 2), (3, 2) and (2, -1).