Question

quadrilateral rstu has vertices at r(3, 2), s(5, -5), t(-1, -7), and u(-3, 0). is rstu a parallelogram? justify your answer. yes, \\(\overline{ru}\\) and \\(\overline{st}\\) are parallel. yes, \\(\overline{rs}\\) is parallel to \\(\overline{tu}\\), and \\(\overline{ru}\\) is parallel to \\(\overline{st}\\). no, \\(\overline{ru}\\) and \\(\overline{st}\\) are not parallel. no, \\(\overline{rs}\\) and \\(\overline{tu}\\) are not parallel.

Explanation:

Step1: Calcular la pendiente de $\overline{RS}$

La fórmula de la pendiente entre dos puntos $(x_1,y_1)$ y $(x_2,y_2)$ es $m = \frac{y_2 - y_1}{x_2 - x_1}$. Para $R(3,2)$ y $S(5,-5)$:
$m_{RS} = \frac{-5 - 2}{5 - 3} = \frac{-7}{2} = -\frac{7}{2}$

Step2: Calcular la pendiente de $\overline{TU}$

Para $T(-1,-7)$ y $U(-3,0)$:
$m_{TU} = \frac{0 - (-7)}{-3 - (-1)} = \frac{7}{-2} = -\frac{7}{2}$

Step3: Calcular la pendiente de $\overline{RU}$

Para $R(3,2)$ y $U(-3,0)$:
$m_{RU} = \frac{0 - 2}{-3 - 3} = \frac{-2}{-6} = \frac{1}{3}$

Step4: Calcular la pendiente de $\overline{ST}$

Para $S(5,-5)$ y $T(-1,-7)$:
$m_{ST} = \frac{-7 - (-5)}{-1 - 5} = \frac{-2}{-6} = \frac{1}{3}$

Step5: Comparar pendientes

$m_{RS} = m_{TU} = -\frac{7}{2}$ y $m_{RU} = m_{ST} = \frac{1}{3}$. Entonces, $\overline{RS} \parallel \overline{TU}$ y $\overline{RU} \parallel \overline{ST}$.

Answer:

Yes, $\overline{RS}$ is parallel to $\overline{TU}$, and $\overline{RU}$ is parallel to $\overline{ST}$.