Question

the vertices of quadrilateral pqrs are listed.
p(3,7), q(6,-2), r(0,-4), s(-3,5)
which of the following is the strongest classification that identifies quadrilateral pqrs?
a. quadrilateral pqrs is a rectangle.
b. quadrilateral pqrs is a parallelogram.
c. quadrilateral pqrs is a trapezoid.
d. quadrilateral pqrs is a square.

Explanation:

Step1: Calculate the slopes of the sides

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
For \(PQ\) with \(P(3,7)\) and \(Q(6,- 2)\):
\(m_{PQ}=\frac{-2 - 7}{6 - 3}=\frac{-9}{3}=-3\)
For \(QR\) with \(Q(6,-2)\) and \(R(0,-4)\):
\(m_{QR}=\frac{-4+2}{0 - 6}=\frac{-2}{-6}=\frac{1}{3}\)
For \(RS\) with \(R(0,-4)\) and \(S(-3,5)\):
\(m_{RS}=\frac{5 + 4}{-3-0}=\frac{9}{-3}=-3\)
For \(SP\) with \(S(-3,5)\) and \(P(3,7)\):
\(m_{SP}=\frac{7 - 5}{3+3}=\frac{2}{6}=\frac{1}{3}\)

Step2: Check the properties of parallelogram

In a parallelogram, opposite sides are parallel. That is \(m_{PQ}=m_{RS}=-3\) and \(m_{QR}=m_{SP}=\frac{1}{3}\)

Step3: Check for rectangle and square

For a rectangle (and square), adjacent sides should be perpendicular (\(m_1\times m_2=- 1\)).
\(m_{PQ}\times m_{QR}=(-3)\times\frac{1}{3}=-1\), but we also need to check the lengths.
The length formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(d_{PQ}=\sqrt{(6 - 3)^2+(-2 - 7)^2}=\sqrt{9 + 81}=\sqrt{90}\)
\(d_{QR}=\sqrt{(0 - 6)^2+(-4 + 2)^2}=\sqrt{36+4}=\sqrt{40}\)
Since \(d_{PQ}
eq d_{QR}\), it is not a rectangle (and thus not a square)

Step4: Check for trapezoid

A trapezoid has only one pair of parallel sides. But here we have two pairs of parallel sides

Answer:

B. Quadrilateral \(PQRS\) is a parallelogram.