Question

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

1. Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot only three points per line.
2. Write down the coordinates of the point of intersection of the lines.
3. Measure and record the distance of the point of intersection of the lines from the origin in cm.

Answer 1

(i) x + y + 3 = 0

⇒ x = - 3 - y

The table for x + y = 3 = 0 is

X	1	0	- 2
Y	- 4	- 3	- 1

Also, we have

3x - 2y + 4 = 0

⇒ 3x = 2y - 4

⇒ x = `(2y - 4)/(3)`

The table for 3x - 2y + 4 = 0 is

X	0	- 2	`- (2)/(3)`
Y	2	- 1	1

Plotting the above points we get the following required graph:

(ii) From the above graph, it is clear that the two lines x + y + 3 = 0 and 3x - 2y + 4 = 0 intersect at the point (-2, -1)

(iii) Applying Pythagoras Theorem,
the distance from the origin

= `sqrt((-2-0)^2 + (-1-0)^2)`

= `sqrt(2^2 + 1^2)`

= `sqrt(4 + 1)`

= `sqrt(5)`

= 2.2 cm (approx).