Question

Help with geometry...

Answer 1

So... doing the distances from ABC to GHI

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ 1}}\quad ,&{{ 2}})\quad % (c,d) G&({{ -4}}\quad ,&{{ 2}})\\ B&({{ 2}}\quad ,&{{ 3}})\quad % (c,d) H&({{ -3}}\quad ,&{{ 3}})\\ C&({{ 3}}\quad ,&{{ 1}})\quad % (c,d) G&({{ -2}}\quad ,&{{ 1}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf -------------------------------\\\\ AG=\sqrt{(-4-1)^2+(2-2)^2} \\\\\\ BH=\sqrt{(-3-2)^2+(3-3)^2} \\\\\\ CG=\sqrt{(-2-3)^2+(1-1)^2}$



and doing the distances from ABC to DEF

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ 1}}\quad ,&{{ 2}})\quad % (c,d) D&({{ 1}}\quad ,&{{ -1}})\\ B&({{ 2}}\quad ,&{{ 3}})\quad % (c,d) E&({{ 2}}\quad ,&{{ 0}})\\ C&({{ 3}}\quad ,&{{ 1}})\quad % (c,d) F&({{ 3}}\quad ,&{{ -2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf -------------------------------\\\\ AD=\sqrt{(1-1)^2+(-1-2)^2} \\\\\\ BE=\sqrt{(2-2)^2+(0-3)^2} \\\\\\ CF=\sqrt{(3-3)^2+(-2-1)^2}$

now... the ABC to JKL... surely you'd know how to do... the same way, just use the distance formula