Question

1. an enclosure area like a quadrilateral wxyz is constructed in a field. imagine that the field’s area is drafted on a coordinate plane; the posts of the enclosure can be represented by vertices w(-1,-3), x(0,-3), y(5,2), and z(-2,2). given that overline{wz} is congruent to overline{xy}, is the figure a parallelogram? explain how you determined your response.

Explanation:

Step1: Calcular longitud de \( \overline{WZ} \) y \( \overline{XY} \)

Para \( \overline{XY} \): \( X(0, -3) \), \( Y(5, 2) \). Usar fórmula de distancia: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \overline{XY} \): \( \sqrt{(5 - 0)^2 + (2 - (-3))^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \)
Para \( \overline{WZ} \): \( W(-1, -3) \), \( Z(-2, 2) \)
\( \overline{WZ} \): \( \sqrt{(-2 - (-1))^2 + (2 - (-3))^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} \) (Wait, error: Wait, the problem says \( \overline{WZ} \cong \overline{XY} \), maybe check coordinates again. Wait, \( W(-1,-3) \), \( X(0,-3) \): \( \overline{WX} \) es horizontal? \( W(-1,-3) \), \( X(0,-3) \): longitud \( 1 \). \( Y(5,2) \), \( Z(-2,2) \): \( \overline{YZ} \)? Wait, maybe check \( \overline{WZ} \) and \( \overline{XY} \) again. Wait, the problem says \( \overline{WZ} \cong \overline{XY} \), so let's recalculate \( \overline{WZ} \): \( W(-1,-3) \), \( Z(-2,2) \): \( \Delta x = -2 - (-1) = -1 \), \( \Delta y = 2 - (-3) = 5 \), distancia \( \sqrt{(-1)^2 + 5^2} = \sqrt{26} \). \( \overline{XY} \): \( X(0,-3) \), \( Y(5,2) \): \( \Delta x = 5 - 0 = 5 \), \( \Delta y = 2 - (-3) = 5 \), distancia \( \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \approx 7.07 \), \( \sqrt{26} \approx 5.1 \). Wait, maybe I misread coordinates. Wait, \( Z(-2,2) \), \( W(-1,-3) \): no, maybe \( Z \) is \( (5,2) \)? No, the problem says \( Z(-2,2) \). Wait, maybe check \( \overline{WX} \) and \( \overline{YZ} \). \( W(-1,-3) \), \( X(0,-3) \): longitud \( 1 \) (horizontal, \( y \) same). \( Y(5,2) \), \( Z(-2,2) \): \( y \) same, longitud \( 5 - (-2) = 7 \)? No, \( 5 - (-2) = 7 \), \( \Delta x = -2 - 5 = -7 \), longitud \( 7 \). Wait, no, \( WX \): \( x \) de -1 a 0, \( y \) -3, so longitud 1. \( YZ \): \( x \) de 5 a -2, \( y \) 2, longitud 7. Not equal. Wait, maybe the figure is a parallelogram if \( \overline{WX} \parallel \overline{YZ} \) and \( \overline{WZ} \parallel \overline{XY} \), or \( \overline{WX} \cong \overline{YZ} \) and \( \overline{WZ} \cong \overline{XY} \). Wait, let's check slopes. Slope of \( \overline{WX} \): \( W(-1,-3) \), \( X(0,-3) \): \( m = \frac{-3 - (-3)}{0 - (-1)} = 0 \) (horizontal line). Slope of \( \overline{YZ} \): \( Y(5,2) \), \( Z(-2,2) \): \( m = \frac{2 - 2}{-2 - 5} = 0 \) (horizontal line). So \( \overline{WX} \parallel \overline{YZ} \). Now check \( \overline{WZ} \) and \( \overline{XY} \): slope of \( \overline{XY} \): \( X(0,-3) \), \( Y(5,2) \): \( m = \frac{2 - (-3)}{5 - 0} = 1 \). Slope of \( \overline{WZ} \): \( W(-1,-3) \), \( Z(-2,2) \): \( m = \frac{2 - (-3)}{-2 - (-1)} = \frac{5}{-1} = -5 \). Not equal. Wait, but the problem says \( \overline{WZ} \cong \overline{XY} \). Wait, maybe coordinates are \( Z(5,2) \)? No, the problem says \( Z(-2,2) \). Wait, maybe I made a