As shown in the figure, line l1: y = 2x + 3 intersects with l2: y = —x + 6 at point B. Line l1 intersects with the x-axis at point D and with the y-axis at point E; line l2 intersects with the x-axis at point C and with the y-axis at point A. The area of the quadrilateral OCBE is

To find the area of the quadrilateral \(OCBE\) formed by the intersection points of the lines \(l_1: y = 2x + 3\) and \(l_2: y = -x + 6\) with the axes, let's follow these steps:

### Step 1: Find the intersection point \(B\) of \(l_1\) and \(l_2\).
Since \(B\) lies on both lines \(y = 2x + 3\) and \(y = -x + 6\), equate the two equations:

\[
2x + 3 = -x + 6
\]
Solve for \(x\):
\[
2x + x = 6 - 3 \quad \Rightarrow \quad 3x = 3 \quad \Rightarrow \quad x = 1
\]
Now, substitute \(x = 1\) into either equation (say \(y = 2x + 3\)):
\[
y = 2(1) + 3 = 5
\]
Thus, the coordinates of \(B\) are:
\[
B(1, 5)
\]

---

### Step 2: Find the intercepts of the lines with the axes.

1. **For line \(l_1: y = 2x + 3\):**
   - **x-intercept \(D\)**: Set \(y = 0\).
     \[
     0 = 2x + 3 \quad \Rightarrow \quad 2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2}
     \]
     So, \(D\left(-\frac{3}{2}, 0\right)\).
   - **y-intercept \(E\)**: Set \(x = 0\).
     \[
     y = 2(0) + 3 = 3
     \]
     So, \(E(0, 3)\).

2. **For line \(l_2: y = -x + 6\):**
   - **x-intercept \(C\)**: Set \(y = 0\).
     \[
     0 = -x + 6 \quad \Rightarrow \quad x = 6
     \]
     So, \(C(6, 0)\).
   - **y-intercept \(A\)**: Set \(x = 0\).
     \[
     y = -0 + 6 = 6
     \]
     So, \(A(0, 6)\).

---

### Step 3: Find the vertices of the quadrilateral \(OCBE\).
- **O**: Origin \((0, 0)\)
- **C**: \((6, 0)\)
- **B**: \((1, 5)\)
- **E**: \((0, 3)\)

---

### Step 4: Use the Shoelace Formula to find the area.

The Shoelace Formula for the area of a polygon with vertices \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\) is:

\[
\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
\]

For the quadrilateral \(OCBE\) with vertices \(O(0, 0)\), \(C(6, 0)\), \(B(1, 5)\), and \(E(0, 3)\), the formula becomes:

\[
\text{Area} = \frac{1}{2} \left| (0 \cdot 0 + 6 \cdot 5 + 1 \cdot 3 + 0 \cdot 0) - (0 \cdot 6 + 0 \cdot 1 + 5 \cdot 0 + 3 \cdot 0) \right|
\]

Simplify:

\[
\text{Area} = \frac{1}{2} \left| 0 + 30 + 3 + 0 - (0 + 0 + 0 + 0) \right| = \frac{1}{2} \left| 33 \right| = \frac{33}{2}
\]

---

### Step 5: Final Answer

The area of the quadrilateral \(OCBE\) is:

\[
\boxed{\frac{33}{2}} \, \text{square units}.
\]

---

Do you want further details on any step? 

Here are 5 related questions:
1. How do we determine the area of polygons with more than four sides?
2. Can the Shoelace Theorem be applied to 3D objects?
3. How does the area change if one of the lines has a different slope?
4. What are the conditions for two lines to be perpendicular?
5. How do we find the point of intersection between two nonlinear curves?

**Tip:** The Shoelace Formula works only when vertices are arranged in a cyclic order (clockwise or counterclockwise).

Discover how to find the area of the quadrilateral OCBE formed by the intersection of two lines (y = 2x + 3 and y = -x + 6) with the x-axis and y-axis. The solution involves linear equations, coordinate geometry, and the application of the Shoelace Formula. This problem is ideal for students in Grades 9-12.

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