Question

Classifique quanto ao comprimento dos lados o triangulo [MAR] sendo M=(-2;1), A=(5;1), R=(5;4).

Answer 1

$\texttt{Dist\^ancia entre dois pontos:}\\\\\\d=\left | \overrightarrow{AB} \right |=\left | B-A \right |\\\\\\d= \sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\\\\\\\texttt{Essas coordenadas que o exerc\'icio fornece, nada mais s\~ao}\\\texttt{que os v\'ertices desse tri\^angulo. Como o exerc\'icio quer}\\\texttt{a medida dos lados, basta calcularmos a dist\^ancia entre}\\\texttt{os v\'ertices.}$


$\texttt{Dist\^ancia entre o v\'ertice M e o v\'ertice A:}\\\\\\d= \sqrt{(5-(-2))^2+(1-1)^2}\\\\\\d=\sqrt{(5+2)^2+0^2}\\\\\\d=\sqrt{7^2+0}\\\\\\d= \sqrt{49}\\\\\\ d=7$


$\texttt{Dist\^ancia entre o v\'ertice M e o v\'ertice R:}\\\\\\d=\sqrt{(5-(-2))^2+(4-1)^2}\\\\\\d= \sqrt{(5+2)^2+3^2}\\\\\\d= \sqrt{7^2+3^2}\\\\\\d= \sqrt{49+9}\\\\\\d=\sqrt{58}$


$\texttt{Dist\^ancia entre o v\'ertice R e o v\'ertice A:}\\\\\\d=\sqrt{(5-5)^2+(1-4)^2}\\\\\\d=\sqrt{0^2+3^2}\\\\\\d=\sqrt{0+9}\\\\\\d=\sqrt{9}\\\\\\d=3$



$\large\texttt{Como as medidas dos tr\^es lados s\~ao diferentes entre}\\\texttt{s\'i, o tri\^angulo \'e classificado como escaleno.}$