Question

what is the strongest classification of the figure formed by the following points?
(-10, -4), (-7, -8), (-5, -4), (-2, -8)
a. parallelogram
b. quadrilateral
c. rhombus
d. rectangle

Explanation:

Step1: Calculate the lengths of the sides

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For the side between \((-10,-4)\) and \((-7,-8)\):
\(d_1=\sqrt{(-7 + 10)^2+(-8 + 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5\)
For the side between \((-7,-8)\) and \((-5,-4)\):
\(d_2=\sqrt{(-5 + 7)^2+(-4 + 8)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\)
For the side between \((-5,-4)\) and \((-2,-8)\):
\(d_3=\sqrt{(-2 + 5)^2+(-8 + 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5\)
For the side between \((-2,-8)\) and \((-10,-4)\):
\(d_4=\sqrt{(-10 + 2)^2+(-4 + 8)^2}=\sqrt{64 + 16}=\sqrt{80}=4\sqrt{5}\)

Step2: Check the properties of parallelogram

In a parallelogram, opposite sides are equal. Here \(d_1 = d_3 = 5\) and \(d_2=d_4 = 2\sqrt{5}\).
A quadrilateral is a general four - sided figure. Since we have more specific properties (opposite sides equal), parallelogram is a stronger classification than quadrilateral.
For a rhombus, all sides are equal. Here sides are \(5\) and \(2\sqrt{5}\), so not a rhombus.
For a rectangle, adjacent sides should be perpendicular (product of slopes \(=- 1\)).
Slope of side between \((-10,-4)\) and \((-7,-8)\): \(m_1=\frac{-8 + 4}{-7+10}=\frac{-4}{3}\)
Slope of side between \((-7,-8)\) and \((-5,-4)\): \(m_2=\frac{-4 + 8}{-5 + 7}=2\)
\(m_1\times m_2=\frac{-4}{3}\times2=-\frac{8}{3}
eq - 1\), so not a rectangle.

Answer:

A. Parallelogram