mistake. Let's recalculate \( \overline{WZ} \): \( W(-1,-3) \), \( Z(-2,2) \): \( \sqrt{(-2 + 1)^2 + (2 + 3)^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{26} \). \( \overline{XY} \): \( X(0,-3) \), \( Y(5,2) \): \( \sqrt{(5 - 0)^2 + (2 + 3)^2} = \sqrt{25 + 25} = \sqrt{50} \). Not congruent. Wait, maybe the problem has a typo, but assuming \( \overline{WZ} \cong \overline{XY} \) and \( \overline{WX} \cong \overline{YZ} \) (since slopes of \( WX \) and \( YZ \) are 0, so they are parallel and if we check lengths: \( WX \): 1, \( YZ \): 7. No. Wait, maybe the figure is a parallelogram because \( \overline{WX} \parallel \overline{YZ} \) (both horizontal) and we need to check if \( \overline{WZ} \parallel \overline{XY} \). Wait, slope of \( XY \) is 1, slope of \( WZ \) is -5. Not parallel. Wait, maybe the vertices are \( W(-1,-3) \), \( X(0,-3) \), \( Y(5,2) \), \( Z(4,2) \)? Then \( WZ \) would be \( (-1,-3) \) to \( (4,2) \): \( \Delta x = 5 \), \( \Delta y = 5 \), same as \( XY \). But the problem says \( Z(-2,2) \). Maybe I misread \( Z \) as \( (4,2) \). Alternatively, maybe the key is that \( \overline{WX} \) and \( \overline{YZ} \) are both horizontal (same y-coordinate: \( WX \) has y=-3, \( YZ \) has y=2? No, \( Y(5,2) \), \( Z(-2,2) \) has y=2, so \( YZ \) is horizontal, \( WX \) is horizontal (y=-3). So they are parallel (both horizontal, different y). Now, for a parallelogram, we need another pair of parallel sides. Let's check \( \overline{WZ} \) and \( \overline{XY} \). \( \overline{XY} \): from (0,-3) to (5,2): direction vector (5,5). \( \overline{WZ} \): from (-1,-3) to (-2,2): direction vector (-1,5). Not same. But the problem says \( \overline{WZ} \cong \overline{XY} \). Wait, maybe the figure is a parallelogram because \( \overline{WX} \parallel \overline{YZ} \) (both horizontal) and \( \overline{WZ} \cong \overline{XY} \), so even if slopes of \( WZ \) and \( XY \) are different, but since \( WX \parallel YZ \) and \( WZ \cong XY \), is that enough? No, in a parallelogram, both pairs of opposite sides are parallel and congruent. Wait, maybe I made a mistake in distance calculation. Let's recalculate \( \overline{WZ} \): \( W(-1,-3) \), \( Z(-2,2) \): \( \sqrt{(-2 - (-1))^2 + (2 - (-3))^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} \). \( \overline{XY} \): \( X(0,-3) \), \( Y(5,2) \): \( \sqrt{(5 - 0)^2 + (2 - (-3))^2} = \sqrt{25 + 25} = \sqrt{50} \). Not congruent. So maybe the problem has an error, but assuming that \( \overline{WX} \parallel \overline{YZ} \) (both horizontal) and we are told \( \overline{WZ} \cong \overline{XY} \), then by the definition of a parallelogram (if one pair of opposite sides is congruent and parallel, or both pairs are congruent/parallel), but here \( WX \) and \( YZ \) are parallel (both horizontal) and if we check their lengths: \( WX \): 1, \( YZ \): 7. Not congruent. Wait, \( WX \): from -1 to 0 in x, y=-3: length 1. \( YZ \): from 5 to -2 in x, y=2: length 7. Not congruent. So maybe the figure is not a parallelogram? But the question is "is the figure a parallelogram? Explain". Wait, maybe I made a mistake in coordinates. Let's list coordinates:
- \( W(-1, -3) \)
- \( X(0, -3) \)
- \( Y(5, 2) \)
- \( Z(-2, 2) \)

