the vertices of triangle rst are r(-1,-2), s(2,-1), and t(4,2). if △jkl ~ △rst and the length of kl is √13/2 units, what is the length of jk? √10 √13 2√13 √10/2

Question

Accepted Answer

# Explanation:
## Step1: Calculate the length of $ \overline{RS} $
Use the distance formula $ d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} $. For $ R(-1,-2) $ and $ S(2,-1) $, $ x_1=-1,y_1 = - 2,x_2=2,y_2=-1 $.
$$
\begin{align*}
d_{RS}&=\sqrt{(2-(-1))^2+(-1 - (-2))^2}\
&=\sqrt{(2 + 1)^2+(-1+2)^2}\
&=\sqrt{3^2+1^2}\
&=\sqrt{9 + 1}\
&=\sqrt{10}
\end{align*}
$$
## Step2: Use the property of similar triangles
Since $ \triangle JKL\sim\triangle RST $, the ratio of corresponding sides is equal. Let $ \frac{JK}{RS}=\frac{KL}{ST} $. We know $ KL=\frac{\sqrt{13}}{2} $. First, find $ ST $:
For $ S(2,-1) $ and $ T(4,2) $, $ x_1 = 2,y_1=-1,x_2=4,y_2 = 2 $
$$
\begin{align*}
d_{ST}&=\sqrt{(4 - 2)^2+(2-(-1))^2}\
&=\sqrt{2^2+3^2}\
&=\sqrt{4+9}\
&=\sqrt{13}
\end{align*}
$$
From $ \frac{JK}{RS}=\frac{KL}{ST} $, substitute $ RS = \sqrt{10},KL=\frac{\sqrt{13}}{2},ST=\sqrt{13} $
$$
\begin{align*}
\frac{JK}{\sqrt{10}}&=\frac{\frac{\sqrt{13}}{2}}{\sqrt{13}}\
\frac{JK}{\sqrt{10}}&=\frac{1}{2}\
JK&=\frac{\sqrt{10}}{2}
\end{align*}
$$

# Answer:
$\frac{\sqrt{10}}{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of \( \overline{RS} \)

Step2: Use the property of similar triangles

Answer: