check the picture below.
now, the distance form -4,2 to 2,2 you can pretty much get it off the grid by counting the units.
and the distance from 2,2 to 2, -1, you also can get it off the grid by just counting.
now, let's check the other lengths,
from -4, 2 to -5, -2
from -5, -2 to -2, -3
from -2, -3 to 2, -1
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -4}}\quad ,&{{ 2}})\quad
% (c,d)
&({{ -5}}\quad ,&{{ -2}})
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{[-5-(-4)]^2+[-2-2]^2}\implies d=\sqrt{(-5+4)^2+(-4)^2}
\\\\\\
d=\sqrt{(-1)^2+(-4)^2}\implies d=\sqrt{1+16}\implies \boxed{d=\sqrt{17}}\\\\
-------------------------------[/tex]
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (c,d)
&({{ -5}}\quad ,&{{ -2}})
% (a,b)
&({{ -2}}\quad ,&{{ -3}})\quad
\end{array}\qquad
\\\\\\
d=\sqrt{[-2-(-5)]^2+[-3-(-2)]^2}\\\\\\ d=\sqrt{(-2+5)^2+(-3+2)^2}
\\\\\\
d=\sqrt{3^2+(-1)^2}\implies d=\sqrt{9+1}\implies \boxed{d=\sqrt{10}}\\\\
-------------------------------\\\\
[/tex]
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -2}}\quad ,&{{ -3}})\quad
% (c,d)
&({{ 2}}\quad ,&{{ -1}})
\end{array}\qquad
\\\\\\
d=\sqrt{[2-(-2)]^2+[-1-(-3)]^2}\implies d=\sqrt{(2+2)^2+(-1+3)^2}
\\\\\\
d=\sqrt{4^2+2^2}\implies d=\sqrt{16+4}\implies d=\sqrt{20}\implies \boxed{d=2\sqrt{5}}[/tex]
[tex]\bf \textit{thus the perimeter is then}\implies 6~+~3~+~\sqrt{17}~+~\sqrt{10}~+~2\sqrt{5}[/tex]
