Question

Determine o ponto A(x,-3) sendo que ele é distante do ponto b(-6-8) em 10 unidades de medida

Answer 1

$\mathrm{Seja\ a\ dist\hat{a}ncia\ entre\ os\ pontos}\ A(x,-3)\ \text{e}\ B(-6,-8)\ \mathrm{igual\ a\ 10\ u.c.}$

$\boxed{d_{\mathrm{AB}}^2=\delta x^2+\delta y^2}\Longrightarrow 10^2=(-6-x)^2+(-8-(-3))^2$

$\Longrightarrow 100=36+12x+x^2+25\Longrightarrow x^2+12x-39=0$

$\mathrm{Obtemos\ uma\ equa\c{c}\tilde{a}o\ quadr\acute{a}tica\ de\ coeficientes}\ a=1,\ b=12\ \text{e}\ c=-39.$

$\boxed{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\Longrightarrow x=\dfrac{-12\pm10\sqrt{3}}{2}\ \therefore\ \boxed{x=-6\pm5\sqrt{3}}$

$\mathrm{Os\ poss\acute{\i}veis\ valores\ para}\ x\ \mathrm{s\tilde{a}}\text{o:}$

$\boxed{x=\left\{-6-5\sqrt{3},-6+5\sqrt{3}\right\}}$