Question

Determine os vértices C e D no quadrado OBCD, dados O: origem e B=(2,3)

Answer 1

$\mathrm{V\acute{e}rtices\ \to\ O=(0,0),\ B=(2,3),\ C=(m,n),\ D=(p,q)}\\\\ \mathrm{Lado\ \to\ l=d_{OB}=\sqrt{(3-0)^2+(2-0)^2}=\sqrt{13}\ u.c.}\\\\ \mathrm{Diagonal\ \to\ x=l\sqrt{2}=\sqrt{13}\sqrt{2}=\sqrt{26}\ u.c.}$

$\mathrm{d_{BC}=l\ \to\ \sqrt{(m-2)^2+(n-3)^2}=\sqrt{13}\ \to}\\ \mathrm{\to\ m^2-4m+4+n^2-6n+9=13\ \to\ m^2+n^2-4m-6n=0}\\\\ \mathrm{d_{OC}=x\ \to\ \sqrt{(m-0)^2+(n-0)^2}=\sqrt{26}\ \to\ m^2+n^2=26}\\\\ \mathrm{26-4m-6n=0\ \to\ 2m+3n=13\ \to\ n=\dfrac{13-2m}{3}}\\\\ \mathrm{m^2+\bigg(\dfrac{13-2m}{3}\bigg)^2=26\ \to\ \dfrac{9m^2}{9}+\dfrac{4m^2-52m+169}{9}=26\ \to}\\\\ \mathrm{\to\ 13m^2-52m+169=9(26)\ \to\ m^2-4m+13=18\ \to}\\\\ \mathrm{\to\ m^2-4m-5=0\ \to\ \boxed{\mathrm{m=5}}\ ou\ \boxed{\mathrm{m=-1}}}\\\\ \mathrm{Se\ m=5\ \to\ n=\dfrac{13-10}{3}=\dfrac{3}{3}\ \to\ \boxed{\mathrm{n=1}}}\\\\ \mathrm{Se\ m=-1\ \to\ n=\dfrac{13+2}{3}=\dfrac{15}{3}\ \to\ \boxed{\mathrm{n=5}}}\\\\ \mathrm{C=(m,n)\ \to\ C=\{(5,1);(-1,5)\}}$

$\mathrm{d_{OD}=l\ \to\ \sqrt{(p-0)^2+(q-0)^2}=\sqrt{13}\ \to\ p^2+q^2=13}\\\\ \mathrm{d_{BD}=x\ \to\ \sqrt{(p-2)^2+(q-3)^2}=\sqrt{23}\ \to}\\ \mathrm{\to\ p^2-4p+4+q^2-6q+9=26\ \to\ p^2+q^2-4p-6q=13}\\\\ \mathrm{13-4p-6q=13\ \to\ 2p+3q=0\ \to\ q=-\dfrac{2p}{3}}\\\\ \mathrm{p^2+\bigg(\dfrac{-2p}{3}\bigg)^2=13\ \to\ \dfrac{9p^2}{9}+\dfrac{4p^2}{9}=13\ \to\ 13p^2=9(13)\ \to}\\\\ \mathrm{p^2=9\ \to\ p=\pm\sqrt{9}\ \to\ \boxed{\mathrm{p=3}}\ ou\ \boxed{\mathrm{p=-3}}}\\\\ \mathrm{Se\ p=3\ \to\ q=-\dfrac{2(3)}{3}\ \to\ \boxed{\mathrm{p=-2}}}\\\\ \mathrm{Se\ p=-3\ \to\ q=-\dfrac{2(-3)}{3}\ \to\ \boxed{\mathrm{p=2}}}\\\\ \mathrm{D=(p,q)\ \to\ D=\{(3,-2);(-3,2)\}}$

$\textrm{Combina\c{c}\~oes entre C e D:}\\\\ \mathrm{\boxed{\mathrm{C=(5,1)}}\ e\ \boxed{\mathrm{D=(3,-2)}}}\\\\ \mathrm{\boxed{\mathrm{C=(-1,5)}}\ e\ \boxed{\mathrm{D=(-3,2)}}}$