(1, 1), (3, 4), (5, –2) and (4, –7) are vertices of a quadrilateral then find its area.

Answer

To find the area of the quadrilateral with vertices at (1, 1), (3, 4), (5, -2), and (4, -7), we can divide the quadrilateral into two triangles and calculate the area of each triangle using the formula for the area of a triangle based on its vertices.

### Step-by-Step Solution:

1. **Identify the vertices**:
Let the vertices be:
- A(1, 1)
- B(3, 4)
- C(5, -2)
- D(4, -7)

2. **Divide the quadrilateral into two triangles**:
We can divide the quadrilateral ABCD into two triangles:
- Triangle ABC
- Triangle ACD

3. **Calculate the area of Triangle ABC**:
The formula for the area of a triangle given vertices (x1, y1), (x2, y2), (x3, y3) is:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For triangle ABC:
- A(1, 1), B(3, 4), C(5, -2)
- Substituting the coordinates into the formula:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| 1(4 - (-2)) + 3(-2 - 1) + 5(1 - 4) \right| \]
\[ = \frac{1}{2} \left| 1(6) + 3(-3) + 5(-3) \right| \]
\[ = \frac{1}{2} \left| 6 - 9 - 15 \right| \]
\[ = \frac{1}{2} \left| 6 - 24 \right| = \frac{1}{2} \left| -18 \right| = \frac{1}{2} \times 18 = 9 \]

4. **Calculate the area of Triangle ACD**:
For triangle ACD:
- A(1, 1), C(5, -2), D(4, -7)
- Substituting the coordinates into the formula:
\[ \text{Area}_{ACD} = \frac{1}{2} \left| 1(-2 - (-7)) + 5(-7 - 1) + 4(1 - (-2)) \right| \]
\[ = \frac{1}{2} \left| 1(5) + 5(-8) + 4(3) \right| \]
\[ = \frac{1}{2} \left| 5 - 40 + 12 \right| \]
\[ = \frac{1}{2} \left| 5 - 40 + 12 \right| = \frac{1}{2} \left| -23 \right| = \frac{1}{2} \times 23 = 11.5 \]

5. **Calculate the total area of the quadrilateral**:
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = 9 + 11.5 = 20.5 \]

### Final Answer:
The area of the quadrilateral is **20.5 square units**.