Answer
$\left(5+\sqrt{13}+\sqrt{58}\right)$ units
Work Step by Step
The Distance Formula, $d$, between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $
d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
$.
Using the Distance Formula, the distance, $d_1$, between $(1,3)$ and $(1,-2)$
is
\begin{align*}\require{cancel}
d_1&=
\sqrt{(1-1)^2+(3-(-2))^2}
\\&=
\sqrt{0^2+(3+2)^2}
\\&=
\sqrt{0+25}
\\&=
\sqrt{25}
\\&=
5
.\end{align*}
Using the Distance Formula, the distance, $d_2$, between $(1,-2)$ and $(-2,-4)$
is
\begin{align*}\require{cancel}
d_2&=
\sqrt{(1-(-2))^2+(-2-(-4))^2}
\\&=
\sqrt{(1+2)^2+(-2+4)^2}
\\&=
\sqrt{3^2+2^2}
\\&=
\sqrt{9+4}
\\&=
\sqrt{13}
.\end{align*}
Using the Distance Formula, the distance, $d_3$, between $(-2,-4)$ and $(1,3)$
is
\begin{align*}\require{cancel}
d_3&=
\sqrt{(-2-1)^2+(-4-3)^2}
\\&=
\sqrt{(-3)^2+(-7)^2}
\\&=
\sqrt{9+49}
\\&=
\sqrt{58}
.\end{align*}
Adding all the distances, then the perimeter, $P$, of the given triangle is
\begin{align*}\require{cancel}
P&=
d_1+d_2+d_3
\\&=
5+\sqrt{13}+\sqrt{58}
.\end{align*}
Hence, the perimeter of the triangle is $\left(5+\sqrt{13}+\sqrt{58}\right)$ units.
