△pqr has vertices at p(-7,1), q(-4,7), and r(-1,1). is △pqr an equilateral triangle? justify your answer. yes, all sides are congruent. yes, (overline{pq}) is congruent to (overline{qr}). no, (overline{pq}) is not congruent to (overline{pr}). no, (overline{pq}) is not congruent to (overline{qr}).

Question

Accepted Answer

# Explanation:
## Step1: Calculate the length of $PQ$
Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $P(-7,1)$ and $Q(-4,7)$, $x_1=-7,y_1 = 1,x_2=-4,y_2 = 7$.
$$
\begin{align*}
PQ&=\sqrt{(-4+7)^2+(7 - 1)^2}\
&=\sqrt{3^2+6^2}\
&=\sqrt{9 + 36}\
&=\sqrt{45}=3\sqrt{5}
\end{align*}
$$
## Step2: Calculate the length of $PR$
For $P(-7,1)$ and $R(-1,1)$, $x_1=-7,y_1 = 1,x_2=-1,y_2 = 1$.
$$
\begin{align*}
PR&=\sqrt{(-1 + 7)^2+(1-1)^2}\
&=\sqrt{6^2+0^2}\
&=6
\end{align*}
$$
## Step3: Calculate the length of $QR$
For $Q(-4,7)$ and $R(-1,1)$, $x_1=-4,y_1 = 7,x_2=-1,y_2 = 1$.
$$
\begin{align*}
QR&=\sqrt{(-1 + 4)^2+(1 - 7)^2}\
&=\sqrt{3^2+(-6)^2}\
&=\sqrt{9+36}\
&=\sqrt{45}=3\sqrt{5}
\end{align*}
$$

Since $PQ = 3\sqrt{5}$, $PR=6$, and $PQ
eq PR$

# Answer:
No, $\overline{PQ}$ is not congruent to $\overline{PR}$.

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of \(PQ\)

Step2: Calculate the length of \(PR\)

Step3: Calculate the length of \(QR\)

Answer: