The area of the triangle formed by the graphs of the equations `x = 0, 2x + 3y = 6` and `x + y = 3` is :

Options

A. 3 sq. unit
B. `4 1/2 sq. unit`
C. `1 1/2 sq. unit`
D. 1 sq. unit

Answer

To find the area of the triangle formed by the graphs of the equations \(x = 0\), \(2x + 3y = 6\), and \(x + y = 3\), we will follow these steps:

### Step 1: Identify the equations and their intersections
We have three equations:
1. \(x = 0\) (the y-axis)
2. \(2x + 3y = 6\)
3. \(x + y = 3\)

### Step 2: Find the intersection points of the lines
- **Intersection of \(x = 0\) and \(2x + 3y = 6\)**:
Substitute \(x = 0\) into \(2x + 3y = 6\):
\[ 2(0) + 3y = 6 \implies 3y = 6 \implies y = 2 \]
So, the point is \((0, 2)\).

- **Intersection of \(x = 0\) and \(x + y = 3\)**:
Substitute \(x = 0\) into \(x + y = 3\):
\[ 0 + y = 3 \implies y = 3 \]
So, the point is \((0, 3)\).

- **Intersection of \(2x + 3y = 6\) and \(x + y = 3\)**:
We can solve these two equations simultaneously. From \(x + y = 3\), we can express \(y\) in terms of \(x\):
\[ y = 3 - x \]
Substitute \(y\) into \(2x + 3y = 6\):
\[ 2x + 3(3 - x) = 6 \implies 2x + 9 - 3x = 6 \implies -x + 9 = 6 \implies -x = -3 \implies x = 3 \]
Now substitute \(x = 3\) back into \(y = 3 - x\):
\[ y = 3 - 3 = 0 \]
So, the point is \((3, 0)\).

### Step 3: Identify the vertices of the triangle
The vertices of the triangle formed by the intersection points are:
1. \(A(0, 2)\)
2. \(B(0, 3)\)
3. \(C(3, 0)\)

### Step 4: Calculate the area of the triangle
The area \(A\) of a triangle formed by the vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the points \(A(0, 2)\), \(B(0, 3)\), and \(C(3, 0)\):
\[ A = \frac{1}{2} \left| 0(3 - 0) + 0(0 - 2) + 3(2 - 3) \right| \]
\[ = \frac{1}{2} \left| 0 + 0 + 3(-1) \right| = \frac{1}{2} \left| -3 \right| = \frac{3}{2} \]

### Final Answer
The area of the triangle is \(\frac{3}{2}\) square units.