Question

note that wxyz has vertices w(-4, -4), x(-2, 3), y(y, 2), and z(-1, -5). complete the following to determine if wxyz is a parallelogram. (a) find the length of \\(\overline{wz}\\) and the length of \\(\overline{xy}\\). give exact answers (not decimal approximations). length of \\(\overline{wz}\\): \\(-\frac{1}{3}\\) length of \\(\overline{xy}\\): \\(-\frac{1}{3}\\) (b) find the length of \\(\overline{yz}\\) and the length of \\(\overline{wx}\\). give exact answers (not decimal approximations). length of \\(\overline{yz}\\): \\(-\frac{7}{2}\\) length of \\(\overline{wx}\\): \\(-\frac{7}{2}\\) (c) from parts (a) and (b), what can we conclude?

Explanation:

Step1: Recall Distance Formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Step2: Find Length of \(\overline{WZ}\)

Points \(W(-4, -4)\) and \(Z(-1, -5)\).
Substitute into formula:
\(x_1 = -4, y_1 = -4\); \(x_2 = -1, y_2 = -5\).
\(d_{WZ} = \sqrt{(-1 - (-4))^2 + (-5 - (-4))^2}\)
\(= \sqrt{(3)^2 + (-1)^2}\)
\(= \sqrt{9 + 1}\)
\(= \sqrt{10}\).

Step3: Find Length of \(\overline{XY}\)

Assume \(Y\) has coordinates (let's correct the typo, likely \(Y(1, -2)\) from context). Points \(X(-2, 3)\) and \(Y(1, -2)\).
\(x_1 = -2, y_1 = 3\); \(x_2 = 1, y_2 = -2\).
\(d_{XY} = \sqrt{(1 - (-2))^2 + (-2 - 3)^2}\)
\(= \sqrt{(3)^2 + (-5)^2}\)
\(= \sqrt{9 + 25}\)
\(= \sqrt{34}\). Wait, maybe \(Y\) is \((0, -2)\)? Let's check again. If \(Y(0, -2)\):
\(d_{XY} = \sqrt{(0 - (-2))^2 + (-2 - 3)^2} = \sqrt{4 + 25} = \sqrt{29}\). Wait, original problem might have a typo, but using correct distance formula: length can't be negative or fraction. The given answers are wrong. Let's redo with correct approach.

Step4: Correcting (a)

For \(\overline{WZ}\): \(W(-4, -4)\), \(Z(-1, -5)\):
\(\Delta x = -1 - (-4) = 3\), \(\Delta y = -5 - (-4) = -1\).
Length: \(\sqrt{3^2 + (-1)^2} = \sqrt{10}\).

For \(\overline{XY}\): Assume \(X(-2, 3)\), \(Y(1, -2)\) (common typo fix):
\(\Delta x = 1 - (-2) = 3\), \(\Delta y = -2 - 3 = -5\).
Length: \(\sqrt{3^2 + (-5)^2} = \sqrt{34}\). But if \(Y\) is \((0, -2)\):
\(\Delta x = 0 - (-2) = 2\), \(\Delta y = -2 - 3 = -5\).
Length: \(\sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}\).

Step5: Part (b) - Length of \(\overline{YZ}\) and \(\overline{WX}\)

\(\overline{YZ}\): \(Y\) (say \(Y(1, -2)\)), \(Z(-1, -5)\):
\(\Delta x = -1 - 1 = -2\), \(\Delta y = -5 - (-2) = -3\).
Length: \(\sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\).

\(\overline{WX}\): \(W(-4, -4)\), \(X(-2, 3)\):
\(\Delta x = -2 - (-4) = 2\), \(\Delta y = 3 - (-4) = 7\).
Length: \(\sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53}\).

But the key is using distance formula correctly: length is positive, square root, no negative/fraction.

Answer:

(a) Length of \(\overline{WZ}\): \(\boldsymbol{\sqrt{10}}\), Length of \(\overline{XY}\): \(\boldsymbol{\sqrt{34}}\) (or correct based on \(Y\)’s actual coordinates).
(b) Length of \(\overline{YZ}\): \(\boldsymbol{\sqrt{13}}\) (example), Length of \(\overline{WX}\): \(\boldsymbol{\sqrt{53}}\) (example).
(c) If both pairs of opposite sides are equal, \(WXYZ\) is a parallelogram. (Since in a parallelogram, opposite sides are congruent.)