Question

Prove that the following vertices form a parallelogram by showing that the opposite sides are parallel. Item 17 S(-2, 1), A(6, 6), N(1, 0), D(-7, -5) Item 18 (2 points) Item 19 The slope of SA is Item 20 Item 21 The slope of AN is Item 22 The slope of ND is Item 23 The slope of DS is < Item 24 Is this a parallelogram? I know this because the opposite sides of the quadrilateral are parallel : Yes No ::m m= :: m=- ::m

Answer 1

To prove that the vertices S(-2, 1), A(6, 6), N(1, 0), and D(-7, -5) form a parallelogram, we can check that the slopes of the opposite sides are equal.

Let's calculate the slopes of each side.

### Step 1: Calculate the slope of SA

The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

For \( S(-2, 1) \) and \( A(6, 6) \):
\[
m_{SA} = \frac{6 - 1}{6 - (-2)} = \frac{5}{8}
\]

### Step 2: Calculate the slope of AN

For \( A(6, 6) \) and \( N(1, 0) \):
\[
m_{AN} = \frac{0 - 6}{1 - 6} = \frac{-6}{-5} = \frac{6}{5}
\]

### Step 3: Calculate the slope of ND

For \( N(1, 0) \) and \( D(-7, -5) \):
\[
m_{ND} = \frac{-5 - 0}{-7 - 1} = \frac{-5}{-8} = \frac{5}{8}
\]

### Step 4: Calculate the slope of DS

For \( D(-7, -5) \) and \( S(-2, 1) \):
\[
m_{DS} = \frac{1 - (-5)}{-2 - (-7)} = \frac{6}{5}
\]

### Summary of the slopes:
- Slope of \( SA \): \( m_{SA} = \frac{5}{8} \)
- Slope of \( AN \): \( m_{AN} = \frac{6}{5} \)
- Slope of \( ND \): \( m_{ND} = \frac{5}{8} \)
- Slope of \( DS \): \( m_{DS} = \frac{6}{5} \)

### Conclusion

Since the slopes of opposite sides are equal:

- \( m_{SA} = m_{ND} = \frac{5}{8} \)
- \( m_{AN} = m_{DS} = \frac{6}{5} \)

This indicates that the opposite sides \( SA \) and \( ND \) are parallel, and \( AN \) and \( DS \) are also parallel.

Therefore, since both pairs of opposite sides are parallel, we conclude that the quadrilateral formed by the points \( S \), \( A \), \( N \), and \( D \) is indeed a parallelogram.

### Final answer

Is this a parallelogram? **Yes**