What is the area bounded by the lines `x=0,y=0` and `x+y+2=0`?

Options

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

Answer

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

### Step 1: Identify the lines
The lines given are:
1. \(x = 0\) (the y-axis)
2. \(y = 0\) (the x-axis)
3. \(x + y + 2 = 0\) (which can be rewritten as \(y = -x - 2\))

### Step 2: Find the intercepts of the line \(x + y + 2 = 0\)
To find the points where the line intersects the axes:
- **For the y-intercept** (set \(x = 0\)):
\[ y = -0 - 2 = -2 \quad \text{(Point: (0, -2))} \]
- **For the x-intercept** (set \(y = 0\)):
\[ 0 = -x - 2 \implies x = -2 \quad \text{(Point: (-2, 0))} \]

### Step 3: Plot the points and the lines
We have the points:
- \(A(0, -2)\)
- \(B(-2, 0)\)
- The origin \(O(0, 0)\)

Now, we can plot these points on the Cartesian plane. The line \(x + y + 2 = 0\) will pass through points \(A\) and \(B\).

### Step 4: Determine the shape formed
The area bounded by these lines and the axes forms a right triangle with vertices at \(O(0, 0)\), \(A(0, -2)\), and \(B(-2, 0)\).

### Step 5: Calculate the area of the triangle
The area \(A\) of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case:
- The base is the distance along the x-axis from \(O\) to \(B\), which is \(2\) units.
- The height is the distance along the y-axis from \(O\) to \(A\), which is also \(2\) units.

Substituting these values into the area formula:
\[ \text{Area} = \frac{1}{2} \times 2 \times 2 = \frac{1}{2} \times 4 = 2 \text{ square units} \]

### Conclusion
The area bounded by the lines \(x=0\), \(y=0\), and \(x+y+2=0\) is \(2\) square units.

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