A quadrilateral has the vertices at the points (-4, 2), (2, 6), (8, 5) and (9, -7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram

Options

A. Rectangle
B. Square
C. Parallelogram
D. Rhombus

Answer

To show that the midpoints of the sides of the quadrilateral formed by the vertices (-4, 2), (2, 6), (8, 5), and (9, -7) are the vertices of a parallelogram, we will follow these steps:

### Step 1: Identify the vertices of the quadrilateral
Let the vertices of the quadrilateral be:
- A = (-4, 2)
- B = (2, 6)
- C = (8, 5)
- D = (9, -7)

### Step 2: Calculate the midpoints of each side
The midpoints of the sides can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

1. **Midpoint of AB (M1)**:
\[ M1 = \left( \frac{-4 + 2}{2}, \frac{2 + 6}{2} \right) = \left( \frac{-2}{2}, \frac{8}{2} \right) = (-1, 4) \]

2. **Midpoint of BC (M2)**:
\[ M2 = \left( \frac{2 + 8}{2}, \frac{6 + 5}{2} \right) = \left( \frac{10}{2}, \frac{11}{2} \right) = (5, \frac{11}{2}) \]

3. **Midpoint of CD (M3)**:
\[ M3 = \left( \frac{8 + 9}{2}, \frac{5 + (-7)}{2} \right) = \left( \frac{17}{2}, \frac{-2}{2} \right) = \left( \frac{17}{2}, -1 \right) \]

4. **Midpoint of DA (M4)**:
\[ M4 = \left( \frac{9 + (-4)}{2}, \frac{-7 + 2}{2} \right) = \left( \frac{5}{2}, \frac{-5}{2} \right) \]

### Step 3: List the midpoints
The midpoints are:
- M1 = (-1, 4)
- M2 = (5, 5.5)
- M3 = (8.5, -1)
- M4 = (2.5, -2.5)

### Step 4: Show that the midpoints form a parallelogram
To show that the midpoints form a parallelogram, we need to prove that the opposite sides are equal in length.

1. **Distance between M1 and M2 (d1)**:
\[ d1 = \sqrt{(5 - (-1))^2 + \left(\frac{11}{2} - 4\right)^2} = \sqrt{(6)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{36 + \frac{9}{4}} = \sqrt{\frac{144 + 9}{4}} = \sqrt{\frac{153}{4}} = \frac{\sqrt{153}}{2} \]

2. **Distance between M3 and M4 (d2)**:
\[ d2 = \sqrt{\left(2.5 - \frac{17}{2}\right)^2 + \left(-2.5 - (-1)\right)^2} = \sqrt{\left(-\frac{12}{2}\right)^2 + \left(-1.5\right)^2} = \sqrt{36 + 2.25} = \sqrt{38.25} \]

3. **Distance between M1 and M4 (d3)**:
\[ d3 = \sqrt{\left(2.5 - (-1)\right)^2 + \left(-2.5 - 4\right)^2} = \sqrt{(3.5)^2 + (-6.5)^2} = \sqrt{12.25 + 42.25} = \sqrt{54.5} \]

4. **Distance between M2 and M3 (d4)**:
\[ d4 = \sqrt{\left(8.5 - 5\right)^2 + \left(-1 - \frac{11}{2}\right)^2} = \sqrt{(3.5)^2 + (-6.5)^2} = \sqrt{12.25 + 42.25} = \sqrt{54.5} \]

### Conclusion
Since the lengths of opposite sides (d1 = d2 and d3 = d4) are equal, we conclude that the midpoints of the sides of the quadrilateral form a parallelogram.