1. **State the problem:** Determine if quadrilateral ABCD with vertices A(-3,0), B(3,2), C(4,-1), and D(-2,-3) is a parallelogram. 2. **Recall the definition:** A quadrilateral is a parallelogram if both pairs of opposite sides are parallel, which means their slopes are equal. 3. **Calculate slopes of all sides:** - Slope of AB = \frac{2 - 0}{3 - (-3)} = \frac{2}{6} = \frac{1}{3} - Slope of BC = \frac{-1 - 2}{4 - 3} = \frac{-3}{1} = -3 - Slope of CD = \frac{-3 - (-1)}{-2 - 4} = \frac{-2}{-6} = \frac{1}{3} - Slope of DA = \frac{0 - (-3)}{-3 - (-2)} = \frac{3}{-1} = -3 4. **Check opposite sides:** - AB and CD slopes: both \frac{1}{3} (equal, so AB || CD) - BC and DA slopes: both -3 (equal, so BC || DA) 5. **Conclusion:** Since both pairs of opposite sides are parallel, ABCD is a parallelogram. 6. **Additional info:** Midpoints of diagonals AC and BD are: - Midpoint AC = \left(\frac{-3 + 4}{2}, \frac{0 + (-1)}{2}\right) = \left(\frac{1}{2}, -\frac{1}{2}\right) - Midpoint BD = \left(\frac{3 + (-2)}{2}, \frac{2 + (-3)}{2}\right) = \left(\frac{1}{2}, -\frac{1}{2}\right) Since midpoints of diagonals are the same, diagonals bisect each other, confirming ABCD is a parallelogram. 7. **Lengths of diagonals:** - Length AC = \sqrt{(4 - (-3))^2 + (-1 - 0)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} - Length BD = \sqrt{(-2 - 3)^2 + (-3 - 2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} 8. **Is ABCD a rhombus?** A rhombus has all sides equal. Calculate side lengths: - AB = \sqrt{(3 - (-3))^2 + (2 - 0)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} - BC = \sqrt{(4 - 3)^2 + (-1 - 2)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} - CD = \sqrt{(-2 - 4)^2 + (-3 - (-1))^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} - DA = \sqrt{(-3 - (-2))^2 + (0 - (-3))^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} Since sides are not all equal, ABCD is not a rhombus. **Final answers:** - ABCD is a parallelogram. - ABCD is not a rhombus. - Length of BD = $\sqrt{50}$.