Theorem 2.12

Theorem 2.12: Let A, B and C be three noncolinear points, and let

D = qA + rB + sC

where

q + r + s = 1

be a point in the plane. Then D is on the same side of BC as A if and only if q is positive.

Proof: Let

A = (x₀, y₀)

B = (x₁, y₁)

C = (x₂, y₂)

and

D = (x, y)

Then

D = qA + rB + sC

= q(x₀, y₀) + r (x₁, y₁) + s (x₂, y₂)

= (qx₀ + rx₁ + sx₂, qy₀ + ry₁ + sy₂)

Finally let

ax + by = c

be the equation for the line determined by BC. If we check the coordinates for the point D in the left side of this equation we get

a(qx₀ + rx₁ + sx₂) + b(qy₀ + ry₁ + sy₂)

Remove parentheses

= aqx₀ + arx₁ + asx₂ + bqy₀ + bry₁ + bsy₂

These terms can be rearranged as

= q(ax₀ + by₀) + r(ax₁ + by₁) + s(ax₂ + by₂)

Since B and C are on the line,

ax₁ + by₁ = c

and

ax₂ + by₂ = c

If we substitute these into the last equation, we get

= q(ax₀ + by₀) + rc + sc

= q(ax₀ + by₀) + (r + s)c

However, since q + r + s = 1,

r + s = 1 - q

Our equation then becomes

= q(ax₀ + by₀) + (1 - q)c

= q(ax₀ + by₀ - c) + c

This will satisfy the same inequality as the point A if and only if q is positive.