# Quadrance

**For the beginner: Quadrance is just distance squared but for the advanced this is not the appropriate notion.**

This section begins with the concept of the \(quadrance\) between points, and then examines null lines, midpoints, the Triple Quad Formula, Pythegoras' theorem, the quadrea of a triangle, and the generalization of the classical Heron's formula called -- in this wiki--Archimedes' formula. Perpendicular bisectors and quadrance from a point to a line are discussed. Archimedes' function and the Quadruple quad formula are defined. All these topics hold in an arbitrary field F not of characteristic two (from now on this will not necessarily be repeated).

The

quadrance\(Q(A_1,A_2)\)between the points \(A_1\equiv [x_1,y_1]\) and \(A_2 \equiv [x_1,y_2]\) is the number \[Q(A_1,A_2) \equiv (x_2-x_1)^2+(y_2-y_1)^2\]

Sometimes \(Q(A_1,A_2)\) will be called the quadrance **from \(A_1\)to\(A_2\)** or the **quadrance of the side \(\overline{A_1A_2}\)**. Clearly \[Q(A_1,A_2)=Q(A_2,A_1).\]

In rational or decimal number fields \(Q(A_1,A_2)\) is always positive, and is zero precisely when \(A_1=A_2\). This is not necessarily the case of other fields.

In the complex number field with \(A_2 \equiv [0,0]\) and \(A_2 \equiv [1,i]\)

\[Q(A_1,A_2) = 1^2 +i^2=0. \diamond\]

(Null line)If \(A_1\) and \(A_2\) are distant points, then \(A_1A_2\) is a null line precisely when \(Q(A_1,A_2)=0.\)

If \(A_1\equiv [x_1,y_1]\) and \(A_2\equiv [x_2,y_2]\) are distinct points, then by the Line through two points theorem

\[A_1 A_2 = \langle y_1-y_2: x_2-x_1:x_1y_2-x_1y_2-x_2y_1\rangle \]

This is a null line precisely when \[(y_1-y_2)^2 + (x_2-x_1)^2=0\] which is exactly the condition that \(Q(A_1,A_2)=0. \blacksquare\)

(MidpointIf A1A2 is a non-null line, then there is a unique point \(A\) lying on \(A_1A_2\) which satisfies \[Q(A_1,A) = Q(A,A_2).\] This is the midpoint \(M\equiv \frac{1}{2}A_1 + \frac{1}{2}A_2\) of the side\(\overline{A_1A_2}\).Furthermore \[Q(A_1,M) = Q(M,A_2) = \frac{Q(A_1,A_2)}{4}.\]

If \(A_1\equiv [x_1,y_1]\) and\( A_2 \equiv [x_2,y_2]\) are distinct points, then by the Affine combination theorem(link required) any point on the line \(A_1A_2\), has the form

\[A \equiv \lambda A_1 +(1-\lambda)A_2 = [\lambda x_1 + (1-\lambda) x_2, \lambda y_1 + (1-\lambda) y_2 )^2\] \[ = (\lambda x_1 +(-\lambda)x_2)^2 + (\lambda y_1 +(-\lambda)y_2)^2.\] .

Rewrite this as \[\left( (1-\lambda)^2 -\lambda^2 \right) \left( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right) = 0 \]

By assumption \(A_1A_2\) is a non-null line, so by the previous Null line theorem, \(x_2-x_1)^2 +(y_2-y_1)^2 \not = 0\). Thus \(\lambda = \frac{1}{2}\), and \(A\) is the midpoint

\[ M \equiv \frac{1}{2} A_1 + \frac{1}{2}A_2 = \left[ \frac{(x_1 +x_2)}{2}, \frac{(y_1 +y_2)}{2} \right] \]

of the side \( \overline{A_1A_2}\). Then

\[ Q(A_1,M)= \left( \frac{x_1-x_2}{2} \right) + \left( \frac{y_1-y_2}{2} \right) = Q(M,A_2) \] \[ = \frac{Q(A_1,A_2)}{4} \blacksquare \]

For \( \triangle A_1A_2A_3\) , the numbers \(Q_1 \equiv Q(A_2,A_3), Q_2\equiv Q(A_1,A_3)\) and \(Q_3 \equiv Q(A_1,A_2)\) are the qudarnces of the triangle, with \(Q_1\) the

quadrance of the side \(\overline{A_2A_3}\), or the **quadrance opposite the vertex\(\overline{l_1l_2}\), and similarly for the other quadrances.