Question

Given the vertices, determine the most specific classification: Quadrilateral, Parallelogram, Rectangle, Rhombus or Square. K(3, 2), I(9, 4), N(6, 0), G(1, -2) (1 point) Item 19 Item 20 Item 21 Item 22 Item 23 Item 24 < < Next Quadrilateral Parallelogram Rectangle Rhombus Square

Answer 1

To classify the quadrilateral formed by the vertices K(3, 2), I(9, 4), N(6, 0), and G(1, -2), we need to analyze the sides and angles of the shape.

1. **Calculate the length of each side:**
- Distance KI:
\[
d(KI) = \sqrt{(9-3)^2 + (4-2)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}
\]
- Distance IN:
\[
d(IN) = \sqrt{(6-9)^2 + (0-4)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
- Distance NG:
\[
d(NG) = \sqrt{(1-6)^2 + (-2-0)^2} = \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}
\]
- Distance GK:
\[
d(GK) = \sqrt{(3-1)^2 + (2-(-2))^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
\]

2. **Check if opposite sides are equal (necessary for parallelogram):**
- Compare sides KI and NG:
\[
d(KI) = 2\sqrt{10} \quad \text{and} \quad d(NG) = \sqrt{29} \quad (not\ equal)
\]
- Compare sides IN and GK:
\[
d(IN) = 5 \quad \text{and} \quad d(GK) = 2\sqrt{5} \quad (not\ equal)
\]

3. **Determine if it is a trapezoid:**
- Check if slopes of opposite sides are equal for parallelogram:
\[
\text{slope of } KI = \frac{4-2}{9-3} = \frac{2}{6} = \frac{1}{3}
\]
\[
\text{slope of } NG = \frac{-2-0}{1-6} = \frac{-2}{-5} = \frac{2}{5}
\]
\[
\text{slope of } IN = \frac{0-4}{6-9} = \frac{-4}{-3} = \frac{4}{3}
\]
\[
\text{slope of } GK = \frac{2-(-2)}{3-1} = \frac{4}{2} = 2
\]

Since neither pairs of opposite sides are equal nor parallel, this quadrilateral does not classify as a parallelogram.

**Conclusion:**
Based on these calculations, the figure cannot be classified as a quadrilateral, parallelogram, rectangle, rhombus, or square. Thus, the most specific classification of the given quadrilateral is just a **Quadrilateral**.