Draw the graphs of the following equations on the same graph paper :
3x - 2y = 9, 4y - 6x + 12 = 0
Find the area of trapezium formed by these two lines and the axes.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Convert the equations into slope-intercept form (if necessary) and find intercepts.

1. **For the first equation:**
\[ 3x - 2y = 9 \]
Rearranging gives:
\[ -2y = -3x + 9 \implies y = \frac{3}{2}x - \frac{9}{2} \]

- **Finding intercepts:**
- When \(x = 0\):
\[ y = -\frac{9}{2} = -4.5 \]
So, the y-intercept is \( (0, -4.5) \).
- When \(y = 0\):
\[ 0 = \frac{3}{2}x - \frac{9}{2} \implies \frac{3}{2}x = \frac{9}{2} \implies x = 3 \]
So, the x-intercept is \( (3, 0) \).

2. **For the second equation:**
\[ 4y - 6x + 12 = 0 \implies 4y = 6x - 12 \implies y = \frac{3}{2}x - 3 \]

- **Finding intercepts:**
- When \(x = 0\):
\[ y = -3 \]
So, the y-intercept is \( (0, -3) \).
- When \(y = 0\):
\[ 0 = \frac{3}{2}x - 3 \implies \frac{3}{2}x = 3 \implies x = 2 \]
So, the x-intercept is \( (2, 0) \).

### Step 2: Plot the points on the graph.

- The points to plot are:
- From the first equation: \( (0, -4.5) \) and \( (3, 0) \)
- From the second equation: \( (0, -3) \) and \( (2, 0) \)

### Step 3: Draw the lines.

- Draw a line through the points \( (0, -4.5) \) and \( (3, 0) \) for the first equation.
- Draw another line through the points \( (0, -3) \) and \( (2, 0) \) for the second equation.

### Step 4: Identify the trapezium formed.

- The trapezium is formed by the two lines and the x-axis. The vertices of the trapezium are:
- \( A(0, -3) \)
- \( B(2, 0) \)
- \( C(3, 0) \)
- \( D(0, -4.5) \)

### Step 5: Calculate the area of the trapezium.

- The area of the trapezium can be calculated by finding the area of the larger triangle \( ODC \) and subtracting the area of the smaller triangle \( OAB \).

1. **Area of triangle \( ODC \)**:
- Base \( OC = 3 \)
- Height \( OD = 4.5 \)
\[ \text{Area}_{ODC} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \times 4.5 = \frac{13.5}{2} = 6.75 \]

2. **Area of triangle \( OAB \)**:
- Base \( AB = 2 \)
- Height \( OA = 3 \)
\[ \text{Area}_{OAB} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 3 = 3 \]

3. **Area of the trapezium**:
\[ \text{Area}_{\text{trapezium}} = \text{Area}_{ODC} - \text{Area}_{OAB} = 6.75 - 3 = 3.75 \]

### Final Answer:
The area of the trapezium formed by the two lines and the axes is \( 3.75 \) square units.

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