Use graph paper for this question. Take 2 cm = 1 unit on both the axes.

(i) Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot only three points per line.

(ii) Write down the co-ordinates of the point of intersection of the lines.

(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.

(i)

First equation: x + y + 3 = 0

Step 1:

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

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

Let x = -4, then -4 + y + 3 = 0 ⇒ y = 1

Let x = -6, then -6 + y + 3 = 0 ⇒ y = 3

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 = 0

Step 1:

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

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

Let x = 2, then 3 $\times$ 2 - 2y + 4 = 0 ⇒ y = 5

Let x = 4, then 3 $\times$ 4 - 2y + 4 = 0 ⇒ y = 8

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.

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

Hence, co-ordinates of the point of intersection of the lines are (-2, -1).

(iii) In triangle OAB,

Using pythagoras theorem,

OA² = AB² + OB²

= 2² + 1²

= 4 + 1

= 5

OA = $\sqrt5$

OA = 2.2 cm

Hence, the distance of the point of intersection of the lines from the origin is 2.2 cm.