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

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

(ii) Write down the coordinates 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) Given,

⇒ x + y + 3 = 0

⇒ y = -(3 + x) ………(1)

When x = -1, y = -[3 + (-1)] = -(3 - 1) = -2,

x = 0, y = -(3 + 0) = -3,

x = 1, y = -(3 + 1) = -4.

Table of values for equation (1)

Steps of construction :

1. Plot the points (-1, -2), (0, -3) and (1, -4) on graph paper.

2. Connect points by straight line.

Given,

⇒ 3x - 2y + 4 = 0

⇒ 2y = 3x + 4

⇒ y = $\dfrac{3x + 4}{2}$

When x = -2, y = $\dfrac{3 \times -2 + 4}{2} = \dfrac{-6 + 4}{2} = \dfrac{-2}{2}$ = -1,

x = 0, y = $\dfrac{3 \times 0 + 4}{2} = \dfrac{4}{2}$ = 2,

x = 2, y = $\dfrac{3 \times 2 + 4}{2} = \dfrac{6 + 4}{2} = \dfrac{10}{2}$ = 5.

Table of values for equation (2)

Steps of construction :

1. Plot the points (-2, -1), (0, 2) and (2, 5) on graph paper.

2. Connect points by straight line.

(ii) From graph,

P(-2, -1) is the intersection of lines.

Hence, coordinates of point of intersection = (-2, -1).

(iii) From P, draw a perpendicular to y-axis.

As, 1 unit = 2 cm.

So, PQ = 2 unit = 4 cm, OQ = 1 unit = 2 cm.

In right angle triangle,

By pythagoras theorem,

⇒ OP² = PQ² + OQ²

⇒ OP² = 4² + 2²

⇒ OP² = 16 + 4

⇒ OP² = 20

⇒ OP = $\sqrt{20}$ = 4.5 cm

Hence, distance of the point of intersection of the lines from the origin = 4.5 cm.