Question

the vertices of a quadrilateral are listed below. l(3,5), m(11,-3), n(3,-11), o(-5,-3) which of the following is the strongest classification that identifies this quadrilateral? a. the quadrilateral is a square. b. the quadrilateral is a rhombus. c. the quadrilateral is a rectangle. d. the quadrilateral is a trapezoid.

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 \(LM\): \(x_1 = 3,y_1 = 5,x_2 = 11,y_2=-3\)
\(LM=\sqrt{(11 - 3)^2+(-3 - 5)^2}=\sqrt{64 + 64}=\sqrt{128}=8\sqrt{2}\)
For \(MN\): \(x_1 = 11,y_1=-3,x_2 = 3,y_2=-11\)
\(MN=\sqrt{(3 - 11)^2+(-11+ 3)^2}=\sqrt{64 + 64}=\sqrt{128}=8\sqrt{2}\)
For \(NO\): \(x_1 = 3,y_1=-11,x_2=-5,y_2=-3\)
\(NO=\sqrt{(-5 - 3)^2+(-3 + 11)^2}=\sqrt{64 + 64}=\sqrt{128}=8\sqrt{2}\)
For \(OL\): \(x_1=-5,y_1=-3,x_2 = 3,y_2 = 5\)
\(OL=\sqrt{(3 + 5)^2+(5 + 3)^2}=\sqrt{64 + 64}=\sqrt{128}=8\sqrt{2}\)

Step2: Calculate the slopes of the sides

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\)
Slope of \(LM\): \(m_{LM}=\frac{-3 - 5}{11 - 3}=\frac{-8}{8}=-1\)
Slope of \(MN\): \(m_{MN}=\frac{-11+ 3}{3 - 11}=\frac{-8}{-8}=1\)
Slope of \(NO\): \(m_{NO}=\frac{-3 + 11}{-5 - 3}=\frac{8}{-8}=-1\)
Slope of \(OL\): \(m_{OL}=\frac{5 + 3}{3 + 5}=\frac{8}{8}=1\)

Step3: Check the properties

Since all sides \(LM = MN=NO=OL = 8\sqrt{2}\) (equal - length) and the product of adjacent - side slopes: \(m_{LM}\times m_{MN}=(-1)\times1=-1\) (perpendicular), \(m_{MN}\times m_{NO}=1\times(-1)=-1\) (perpendicular), \(m_{NO}\times m_{OL}=(-1)\times1=-1\) (perpendicular), \(m_{OL}\times m_{LM}=1\times(-1)=-1\) (perpendicular). A square has all sides equal and all angles equal to \(90^{\circ}\) (adjacent sides are perpendicular). A rhombus has all sides equal but not necessarily right - angles. A rectangle has opposite sides equal and all angles \(90^{\circ}\). A trapezoid has only one pair of parallel sides.

Answer:

A. The quadrilateral is a square.