Question

given the verticies determine the most specific classification: Quadrilateral, parallelogram, rectangle, rhombus, or square. L(-4,4), A(3,2), K(5,-5), E(-2,-3)

Answer 1

To classify the shape formed by the vertices L(-4, 4), A(3, 2), K(5, -5), and E(-2, -3), we can follow these steps:

1. **Check if it's a Quadrilateral**: A quadrilateral has four vertices, which we have.

2. **Calculate the distances between the points**:
\[
\text{Distance (d)} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

- Distance LA:
\[
d_{LA} = \sqrt{(3 - (-4))^2 + (2 - 4)^2} = \sqrt{(3 + 4)^2 + (2 - 4)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53}
\]
- Distance AK:
\[
d_{AK} = \sqrt{(5 - 3)^2 + (-5 - 2)^2} = \sqrt{(2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53}
\]
- Distance KE:
\[
d_{KE} = \sqrt{(-2 - 5)^2 + (-3 - (-5))^2} = \sqrt{(-7)^2 + (2)^2} = \sqrt{49 + 4} = \sqrt{53}
\]
- Distance EL:
\[
d_{EL} = \sqrt{(-4 - (-2))^2 + (4 - (-3))^2} = \sqrt{(-2)^2 + (7)^2} = \sqrt{4 + 49} = \sqrt{53}
\]

Since all sides are equal (\(d_{LA} = d_{AK} = d_{KE} = d_{EL} = \sqrt{53}\)), we can conclude the shape is at least a rhombus.

3. **Check the diagonals**: Calculate the distances of the diagonals to further classify the shape.
- Diagonal LK:
\[
d_{LK} = \sqrt{(5 - (-4))^2 + (-5 - 4)^2} = \sqrt{(5 + 4)^2 + (-9)^2} = \sqrt{9^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}
\]
- Diagonal AE:
\[
d_{AE} = \sqrt{(-2 - 3)^2 + (-3 - 2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
\]

4. **Check for right angles**: For it to be a rectangle or a square, it must have right angles. We can check if any adjacent sides are perpendicular:
- Vectors LA and AK:
\[
\text{Vector LA} = (3 - (-4), 2 - 4) = (7, -2)
\]
\[
\text{Vector AK} = (5 - 3, -5 - 2) = (2, -7)
\]
- The dot product should be zero for perpendicular vectors:
\[
LA \cdot AK = 7 \times 2 + (-2) \times (-7) = 14 + 14 = 28 \quad (\text{not perpendicular})
\]

Since all sides are equal, and the diagonals are not the same length, and the angles are not right angles, we can conclude that the quadrilateral is a **rhombus**.

### Conclusion
The most specific classification for the shape formed by the vertices L(-4, 4), A(3, 2), K(5, -5), and E(-2, -3) is a **rhombus**.