Question

JÁ RESPONDI AS 1,2,3 SÓ FALTA 4 E 5...

Answer 1

Explicação passo-a-passo:

4) O baricentro $G(x_G,y_G)$ de um triângulo de vértices $A(x_A,y_A)$, $B(x_B,y_B)$ e $C(x_C,y_C)$ tem coordenadas:

$x_G=\dfrac{x_A+x_B+x_C}{3}$

$y_G=\dfrac{y_A+y_B+y_C}{3}$

Como $A(\frac{1}{2},-\frac{3}{2})$, $B(-\frac{1}{2},\frac{2}{3})$ e $C(\frac{1}{2},\frac{1}{3})$, temos que:

$x_G=\dfrac{\frac{1}{2}-\frac{1}{3}+\frac{1}{2}}{3}=\dfrac{1-\frac{1}{3}}{3}=\dfrac{\frac{2}{3}}{3}=\dfrac{2}{9}$

$y_G=\dfrac{-\frac{3}{2}+\frac{2}{3}+\frac{1}{3}}{3}=\dfrac{1-\frac{3}{2}}{3}=\dfrac{-\frac{1}{2}}{3}=\dfrac{-1}{6}$

Logo, $G(\frac{2}{9},-\frac{1}{6})$

5) $A(\frac{2}{3},\frac{1}{4})$, $B(\frac{4}{3},-\frac{3}{4})$ e $C(-\frac{5}{3},\frac{5}{4})$

$x_G=\dfrac{\frac{2}{3}+\frac{4}{3}-\frac{5}{3}}{3}=\dfrac{\frac{1}{3}}{3}=\dfrac{1}{9}$

$y_G=\dfrac{\frac{1}{4}-\frac{3}{4}+\frac{5}{4}}{3}=\dfrac{\frac{3}{4}}{3}=\dfrac{1}{4}$

Logo, $G(\frac{1}{9},\frac{1}{4})$