To show that the quadrilateral ABCD is a parallelogram, we need to verify that either:

1. Both pairs of opposite sides are parallel, or
2. Both pairs of opposite sides are equal in length.

I'll use the second method here: showing that opposite sides are equal in length.

Step 1: Distance formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 2: Find the lengths of sides AB, BC, CD, and DA.

Length of AB:

The coordinates of A and B are $(-2, 3)$ and $(6, 1)$, respectively. Applying the distance formula:

$AB = \sqrt{(6 - (-2))^2 + (1 - 3)^2} = \sqrt{(6 + 2)^2 + (-2)^2} = \sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}$

$AB = \sqrt{68} \approx 8.25$

Length of CD:

The coordinates of C and D are $(5, -2)$ and $(-3, 0)$, respectively. Applying the distance formula:

$CD = \sqrt{(5 - (-3))^2 + (-2 - 0)^2} = \sqrt{(5 + 3)^2 + (-2)^2} = \sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}$

$CD = \sqrt{68} \approx 8.25$

So, $AB = CD$.

Length of AD:

The coordinates of A and D are $(-2, 3)$ and $(-3, 0)$, respectively. Applying the distance formula:

$AD = \sqrt{(-3 - (-2))^2 + (0 - 3)^2} = \sqrt{(-3 + 2)^2 + (-3)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$

$AD = \sqrt{10} \approx 3.16$

Length of BC:

The coordinates of B and C are $(6, 1)$ and $(5, -2)$, respectively. Applying the distance formula:

$BC = \sqrt{(5 - 6)^2 + (-2 - 1)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$

$BC = \sqrt{10} \approx 3.16$

So, $AD = BC$.

Step 3: Conclusion

Since opposite sides $AB = CD$ and $AD = BC$, the quadrilateral ABCD is a parallelogram.

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Would you like any further details or have questions? Here are some additional topics to explore:

1. How do you prove a parallelogram is a rhombus?
2. How can you calculate the area of parallelogram