Question

Sabendo que os vértices do triângulo ABC são os pontos A(1,2), B(-1,-1) e C(-1,2), represente-o num plano cartesiano e determine:

a)classificação do triângulo ABC, quanto às medidas dos seus lados

b) O perímetro do triângulo ABC

c) A área do triângulo ABC

d) O ponto médio do segmento BC

e) A inclinação do segmento AB

f) Em que quadrante se encontra os pontos A,B e C.​

Answer 1

Vamos começar calculando a medida de cada lado do triangulo.

Para isso vamos utilizar a equação da distancia entre dois pontos.

Lado AB:

$\overline{AB}~=~Distancia_{\,A,B}\\\\\\\overline{AB}~=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\\\\\\overline{AB}~=\sqrt{(1-(-1))^2+(2-(-1))^2}\\\\\\\overline{AB}~=\sqrt{(2)^2+(3)^2}\\\\\\\overline{AB}~=\sqrt{4+9}\\\\\\\boxed{\overline{AB}~=~\sqrt{13}~~unidades~de~comprimento}$

Lado AC:

$\overline{AC}~=~Distancia_{\,A,C}\\\\\\\overline{AC}~=\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}\\\\\\\overline{AC}~=\sqrt{(1-(-1))^2+(2-2)^2}\\\\\\\overline{AC}~=\sqrt{(2)^2+(0)^2}\\\\\\\overline{AC}~=\sqrt{4+0}\\\\\\\boxed{\overline{AC}~=~2~~unidades~de~comprimento}$

Lado BC:

$\overline{BC}~=~Distancia_{\,B,C}\\\\\\\overline{BC}~=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}\\\\\\\overline{BC}~=\sqrt{(-1-(-1))^2+(-1-2)^2}\\\\\\\overline{BC}~=\sqrt{(0)^2+(-3)^2}\\\\\\\overline{BC}~=\sqrt{0+9}\\\\\\\boxed{\overline{BC}~=~3~~unidades~de~comprimento}$

a)

Como todos lados tem medidas diferentes, o triangulo é classificado como escaleno.

b)

$Perimetro~=~\overline{AB}+\overline{AC}+\overline{BC}\\\\\\Perimetro~=~\sqrt{13}+2+3\\\\\\\boxed{Perimetro~=~(\,5+\sqrt{13}\,)~unidades~de~comprimento}$

c)

Utilizando o método do determinante:

$Area~=~\frac{1}{2}\,.\,\left|\left|\begin{array}{ccc}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|\left|\begin{array}{ccc}1&2&1\\-1&-1&1\\-1&2&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~(1.(-1).1+(-1).2.1+(-1).2.1)~-~(1.(-1).(-1)+1.2.1+1.2.(-1))~\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~(-1-2-2)~-~(1+2-2)~\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~-5-1~\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~-6~\right|$

$Area~=~\frac{1}{2}\,.\,6\\\\\\\boxed{Area~=~3~unidades~de~area}$

d)

$Ponto~Medio_{~\overline{BC}}~=~\left(\frac{x_B+x_C}{2}~,~\frac{y_B+y_C}{2}\right)\\\\\\Ponto~Medio_{~\overline{BC}}~=~\left(\frac{-1+(-1)}{2}~,~\frac{-1+2}{2}\right)\\\\\\Ponto~Medio_{~\overline{BC}}~=~\left(\frac{-2}{2}~,~\frac{1}{2}\right)\\\\\\\boxed{Ponto~Medio_{~\overline{BC}}~=~\left(-1~,~\frac{1}{2}\right)}$

e)

$m_{~\overline{AB}}~=~\frac{y_A-y_B}{x_A-x_B}\\\\\\m_{~\overline{AB}}~=~\frac{2-(-1)}{1-(-1)}\\\\\\\boxed{m_{~\overline{AB}}~=~\frac{3}{2}}\\\\\\Inclinacao~=~tg^{-1}(m)\\\\\\Inclinacao~=~tg^{-1}(\frac{3}{2})\\\\\\\boxed{Inclinacao~\approx~56,31^\circ}$

f)

Respectivamente no 1°, 3° e 2° quadrantes