Question

the vertices of a quadrilateral on the coordinate plane are (2,4), (-4,-2), (-2,4), and (4,-2). what type of quadrilateral has these vertices?
a. rectangle
b. trapezoid
c. square
d. parallelogram

Explanation:

Step1: Calculate the slopes of the sides

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
For the side connecting \((2,4)\) and \((- 2,4)\): \(m_1=\frac{4 - 4}{-2 - 2}=0\).
For the side connecting \((-2,4)\) and \((-4,-2)\): \(m_2=\frac{-2 - 4}{-4+2}=3\).
For the side connecting \((-4,-2)\) and \((4,-2)\): \(m_3=\frac{-2+2}{4 + 4}=0\).
For the side connecting \((4,-2)\) and \((2,4)\): \(m_4=\frac{4 + 2}{2 - 4}=-3\).

Step2: Check the properties of the quadrilateral

Since \(m_1=m_3 = 0\) (horizontal sides) and \(m_2=-m_4\) (sides with slopes \(3\) and \(-3\) are not parallel but the opposite sides are parallel (horizontal sides are parallel and non - horizontal sides have slopes that are negative reciprocals in a sense of non - vertical/non - horizontal parallelism check. Also, using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Distance between \((2,4)\) and \((-2,4)\): \(d_1=\sqrt{( - 2 - 2)^2+(4 - 4)^2}=4\).
Distance between \((-2,4)\) and \((-4,-2)\): \(d_2=\sqrt{(-4 + 2)^2+(-2 - 4)^2}=\sqrt{4 + 36}=\sqrt{40}\).
Distance between \((-4,-2)\) and \((4,-2)\): \(d_3=\sqrt{(4 + 4)^2+(-2 + 2)^2}=8\).
Distance between \((4,-2)\) and \((2,4)\): \(d_4=\sqrt{(2 - 4)^2+(4 + 2)^2}=\sqrt{4+36}=\sqrt{40}\).
Opposite sides are equal (\(d_1 = d_3\) and \(d_2=d_4\)) and opposite sides are parallel (slopes of opposite sides are equal).

Answer:

A. rectangle