show that a quadrilateral with the given vertices is a parallelogram. find the lengths of the sides of the quadrilateral. st = \square tu = \square vu = \square sv = \square (simplify your answers. type exact answers, using radicals as needed)

Question

Accepted Answer

# Explanation:
## Step1: Find coordinates
Let $ S(-1, -1) $, $ V(-1, -4) $, $ U(2, -5) $, $ T(2, -2) $.
## Step2: Calculate $ ST $
Use distance formula $ d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} $.
For $ S(-1, -1) $ and $ T(2, -2) $:
$$
\begin{align*}
ST&=\sqrt{(2 - (-1))^2+(-2 - (-1))^2}\
&=\sqrt{(3)^2+(-1)^2}\
&=\sqrt{9 + 1}\
&=\sqrt{10}
\end{align*}
$$
## Step3: Calculate $ TU $
For $ T(2, -2) $ and $ U(2, -5) $:
$$
\begin{align*}
TU&=\sqrt{(2 - 2)^2+(-5 - (-2))^2}\
&=\sqrt{0+(-3)^2}\
&=3
\end{align*}
$$
## Step4: Calculate $ VU $
For $ V(-1, -4) $ and $ U(2, -5) $:
$$
\begin{align*}
VU&=\sqrt{(2 - (-1))^2+(-5 - (-4))^2}\
&=\sqrt{(3)^2+(-1)^2}\
&=\sqrt{9 + 1}\
&=\sqrt{10}
\end{align*}
$$
## Step5: Calculate $ SV $
For $ S(-1, -1) $ and $ V(-1, -4) $:
$$
\begin{align*}
SV&=\sqrt{(-1 - (-1))^2+(-4 - (-1))^2}\
&=\sqrt{0+(-3)^2}\
&=3
\end{align*}
$$

# Answer:
$ ST=\sqrt{10} $, $ TU = 3 $, $ VU=\sqrt{10} $, $ SV = 3 $

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QUESTION IMAGE

Question

Explanation:

Step1: Find coordinates

Step2: Calculate \( ST \)

Step3: Calculate \( TU \)

Step4: Calculate \( VU \)

Step5: Calculate \( SV \)

Answer: