Question

determine oe valores de m para as quais a distancia entre A(m-1,3) e B(2,-m) é 6.

Answer 1

Temos que lembrar como fazer distância de ponto a ponto, que calculamos da seguinte forma:

$\boxed{d = \sqrt{(X_{b}-X_{a})^{2} + (Y_{b}-Y_{a})^{2}}}$

Agora vamos substituir os valores:

$d = \sqrt{(X_{b}-X_{a})^{2} + (Y_{b}-Y_{a})^{2}} \\\\ \sqrt{(2-(m-1))^{2} + (-m-3)^{2}} = 6 \\\\ \sqrt{(2-m+1)^{2} + (-m-3)^{2}} = 6 \\\\ \sqrt{(3-m+)^{2} + (-m-3)^{2}} = 6 \\\\ elevamos \ ao \ quadrado \ para \ sumir \ com \ a \ raiz \\\\ (\sqrt{(3-m)^{2} + (-m-3)^{2}})^{2} = 6^{2} \\\\ (3-m)^{2} + (-m-3)^{2}} = 36 \\\\ 9 - 6m + m^{2} + m^{2} + 6m + 9 = 36$

$9 - 6m + m^{2} + m^{2} + 6m + 9 -36 = 0 \\\\ 2m^{2} - 18 = 0 \\\\ \Delta = b^{2} - 4 \cdot a \cdot c \\\\ \Delta = 0^{2} - 4 \cdot (2) \cdot (-18) \\\\ \Delta = 0+144 \\\\ \Delta = 144$


$2m^{2} - 18 = 0 \\\\ m = \frac{-b \pm \sqrt{\Delta}}{2 \cdot a} \\\\ m = \frac{0 \pm \sqrt{144}}{2 \cdot 2} \\\\ m = \frac{0 \pm 12}{4} \\\\\\ m' = \frac{0 + 12}{4} = \frac{12}{4} = \boxed{3} \\\\ m'' = \frac{0 - 12}{4} = \frac{-12}{4} = \boxed{-3}$

Vamos voltar e testar os valores. Em ambos, a distância obrigatoriamente deve dar 6.

$\rightarrow m = 3 \\ a(m-1; 3) \ e \ b(2; -m) \\ a(3-1; 3) \ e \ b(2; -3) \\ a(2; 3) \ e \ b(2; -3) \\\\ d = \sqrt{(X_{b}-X_{a})^{2} + (Y_{b}-Y_{a})^{2}} \\\\ d = \sqrt{(2-2)^{2} + (-3-3)^{2}} \\\\ d = \sqrt{0 + (-6)^{2}} \\\\ d = \sqrt{36} \\\\ \boxed{d = 6}$

Testando o segundo:

$\rightarrow m = -3 \\ a(m-1; 3) \ e \ b(2; -m) \\ a(-3-1; 3) \ e \ b(2; -(-3)) \\\ a(-4; 3) \ e \ b(2; 3) \\\\ d = \sqrt{(X_{b}-X_{a})^{2} + (Y_{b}-Y_{a})^{2}} \\\\ d = \sqrt{(2-(-4))^{2} + (3-3)^{2}} \\\\ d = \sqrt{(2+4)^{2} + 0} \\\\ d = \sqrt{(6)^{2}} \\\\ d = \sqrt{36} \\\\ \boxed{d = 6}$


$\therefore \boxed{\boxed{m = 3} \ e \ \boxed{m = -3}}$