Question

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

Explanation:

Step1: Calculate the slopes of the sides

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For the side connecting \((-10,-4)\) and \((-7,-8)\): \(m_1=\frac{-8-(-4)}{-7 - (-10)}=\frac{-4}{3}\).
For the side connecting \((-7,-8)\) and \((-5,-4)\): \(m_2=\frac{-4-(-8)}{-5-(-7)}=\frac{4}{2} = 2\).
For the side connecting \((-5,-4)\) and \((-2,-8)\): \(m_3=\frac{-8-(-4)}{-2-(-5)}=\frac{-4}{3}\).
For the side connecting \((-2,-8)\) and \((-10,-4)\): \(m_4=\frac{-4-(-8)}{-10-(-2)}=\frac{4}{-8}=-\frac{1}{2}\).
Since \(m_1 = m_3\) and \(m_2
eq m_4\), opposite sides are parallel.

Step2: Check for right - angles (using slope product)

For a rectangle, adjacent sides should be perpendicular (\(m_1\times m_2=- 1\)).
\(m_1\times m_2=\frac{-4}{3}\times2=-\frac{8}{3}
eq - 1\).
For a rhombus, all sides should be equal in length. Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Distance between \((-10,-4)\) and \((-7,-8)\): \(d_1=\sqrt{(-7 + 10)^2+(-8 + 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5\).
Distance between \((-7,-8)\) and \((-5,-4)\): \(d_2=\sqrt{(-5 + 7)^2+(-4 + 8)^2}=\sqrt{4 + 16}=\sqrt{20}\).
Since \(d_1
eq d_2\), it's not a rhombus.

Since it has four sides (quadrilateral) and opposite sides are parallel (parallelogram is a more specific classification than quadrilateral)

Answer:

C. Parallelogram