Question

the vertices of a quadrilateral are listed below. q(-8,4), r(8,4), s(7,-4), t(-9,-4) which of the following is the strongest classification that identifies this quadrilateral? a. the quadrilateral is a rectangle. b. the quadrilateral is a parallelogram. c. the quadrilateral is a square. d. the quadrilateral is a rhombus.

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 \(QR\): \(x_1=-8,y_1 = 4,x_2 = 8,y_2=4\), \(d_{QR}=\sqrt{(8 + 8)^2+(4 - 4)^2}=\sqrt{256}=16\)
For \(RS\): \(x_1 = 8,y_1=4,x_2=7,y_2=-4\), \(d_{RS}=\sqrt{(7 - 8)^2+(-4 - 4)^2}=\sqrt{1 + 64}=\sqrt{65}\)
For \(ST\): \(x_1=7,y_1=-4,x_2=-9,y_2=-4\), \(d_{ST}=\sqrt{(-9 - 7)^2+(-4+4)^2}=\sqrt{256}=16\)
For \(TQ\): \(x_1=-9,y_1=-4,x_2=-8,y_2 = 4\), \(d_{TQ}=\sqrt{(-8 + 9)^2+(4 + 4)^2}=\sqrt{1 + 64}=\sqrt{65}\)
Since \(QR = ST\) and \(RS=TQ\), it is a parallelogram.

Step2: Check for right - angles (for rectangle)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\)
Slope of \(QR\): \(m_{QR}=\frac{4 - 4}{8+8}=0\)
Slope of \(RS\): \(m_{RS}=\frac{-4 - 4}{7 - 8}=8\)
Since \(m_{QR}\times m_{RS}=0\times8 = 0
eq - 1\), it is not a rectangle.
For a rhombus, all sides should be equal. Here \(QR = 16\), \(RS=\sqrt{65}\), not all sides are equal.
For a square, all sides equal and right - angles (not the case here)

Answer:

B. The quadrilateral is a parallelogram