Trova l'area del triangolo in parabola # x ^ 2 = 8y #?

L'area di un triangolo data dai suoi vertici #(x_1,y_1),(x_2,y_2),(x_3,y_3)# is

#A=1/2abs(det((1,1,1),(x_1,x_2,x_3),(y_1,y_2,y_3)))#

ora sostituendo #y_k=x_k^2/8# ne ha

#A = 1/2det((1,1,1),(x_1,x_2,x_3),(x_1^2/8,x_2^2/8,x_3^2/8))#

#A = 1/(2cdot 8)det((1,1,1),(x_1,x_2,x_3),(x_1^2,x_2^2,x_3^2))#

ma

#det((1,1,1),(x_1,x_2,x_3),(x_1^2,x_2^2,x_3^2))=(x_1-x_2)(x_2-x_3)(x_3-x_1)#

so

#A = 1/16 abs((x_1-x_2)(x_2-x_3)(x_3-x_1))#