Check \( \overline{WX} \): length \( |0 - (-1)| = 1 \) (since y is same)
\( \overline{YZ} \): length \( |-2 - 5| = 7 \) (y same) → not congruent
\( \overline{XY} \): length \( \sqrt{(5-0)^2 + (2 - (-3))^2} = \sqrt{25 + 25} = \sqrt{50} \)
\( \overline{WZ} \): length \( \sqrt{(-2 - (-1))^2 + (2 - (-3))^2} = \sqrt{1 + 25} = \sqrt{26} \) → not congruent
Slope of \( \overline{XY} \): \( (2 - (-3))/(5 - 0) = 5/5 = 1 \)
Slope of \( \overline{WZ} \): \( (2 - (-3))/(-2 - (-1)) = 5/(-1) = -5 \) → not parallel
Slope of \( \overline{WX} \): 0 (horizontal)
Slope of \( \overline{YZ} \): 0 (horizontal) → parallel

So only one pair of sides is parallel, and the other pair is neither parallel nor congruent. So the figure is not a parallelogram? But the problem says "Given that \( \overline{WZ} \) is congruent to \( \overline{XY} \), is the figure a parallelogram?" Wait, maybe I misread \( Z \)'s coordinate. If \( Z \) is \( (4, 2) \) instead of \( (-2, 2) \), then \( \overline{WZ} \): \( W(-1,-3) \) to \( Z(4,2) \): \( \Delta x = 5 \), \( \Delta y = 5 \), same as \( \overline{XY} \) ( \( X(0,-3) \) to \( Y(5,2) \): \( \Delta x = 5 \), \( \Delta y = 5 \) ). Then \( \overline{WZ} \cong \overline{XY} \) (same length) and \( \overline{WX} \): \( W(-1,-3) \) to \( X(0,-3) \): \( \Delta x = 1 \), \( \Delta y = 0 \); \( \overline{YZ} \): \( Y(5,2) \) to \( Z(4,2) \): \( \Delta x = -1 \), \( \Delta y = 0 \). So \( \overline{WX} \cong \overline{YZ} \) (length 1) and parallel (both horizontal). Then it would be a parallelogram. But the problem says \( Z(-2,2) \). Maybe a typo, but assuming the given coordinates, the figure is not a parallelogram because only one pair of sides is parallel, and the other pair is neither parallel nor congruent.

Step2: Conclusión

Para ser un paralelogramo, se necesita que ambos pares de lados opuestos sean paralelos (o congruentes y paralelos). Aquí, \( \overline{WX} \parallel \overline{YZ} \) (ambos horizontales), pero \( \overline{WZ} \) y \( \overline{XY} \) no son paralelos (pendientes 1 y -5) ni congruentes (longitudes \( \sqrt{50} \) y \( \sqrt{26} \)). Por lo tanto, la figura **no** es un paralelogramo.

Answer:

La figura no es un paralelogramo, ya que solo un par de lados opuestos (\( \overline{WX} \) y \( \overline{YZ} \)) es paralelo, y el otro par (\( \overline{WZ} \) y \( \overline{XY} \)) no es ni paralelo ni congruente (aunque se dice que \( \overline{WZ} \cong \overline{XY} \), el cálculo muestra que no lo son, o hay un error en coordenadas; asumiendo las dadas, no cumple con la definición de paralelogramo).