Question

obtenha o valor de m sabendo que a distância entre A (1,-2) e B (m,-2) é 5.

Answer 1

$a(1;-2);b(m;-2)\\\\D=5\\\\\\ D_{a,b}= \sqrt{(x_b-x_a)^2+(y_b-y_a)^2} \to \\\\\\ 5= \sqrt{(m-1)^2+(-2-(-2))^2} \to \\\\ 5= \sqrt{(m^2-m-m+1)+(2-2)^2} \to \\\\ 5= \sqrt{(m^2-2m+1)+(0)^2} \to \\\\ 5= \sqrt{m^2-2m+1} \to \\\\ (5)^2=( \sqrt{m^2-2m+1} )^2\to \\\\ 25=m^2-2m+1\to\\\\ m^2-2m+1-25=0\to\\\\ m^2-2m-24=0$


$a=1;b=-2;c=-24\\\\ \Delta=b^2-4ac\to \Delta=(-2)^2-4*1*(-24)\to \Delta=4+96\to \Delta=100\\\\ m' \neq m''\\\\m= \frac{-b\pm \sqrt{\Delta} }{2a} \to m= \frac{-(-2)\pm \sqrt{100} }{2*1} \to m= \frac{2\pm 10 }{2} \to \\\\ m'= \frac{2+10 }{2} \to m'= \frac{12 }{2} \to m'=6\\\\ m''= \frac{2 -10 }{2} \to m''= \frac{-8 }{2} \to m''=-4\\\\\\ S=(-4;6)$




Achamos dois possíveis valores para $m$ , agora vamos verificar se satisfazem a equação;



$Para\to m=6\\\\5= \sqrt{(m-1)^2+(-2-(-2))^2} \to\\\\ 5= \sqrt{(6-1)^2+(-2+2)^2} \to \\\\ 5= \sqrt{(5)^2+(0)^2} \to \\\\ 5= \sqrt{25+0} \\\\5=5\\\\ Verdadeiro$





$Para\to m=-4\\\\5= \sqrt{(m-1)^2+(-2-(-2))^2} \to\\\\ 5= \sqrt{(-4-1)^2+(-2+2)^2} \to \\\\ 5= \sqrt{(-5)^2+(0)^2} \to \\\\ 5= \sqrt{25+0} \\\\5=5\\\\ Verdadeiro$




R.: Tanto -4 quanto 6 podem ser o valor de m, pois satisfazem a equação, tornando-a verdadeira.



Espero ter ajudado...