Question

Calcule o valor de X para que a distância entre A (x,1) B (-1,2 ) seja √10

Answer 1

$\sqrt{10} = \sqrt{(x+1)^2 + (1-2) ^2}\\ 10 = (x+1) ^2 +1\\ 10 = x^2+2x+1+1\\ x^2 +2x -8 = 0\\\\ DELTA = 2^2 - 4 * 1 * (-8)\\ DELTA = 4 + 32\\ DELTA = 36 \\ x = \frac{-2 +- \sqrt{36} }{2} \\ x' = \frac{-2 + 6}{ 2} = 2\\ x'' = \frac{-2-6}{2} = -4$
x pode ser 2 ou -4

Answer 2

Pela fórmula da distância:

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


Substituindo valores:

$\sqrt{(X_{b}-X_{a})^{2}+(Y_{b}-Y_{a})^{2}} = d \\\\ \sqrt{(-1-x)^{2}+(2-1)^{2}} = \sqrt{10} \\\\ \sqrt{(-1-x)^{2}+(1)^{2}} = \sqrt{10} \\\\ \sqrt{(-1-x)^{2}+1} = \sqrt{10} \\\\ \text{elevamos os dois lados ao quadrado para sumir com a raiz} \\\\ (\sqrt{(-1-x)^{2}+1})^{2} = (\sqrt{10})^{2} \\\\ (\not{\sqrt{(-1-x)^{2}+1}})^{\not{2}} = (\not{\sqrt{10}})^{\not{2}}$

$(-1-x)^{2}+1 = 10 \\\\ 1+2x+x^{2}+1-10 = 0 \\\\ x^{2}+2x-8 = 0 \\\\ \Delta = b^{2}-4 \cdot a \cdot c \\\\ \Delta = (2)^{2}-4 \cdot (1) \cdot (-8) \\\\ \Delta = 4+32 \\\\ \Delta = 36 \\\\\\\ x = \frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x = \frac{-2 \pm \sqrt{36}}{2 \cdot 1} \\\\ x = \frac{-2 \pm 6}{2} \\\\ \Rightarrow x' = \frac{-2 + 6}{2} = \frac{4}{2} = \boxed{2} \\\\ \Rightarrow x'' = \frac{-2 - 6}{2} = \frac{-8}{2} = \boxed{-4}$


Portanto, este pode ser os dois valores de x.


$\boxed{\boxed{S = \{-4,2\}}}$