Question

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

Explanation:

Step1: Calculate side - lengths

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
Let $A(-10,-4)$, $B(-7,-8)$, $C(-5,-4)$, $D(-2,-8)$.
$AB=\sqrt{(-7 + 10)^2+(-8 + 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.
$BC=\sqrt{(-5 + 7)^2+(-4 + 8)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}$.
$CD=\sqrt{(-2 + 5)^2+(-8 + 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.
$DA=\sqrt{(-10 + 2)^2+(-4 + 8)^2}=\sqrt{64 + 16}=\sqrt{80}=4\sqrt{5}$.

Step2: Analyze the figure

Since it has four sides, it is a quadrilateral. But the side - lengths are not all equal (so not a rhombus), and the adjacent sides are not perpendicular (by checking slopes of adjacent sides, for example, slope of $AB=\frac{-8 + 4}{-7+10}=-\frac{4}{3}$, slope of $BC=\frac{-4 + 8}{-5 + 7}=2$, and $-\frac{4}{3}\times2
eq - 1$), so not a rectangle. Also, since opposite sides are not equal in all cases, it is just a general quadrilateral.

Answer:

A. Quadrilateral