Question

consider parallelogram wxyz below.
note that wxyz has vertices w(1, -5), x(-6, -1), y(-2, 6), and z(5, 2).
answer the following to determine if the parallelogram is a rectangle, rhombus, square, or none of these.
(a) find the length of \\(\overline{xy}\\) and the length of a side adjacent to \\(\overline{xy}\\)
give exact answers (not decimal approximations).
length of \\(\overline{xy}\\)
length of side adjacent to \\(\overline{xy}\\)
(b) find the slope of \\(\overline{xy}\\) and the slope of a side adjacent to \\(\overline{xy}\\)
slope of \\(\overline{xy}\\)
slope of side adjacent to \\(\overline{xy}\\)
(c) from parts (a) and (b), what can you conclude about parallelogram
wxyz? check all that apply.
wxyz is a rectangle.
wxyz is a rhombus.
wxyz is a square.
wxyz is none of these

Explanation:

Step1: Calculate length of $\overline{XY}$

Use distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$X(-6,-1)$, $Y(-2,6)$
$\overline{XY} = \sqrt{(-2-(-6))^2+(6-(-1))^2} = \sqrt{4^2+7^2} = \sqrt{16+49} = \sqrt{65}$

Step2: Calculate length of adjacent side $\overline{XW}$

Use distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$X(-6,-1)$, $W(1,-5)$
$\overline{XW} = \sqrt{(1-(-6))^2+(-5-(-1))^2} = \sqrt{7^2+(-4)^2} = \sqrt{49+16} = \sqrt{65}$

Step3: Calculate slope of $\overline{XY}$

Use slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
$m_{XY} = \frac{6-(-1)}{-2-(-6)} = \frac{7}{4}$

Step4: Calculate slope of adjacent $\overline{XW}$

Use slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
$m_{XW} = \frac{-5-(-1)}{1-(-6)} = \frac{-4}{7}$

Step5: Classify the parallelogram

Check side lengths (equal = rhombus candidate) and slopes: $m_{XY} \times m_{XW} = \frac{7}{4} \times \frac{-4}{7} = -1$, so sides are perpendicular. A rhombus with perpendicular sides is a square.

Answer:

(a)
Length of $\overline{XY}$: $\sqrt{65}$
Length of side adjacent to $\overline{XY}$: $\sqrt{65}$

(b)
Slope of $\overline{XY}$: $\frac{7}{4}$
Slope of side adjacent to $\overline{XY}$: $-\frac{4}{7}$

(c)
WXYZ is a square.