Find the area of triangle in parabola $x^{2} = 8 y$ $x^2=8y$ ?

Question

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ $A(x_1, y_1), B(x_2,y_2),C(x_3,y_3)$ lie on the parabola $x^{2} = 8 y$ $x^2=8y$ .
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

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