Question

Dado dois pontos P(xp.yp) e Q(xq.yq) no plano cartesiano a distância entre eles e definida por: ​

Answer 1

Sabemos que a area do retangulo eh:

$D_{AB}\times D_{BC}$

Logo podemos usar a formula dada para calcular isso:

$D_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\\\D_{AB}=\sqrt{(1-4)^2+(2-8)^2}\\\\D_{AB}=\sqrt{(-3)^2+(-6)^2}\\\\D_{AB}=\sqrt{9+36}\\\\D_{AB}=\sqrt{45}$

$D_{BC}=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}\\\\D_{BC}=\sqrt{(4-3)^2+(8-5)^2}\\\\D_{BC}=\sqrt{(1)^2+(3)^2}\\\\D_{BC}=\sqrt{1+9}\\\\D_{BC}=\sqrt{10}$

Agora aplicamos a formula anterior:

$D_{AB}\times D_{BC}=\sqrt{45}\times\sqrt{10}=\sqrt{450}\approx21.21$