Help with geometry...

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asked Jan 23, 2018 168k views

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Tzach Ovadia · Answer 1 · 2018-01-29T11:37:52+0000

so... doing the distances from ABC to GHI

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

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

and doing the distances from ABC to DEF

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

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

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

Help with geometry...

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