the vertices of triangle abc are a(5,2), b(-1,-6) and c(1,5). if $ \triangle def $ is similar to $ \triangle abc $ and $ \frac{ab}{de}=5 $, which could be the coordinates of vertices d and e?
$ \bigcirc $ d(-4,2) and e(-4,-1)
$ \bigcirc $ d(-3,-1) and e(-1,1)
$ \bigcirc $ d(-5,3) and e(-5,1)
$ \bigcirc $ d(-2,-2) and e(2,0)

Question

Accepted Answer

# Explanation:
## Step1: Calculate the length of $AB$
Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $A(5,2)$ and $B(-1,-6)$, we have $x_1 = 5,y_1 = 2,x_2=-1,y_2=-6$.
$$
\begin{align*}
AB&=\sqrt{(-1 - 5)^2+(-6 - 2)^2}\
&=\sqrt{(-6)^2+(-8)^2}\
&=\sqrt{36 + 64}\
&=\sqrt{100}\
& = 10
\end{align*}
$$
Since $\frac{AB}{DE}=5$, then $DE=\frac{AB}{5}=\frac{10}{5} = 2$

## Step2: Calculate the length of $DE$ for each option
- **Option 1: $D(-4,2)$ and $E(-4,-1)$**
Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, with $x_1=-4,y_1 = 2,x_2=-4,y_2=-1$
$$
\begin{align*}
DE&=\sqrt{(-4+4)^2+(-1 - 2)^2}\
&=\sqrt{0+(-3)^2}\
&=3
\end{align*}
$$
- **Option 2: $D(-3,-1)$ and $E(-1,1)$**
$$
\begin{align*}
DE&=\sqrt{(-1+3)^2+(1 + 1)^2}\
&=\sqrt{2^2+2^2}\
&=\sqrt{4 + 4}\
&=\sqrt{8}\
&=2\sqrt{2}
\end{align*}
$$
- **Option 3: $D(-5,3)$ and $E(-5,1)$**
$$
\begin{align*}
DE&=\sqrt{(-5 + 5)^2+(1 - 3)^2}\
&=\sqrt{0+(-2)^2}\
&=2
\end{align*}
$$
- **Option 4: $D(-2,-2)$ and $E(2,0)$**
$$
\begin{align*}
DE&=\sqrt{(2 + 2)^2+(0 + 2)^2}\
&=\sqrt{4^2+2^2}\
&=\sqrt{16+4}\
&=\sqrt{20}\
&=2\sqrt{5}
\end{align*}
$$

# Answer:
C. $D(-5,3)$ and $E(-5,1)$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of \(AB\)

Step2: Calculate the length of \(DE\) for each option

Answer: