Use graph paper for this question. Take 2cm=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.

Answer

To solve the given question step by step, we will follow the instructions provided:

### Step 1: Write the equations
The equations given are:
1. \( x + y + 3 = 0 \)
2. \( 3x - 2y + 4 = 0 \)

### Step 2: Rearrange the equations to find y
For the first equation:
\[ y = -x - 3 \]

For the second equation:
\[ 3x - 2y + 4 = 0 \implies 2y = 3x + 4 \implies y = \frac{3x + 4}{2} \]

### Step 3: Find three points for each line
**For the first line \( y = -x - 3 \)**:
1. Let \( x = 0 \):
\[ y = -0 - 3 = -3 \implies (0, -3) \]
2. Let \( x = 1 \):
\[ y = -1 - 3 = -4 \implies (1, -4) \]
3. Let \( x = 2 \):
\[ y = -2 - 3 = -5 \implies (2, -5) \]

**For the second line \( y = \frac{3x + 4}{2} \)**:
1. Let \( x = 0 \):
\[ y = \frac{3(0) + 4}{2} = \frac{4}{2} = 2 \implies (0, 2) \]
2. Let \( x = 2 \):
\[ y = \frac{3(2) + 4}{2} = \frac{6 + 4}{2} = \frac{10}{2} = 5 \implies (2, 5) \]
3. Let \( x = 4 \):
\[ y = \frac{3(4) + 4}{2} = \frac{12 + 4}{2} = \frac{16}{2} = 8 \implies (4, 8) \]

### Step 4: Plot the points on graph paper
- For the first line, plot the points \( (0, -3) \), \( (1, -4) \), and \( (2, -5) \).
- For the second line, plot the points \( (0, 2) \), \( (2, 5) \), and \( (4, 8) \).

### Step 5: Draw the lines
- Draw a line through the points of the first equation.
- Draw a line through the points of the second equation.

### Step 6: Find the point of intersection
- The point of intersection can be found where the two lines cross. From the graph, identify the coordinates of the intersection point, which is \( (-2, -1) \).

### Step 7: Measure the distance from the origin to the point of intersection
- Use the distance formula or Pythagorean theorem:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where \( (x_1, y_1) = (0, 0) \) (the origin) and \( (x_2, y_2) = (-2, -1) \):
\[ \text{Distance} = \sqrt{(-2 - 0)^2 + (-1 - 0)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
- Approximating \( \sqrt{5} \approx 2.23 \) cm.

### Final Answers
1. The coordinates of the point of intersection are \( (-2, -1) \).
2. The distance from the origin to the point of intersection is approximately \( 2.23 \) cm.