Triple Quad Formula

The Triple Quad Formula states that three points $$A_1 \equiv [x_1, y_1]$$, $$A_2 \equiv [x_2, y_2]$$, and $$A_3 \equiv [x_3, y_3]$$ lie on the same line exactly when:

$$ (Q_1 + Q_2 + Q_3)^2 = 2(Q_1^2 + Q_2^2 + Q_3^2) $$

where $$Q_1 = Q(A_1,A_3)$$, $$Q_2 = Q(A_3,A_1)$$, and $$Q_3 = Q(A_1,A_3)$$ are the respective quadrances between the points.

Proof
To simplify the next calculation let

$$a_1 \equiv (x_3 - x_2)$$ $$b_1 \equiv (y_3 - y_2)$$ $$a_2 \equiv (x_3 - x_1)$$ $$b_2 \equiv (y_3 - y_1)$$

Thus by the definition of quadrance:

$$Q_1 = a_1^2 + b_1^2$$

$$Q_2 = a_2^2 + b_2^2$$

$$Q_2 = (a_2 - a_1)^2 + (b_2 - b_1)^2$$

Three points are collinear exactly when:

$$(x_1y_2 - x_1y_3 +x_2y_3 - x_3 y_2 + x_3 y_1 - x_2 y_1) = 0$$

becaues $$(x_1y_2 - x_1y_3 +x_2y_3 - x_3 y_2 + x_3 y_1 - x_2 y_1) = (a_2 b_1 - a_1 b_2)$$ the condition is equivalnant to

$$4(a_1 b_2 - a_2 b_1) ^ 2 = 0 $$

now

$$4(a_1 b_2 - a_2 b_1) ^ 2$$

$$= 4((a_1^2 + b_1^2)(a_2^2 + b_2^2) - (a_1a_2 + b_1b_2) ^2)$$

$$= 4Q_1Q_2 - (Q_1 + Q_2 - Q_3)^2$$

$$= (Q_1 + Q_2 + Q_3)^2 - 2(Q_1^2 + Q_2^2 + Q_3^2)$$

the theorem follows from the last stament.