Find the area of triangle in parabola #x^2=8y#?

Question

Let #A(x_1, y_1), B(x_2,y_2),C(x_3,y_3)# lie on the parabola #x^2=8y#.
Show that the area of #DeltaABC# is
#1/16(x_1-x_2)(x_2-x_3)(x_3-x_1)#

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Answer 1 · 2017-03-17T23:56:27

See below.

The area of a triangle given by it's vertices #(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)))#

now substituting #y_k=x_k^2/8# we have

#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))#

but

#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))#