Find the area of the triangular region bonded by x= 0, y = x and x + 2y =6

Answer

To find the area of the triangular region bounded by the lines \(x = 0\), \(y = x\), and \(x + 2y = 6\), we will follow these steps:

### Step 1: Identify the lines and their intersections
1. **Line 1:** \(x = 0\) is the y-axis.
2. **Line 2:** \(y = x\) is a line passing through the origin at a 45-degree angle.
3. **Line 3:** \(x + 2y = 6\) can be rewritten in slope-intercept form as \(y = -\frac{1}{2}x + 3\).

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

2. **Intersection of \(y = x\) and \(x + 2y = 6\):**
- Substitute \(y = x\) into \(x + 2y = 6\):
\[ x + 2x = 6 \implies 3x = 6 \implies x = 2 \]
- Since \(y = x\), \(y = 2\).
- Point of intersection: \((2, 2)\).

3. **Intersection of \(x = 0\) and \(y = x\):**
- Substitute \(x = 0\) into \(y = x\):
\[ y = 0 \]
- Point of intersection: \((0, 0)\).

### Step 3: Identify the vertices of the triangle
The vertices of the triangle formed by the lines are:
- \(A(0, 3)\)
- \(B(2, 2)\)
- \(C(0, 0)\)

### Step 4: Calculate the area of the triangle
The area \(A\) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case:
- The base \(AC\) is the vertical distance from \(A(0, 3)\) to \(C(0, 0)\), which is \(3\) units.
- The height \(BD\) is the horizontal distance from \(B(2, 2)\) to the y-axis (line \(x = 0\)), which is \(2\) units.

Thus, the area is:
\[ A = \frac{1}{2} \times 3 \times 2 = \frac{1}{2} \times 6 = 3 \text{ square units} \]

### Final Answer
The area of the triangular region bounded by the lines \(x = 0\), \(y = x\), and \(x + 2y = 6\) is \(3\) square units.

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