Question

quadrilateral abcd has the following vertices:
. a(-6,-2)
. b(-4,4)
. c(8,1)
. d(6,-6)
also, ∠a is a right angle.
is quadrilateral abcd a rectangle, and why?
choose 1 answer:
a yes, because ab = bc = cd = ad, and ∠a is a right angle.
b yes, because opposite sides are parallel, and ∠a is a right angle.
c no, because ab is not parallel to cd.
d no, because ab is shorter than bc.

Explanation:

Step1: Calculate the slope of \(AB\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For points \(A(-6,-2)\) and \(B(-4,4)\), \(m_{AB}=\frac{4 - (-2)}{-4 - (-6)}=\frac{4 + 2}{-4 + 6}=\frac{6}{2}=3\).

Step2: Calculate the slope of \(AD\)

For points \(A(-6,-2)\) and \(D(6,-6)\), \(m_{AD}=\frac{-6 - (-2)}{6 - (-6)}=\frac{-6 + 2}{6 + 6}=\frac{-4}{12}=-\frac{1}{3}\).

Step3: Calculate the slope of \(BC\)

For points \(B(-4,4)\) and \(C(8,1)\), \(m_{BC}=\frac{1 - 4}{8 - (-4)}=\frac{-3}{12}=-\frac{1}{4}\).

Step4: Calculate the slope of \(CD\)

For points \(C(8,1)\) and \(D(6,-6)\), \(m_{CD}=\frac{-6 - 1}{6 - 8}=\frac{-7}{-2}=\frac{7}{2}\).

Since \(m_{AB}
eq m_{CD}\) ( \(3
eq\frac{7}{2}\) ), \(\overline{AB}\) is not parallel to \(\overline{CD}\).

Answer:

C. No, because \(\overline{AB}\) is not parallel to \(\overline{CD}\).