so... doing the distances from ABC to GHI
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 1}}\quad ,&{{ 2}})\quad
% (c,d)
G&({{ -4}}\quad ,&{{ 2}})\\
B&({{ 2}}\quad ,&{{ 3}})\quad
% (c,d)
H&({{ -3}}\quad ,&{{ 3}})\\
C&({{ 3}}\quad ,&{{ 1}})\quad
% (c,d)
G&({{ -2}}\quad ,&{{ 1}})
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\[/tex]
[tex]\bf -------------------------------\\\\
AG=\sqrt{(-4-1)^2+(2-2)^2}
\\\\\\
BH=\sqrt{(-3-2)^2+(3-3)^2}
\\\\\\
CG=\sqrt{(-2-3)^2+(1-1)^2}[/tex]
and doing the distances from ABC to DEF
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 1}}\quad ,&{{ 2}})\quad
% (c,d)
D&({{ 1}}\quad ,&{{ -1}})\\
B&({{ 2}}\quad ,&{{ 3}})\quad
% (c,d)
E&({{ 2}}\quad ,&{{ 0}})\\
C&({{ 3}}\quad ,&{{ 1}})\quad
% (c,d)
F&({{ 3}}\quad ,&{{ -2}})
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]
[tex]\bf -------------------------------\\\\
AD=\sqrt{(1-1)^2+(-1-2)^2}
\\\\\\
BE=\sqrt{(2-2)^2+(0-3)^2}
\\\\\\
CF=\sqrt{(3-3)^2+(-2-1)^2}[/tex]
now... the ABC to JKL... surely you'd know how to do... the same way, just use the distance formula
