Question

the vertices of quadrilateral efgh are e(-7,3), f(-4,6), g(5,-3), and h(2,-6). what kind of quadrilateral is efgh?
a trapezoid
b square
c rectangle that is not a square
d rhombus that is not 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 \(EF\): \(m_{EF}=\frac{6 - 3}{-4+7}=\frac{3}{3} = 1\)
For \(FG\): \(m_{FG}=\frac{-3 - 6}{5 + 4}=\frac{-9}{9}=-1\)
For \(GH\): \(m_{GH}=\frac{-6+3}{2 - 5}=\frac{-3}{-3}=1\)
For \(HE\): \(m_{HE}=\frac{3 + 6}{-7 - 2}=\frac{9}{-9}=-1\)

Step2: Calculate the lengths of the sides

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(EF\): \(d_{EF}=\sqrt{(-4 + 7)^2+(6 - 3)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\)
For \(FG\): \(d_{FG}=\sqrt{(5 + 4)^2+(-3 - 6)^2}=\sqrt{81+81}=\sqrt{162}=9\sqrt{2}\)
For \(GH\): \(d_{GH}=\sqrt{(2 - 5)^2+(-6 + 3)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\)
For \(HE\): \(d_{HE}=\sqrt{(-7 - 2)^2+(3 + 6)^2}=\sqrt{81+81}=\sqrt{162}=9\sqrt{2}\)

Step3: Analyze the properties

Since the slopes of adjacent sides are negative reciprocals (\(m_{EF}\times m_{FG}=-1\), \(m_{FG}\times m_{GH}=-1\), \(m_{GH}\times m_{HE}=-1\), \(m_{HE}\times m_{EF}=-1\)), the quadrilateral has right angles.
Since \(EF = GH\) and \(FG=HE\), it is a rectangle.
Also, since \(EF
eq FG\), it is not a square.

Answer:

C. rectangle that is not a square